link to page 34 link to page 36 link to page 59 link to page 80 link to page 678 link to page 682 Chapter 2 Four Important Linear Partial Differential Equations In this chapter we introduce four fundamental linear partial differential equa- tions for which various explicit formulas for solutions are available. These are the transport equation 𝑢 𝑡 + 𝑏 ⋅ 𝐷𝑢 = 0 (§2.1), Laplace’s equation Δ𝑢 = 0 (§2.2), the heat equation 𝑢 𝑡 − Δ𝑢 = 0 (§2.3), the wave equation 𝑢 𝑡𝑡 − Δ𝑢 = 0 (§2.4). Before going further, the reader should review the discussions of inequali- ties, integration by parts, Green’s formulas, convolutions, etc., in Appendices B and C and later refer back to these as necessary. 2.1. TRANSPORT EQUATION One of the simplest partial differential equations is the transport equation with constant coefficients. This is the PDE (1) 𝑢 𝑡 + 𝑏 ⋅ 𝐷𝑢 = 0 in ℝ 𝑛 × (0, ∞) , where 𝑏 is a fixed vector in ℝ 𝑛 , 𝑏 = (𝑏 1 , . . . , 𝑏 𝑛 ) , and 𝑢 ∶ ℝ 𝑛 × [0, ∞) → ℝ is the unknown, 𝑢 = 𝑢(𝑥, 𝑡) . Here 𝑥 = (𝑥 1 , . . . , 𝑥 𝑛 ) ∈ ℝ 𝑛 denotes a typical point in space, and 𝑡 ≥ 0 denotes a typical time. We write 𝐷𝑢 = 𝐷 𝑥 𝑢 = (𝑢 𝑥 1 , . . . , 𝑢 𝑥 𝑛 ) for the gradient of 𝑢 with respect to the spatial variables 𝑥 . 17 link to page 34 link to page 34 link to page 34 link to page 35 link to page 35 link to page 35 link to page 35 link to page 35 link to page 35 link to page 35 link to page 147 18 2. Four Important Linear Partial Differential Equations Which functions 𝑢 solve (1)? To answer, let us suppose for the moment we are given some smooth solution 𝑢 and try to compute it. To do so, we first must recognize that the partial differential equation (1) asserts that a particular directional derivative of 𝑢 vanishes. We exploit this insight by fixing any point (𝑥, 𝑡) ∈ ℝ 𝑛 × (0, ∞) and defining 𝑧(𝑠) ≔ 𝑢(𝑥 + 𝑠𝑏, 𝑡 + 𝑠) (𝑠 ∈ ℝ). We then calculate ̇ 𝑧(𝑠) = 𝐷𝑢(𝑥 + 𝑠𝑏, 𝑡 + 𝑠) ⋅ 𝑏 + 𝑢 𝑡 (𝑥 + 𝑠𝑏, 𝑡 + 𝑠) = 0 ( ̇ = 𝑑 𝑑𝑠 ) , the second equality holding owing to (1). Thus 𝑧(⋅) is a constant function of 𝑠 , and consequently for each point (𝑥, 𝑡) , 𝑢 is constant on the line through (𝑥, 𝑡) with the direction (𝑏, 1) ∈ ℝ 𝑛+1 . Hence if we know the value of 𝑢 at any point on each such line, we know its value everywhere in ℝ 𝑛 × (0, ∞) . 2.1.1. Initial-value problem. For definiteness therefore, let us consider the initial-value problem (2) { 𝑢 𝑡 + 𝑏 ⋅ 𝐷𝑢 = 0 in ℝ 𝑛 × (0, ∞) 𝑢 = 𝑔 on ℝ 𝑛 × {𝑡 = 0} . Here 𝑏 ∈ ℝ 𝑛 and 𝑔 ∶ ℝ 𝑛 → ℝ are known, and the problem is to compute 𝑢 . Given (𝑥, 𝑡) as above, the line through (𝑥, 𝑡) with direction (𝑏, 1) is represented parametrically by (𝑥 + 𝑠𝑏, 𝑡 + 𝑠) (𝑠 ∈ ℝ) . This line hits the plane Γ ≔ ℝ 𝑛 × {𝑡 = 0} when 𝑠 = −𝑡 , at the point (𝑥 − 𝑡𝑏, 0) . Since 𝑢 is constant on the line and 𝑢(𝑥 − 𝑡𝑏, 0) = 𝑔(𝑥 − 𝑡𝑏) , we deduce (3) 𝑢(𝑥, 𝑡) = 𝑔(𝑥 − 𝑡𝑏) (𝑥 ∈ ℝ 𝑛 , 𝑡 ≥ 0). So, if (2) has a sufficiently regular solution 𝑢 , it must certainly be given by (3). And conversely, it is easy to check directly that if 𝑔 is 𝐶 1 , then 𝑢 defined by (3) is indeed a solution of (2). Weak solutions. If 𝑔 is not 𝐶 1 , then there is obviously no 𝐶 1 solution of (2). But even in this case formula (3) certainly provides a strong, and in fact the only reasonable, candidate for a solution. We may thus informally declare 𝑢(𝑥, 𝑡) = 𝑔(𝑥 − 𝑡𝑏) (𝑥 ∈ ℝ 𝑛 , 𝑡 ≥ 0) to be a weak solution of (2), even should 𝑔 not be 𝐶 1 . This all makes sense even if 𝑔 and thus 𝑢 are discontinuous. Such a notion, that a nonsmooth or even discontinuous function may sometimes solve a PDE, will come up again later when we study nonlinear transport phenomena in §3.4. link to page 36 link to page 81 link to page 35 link to page 36 link to page 111 link to page 36 link to page 36 link to page 310 2.2. Laplace’s Equation 19 2.1.2. Nonhomogeneous problem. Next let us look at the associated non- homogeneous problem (4) { 𝑢 𝑡 + 𝑏 ⋅ 𝐷𝑢 = 𝑓 in ℝ 𝑛 × (0, ∞) 𝑢 = 𝑔 on ℝ 𝑛 × {𝑡 = 0} . Fix as before (𝑥, 𝑡) ∈ ℝ 𝑛 × (0, ∞) and, inspired by the calculation above, set 𝑧(𝑠) ≔ 𝑢(𝑥 + 𝑠𝑏, 𝑡 + 𝑠) for 𝑠 ∈ ℝ . Then ̇ 𝑧(𝑠) = 𝐷𝑢(𝑥 + 𝑠𝑏, 𝑡 + 𝑠) ⋅ 𝑏 + 𝑢 𝑡 (𝑥 + 𝑠𝑏, 𝑡 + 𝑠) = 𝑓(𝑥 + 𝑠𝑏, 𝑡 + 𝑠). Consequently 𝑢(𝑥, 𝑡) − 𝑔(𝑥 − 𝑡𝑏) = 𝑧(0) − 𝑧(−𝑡) = ∫ 0 −𝑡 ̇ 𝑧(𝑠) 𝑑𝑠 = ∫ 0 −𝑡 𝑓(𝑥 + 𝑠𝑏, 𝑡 + 𝑠) 𝑑𝑠 = ∫ 𝑡 0 𝑓(𝑥 + (𝑠 − 𝑡)𝑏, 𝑠) 𝑑𝑠, and so (5) 𝑢(𝑥, 𝑡) = 𝑔(𝑥 − 𝑡𝑏) + ∫ 𝑡 0 𝑓(𝑥 + (𝑠 − 𝑡)𝑏, 𝑠) 𝑑𝑠 (𝑥 ∈ ℝ 𝑛 , 𝑡 ≥ 0) solves the initial-value problem (4). We will later employ this formula to solve the one-dimensional wave equa- tion, in §2.4.1. Remark. Observe that we have derived our solutions (3), (5) by in effect con- verting the partial differential equations into ordinary differential equations. This procedure is a special case of the method of characteristics , developed later in §3.2. 2.2. LAPLACE’S EQUATION Among the most important of all partial differential equations are undoubtedly Laplace’s equation (1) Δ𝑢 = 0 and Poisson’s equation ∗ (2) −Δ𝑢 = 𝑓. ∗ I prefer to write (2) with the minus sign, to be consistent with the notation for general second-order elliptic operators in Chapter 6. link to page 36 link to page 36 link to page 36 link to page 672 link to page 36 link to page 684 link to page 37 link to page 37 link to page 709 20 2. Four Important Linear Partial Differential Equations In both (1) and (2), 𝑥 ∈ 𝑈 and the unknown is 𝑢 ∶ ̄ 𝑈 → ℝ , 𝑢 = 𝑢(𝑥) , where 𝑈 ⊂ ℝ 𝑛 is a given open set. In (2) the function 𝑓 ∶ 𝑈 → ℝ is also given. Remember from §A.3 that the Laplacian of 𝑢 is Δ𝑢 ≔ 𝑛 ∑ 𝑖=1 𝑢 𝑥 𝑖 𝑥 𝑖 . DEFINITION. A 𝐶 2 function 𝑢 satisfying (1) is called a harmonic function. Physical interpretation. Laplace’s equation comes up in a wide variety of physical contexts. In a typical interpretation 𝑢 denotes the density of some quantity (e.g. a chemical concentration) in equilibrium. Then if 𝑉 is any smooth subregion within 𝑈 , the net flux of 𝑢 through 𝜕𝑉 is zero: ∫ 𝜕𝑉 F ⋅ 𝝂 𝑑𝑆 = 0, F denoting the flux density and 𝝂 the unit outer normal field. In view of the Gauss–Green Theorem (§C.2), we have ∫ 𝑉 div F 𝑑𝑥 = ∫ 𝜕𝑉 F ⋅ 𝝂 𝑑𝑆 = 0, and so (3) div F = 0 in 𝑈, since 𝑉 was arbitrary. In many instances it is physically reasonable to assume the flux F is proportional to the gradient 𝐷𝑢 but points in the opposite direction (since the flow is from regions of higher to lower concentration). Thus (4) F = −𝑎𝐷𝑢 (𝑎 > 0). Substituting into (3), we obtain Laplace’s equation div(𝐷𝑢) = Δ𝑢 = 0. If 𝑢 denotes the ⎧ ⎨ ⎩ chemical concentration temperature electrostatic potential, equation (4) is ⎧ ⎨ ⎩ Fick’s law of diffusion Fourier’s law of heat conduction Ohm’s law of electrical conduction. See Feynman–Leighton–Sands [ F-L-S , Chapter 12] for a discussion of the ubiq- uity of Laplace’s equation in mathematical physics. Laplace’s equation arises link to page 98 link to page 36 2.2. Laplace’s Equation 21 as well in the study of analytic functions and the probabilistic investigation of Brownian motion. 2.2.1. Fundamental solution. a. Derivation of fundamental solution. One good strategy for investigat- ing any partial differential equation is first to identify some explicit solutions and then, provided the PDE is linear, to assemble more complicated solutions out of the specific ones previously noted. Furthermore, in looking for explicit solutions, it is often wise to restrict attention to classes of functions with cer- tain symmetry properties. Since Laplace’s equation is invariant under rotations (Problem 2), it consequently seems advisable to search first for radial solutions, that is, functions of 𝑟 = |𝑥| . Let us therefore attempt to find a solution 𝑢 of Laplace’s equation (1) in 𝑈 = ℝ 𝑛 , having the form 𝑢(𝑥) = 𝑣(𝑟), where 𝑟 = |𝑥| = (𝑥 2 1 + ⋯ + 𝑥 2 𝑛 ) 1/2 and 𝑣 is to be selected (if possible) so that Δ𝑢 = 0 holds. First note for 𝑖 = 1 , . . . , 𝑛 that 𝜕𝑟 𝜕𝑥 𝑖 = 1 2 (𝑥 2 1 + ⋯ + 𝑥 2 𝑛 ) −1/2 2𝑥 𝑖 = 𝑥 𝑖 𝑟 (𝑥 ≠ 0). We thus have 𝑢 𝑥 𝑖 = 𝑣 ′ (𝑟) 𝑥 𝑖 𝑟 , 𝑢 𝑥 𝑖 𝑥 𝑖 = 𝑣 ″ (𝑟) 𝑥 2 𝑖 𝑟 2 + 𝑣 ′ (𝑟) ( 1 𝑟 − 𝑥 2 𝑖 𝑟 3 ) for 𝑖 = 1 , . . . , 𝑛 , and so Δ𝑢 = 𝑣 ″ (𝑟) + 𝑛 − 1 𝑟 𝑣 ′ (𝑟). Hence Δ𝑢 = 0 if and only if (5) 𝑣 ″ + 𝑛 − 1 𝑟 𝑣 ′ = 0. If 𝑣 ′ ≠ 0 , we deduce log(|𝑣 ′ |) ′ = 𝑣 ″ 𝑣 ′ = 1 − 𝑛 𝑟 , and hence 𝑣 ′ (𝑟) = 𝑎 𝑟 𝑛−1 for some constant 𝑎 . Consequently if 𝑟 > 0 , we have 𝑣(𝑟) = { 𝑏 log 𝑟 + 𝑐 (𝑛 = 2) 𝑏 𝑟 𝑛−2 + 𝑐 (𝑛 ≥ 3), where 𝑏 and 𝑐 are constants. These considerations motivate the following link to page 39 link to page 671 link to page 36 link to page 36 link to page 39 link to page 39 link to page 39 link to page 36 22 2. Four Important Linear Partial Differential Equations DEFINITION. The function (6) Φ(𝑥) ≔ { − 1 2𝜋 log |𝑥| (𝑛 = 2) 1 𝑛(𝑛−2)𝛼(𝑛) 1 |𝑥| 𝑛−2 (𝑛 ≥ 3), defined for 𝑥 ∈ ℝ 𝑛 , 𝑥 ≠ 0 , is the fundamental solution of Laplace’s equation. The reason for the particular choices of the constants in (6) will be apparent in a moment. (Recall from §A.2 that 𝛼(𝑛) denotes the volume of the unit ball in ℝ 𝑛 .) We will sometimes slightly abuse notation and write Φ(𝑥) = Φ(|𝑥|) to em- phasize that the fundamental solution is radial. Observe also that we have the estimates (7) |𝐷Φ(𝑥)| ≤ 𝐶 |𝑥| 𝑛−1 , |𝐷 2 Φ(𝑥)| ≤ 𝐶 |𝑥| 𝑛 (𝑥 ≠ 0) for some constant 𝐶 > 0 . b. Poisson’s equation. By construction the function 𝑥 ↦ Φ(𝑥) is harmonic for 𝑥 ≠ 0 . If we shift the origin to a new point 𝑦 , the PDE (1) is unchanged; and so 𝑥 ↦ Φ(𝑥 − 𝑦) is also harmonic as a function of 𝑥 , 𝑥 ≠ 𝑦 . Let us now take 𝑓 ∶ ℝ 𝑛 → ℝ and note that the mapping 𝑥 ↦ Φ(𝑥 − 𝑦)𝑓(𝑦) (𝑥 ≠ 𝑦) is harmonic for each point 𝑦 ∈ ℝ 𝑛 , and thus so is the sum of finitely many such expressions built for different points 𝑦 . This reasoning might suggest that the convolution (8) 𝑢(𝑥) = ∫ ℝ 𝑛 Φ(𝑥 − 𝑦)𝑓(𝑦) 𝑑𝑦 = { − 1 2𝜋 ∫ ℝ 2 log(|𝑥 − 𝑦|)𝑓(𝑦) 𝑑𝑦 (𝑛 = 2) 1 𝑛(𝑛−2)𝛼(𝑛) ∫ ℝ 𝑛 𝑓(𝑦) |𝑥−𝑦| 𝑛−2 𝑑𝑦 (𝑛 ≥ 3) will solve Laplace’s equation (1). However, this is wrong . Indeed, as intimated by estimate (7), 𝐷 2 Φ(𝑥 − 𝑦) is not summable near the singularity at 𝑦 = 𝑥 , and so naive differentiation through the integral sign is unjustified (and incorrect). We must proceed more carefully in calculating Δ𝑢 . Let us for simplicity now assume 𝑓 ∈ 𝐶 2 𝑐 (ℝ 𝑛 ) ; that is, 𝑓 is twice continu- ously differentiable, with compact support. THEOREM 1 (Solving Poisson’s equation) . Define 𝑢 by (8) . Then (i) 𝑢 ∈ 𝐶 2 (ℝ 𝑛 ) , and (ii) −Δ𝑢 = 𝑓 in ℝ 𝑛 . We consequently see that (8) provides us with a formula for a solution of Poisson’s equation (2) in ℝ 𝑛 . link to page 40 link to page 684 2.2. Laplace’s Equation 23 Proof. 1. We have (9) 𝑢(𝑥) = ∫ ℝ 𝑛 Φ(𝑥 − 𝑦)𝑓(𝑦) 𝑑𝑦 = ∫ ℝ 𝑛 Φ(𝑦)𝑓(𝑥 − 𝑦) 𝑑𝑦; hence 𝑢(𝑥 + ℎ𝑒 𝑖 ) − 𝑢(𝑥) ℎ = ∫ ℝ 𝑛 Φ(𝑦) [ 𝑓(𝑥 + ℎ𝑒 𝑖 − 𝑦) − 𝑓(𝑥 − 𝑦) ℎ ] 𝑑𝑦, where ℎ ≠ 0 and 𝑒 𝑖 = (0, . . . , 1, . . . , 0) , the 1 in the 𝑖 th -slot. But 𝑓(𝑥 + ℎ𝑒 𝑖 − 𝑦) − 𝑓(𝑥 − 𝑦) ℎ → 𝑓 𝑥 𝑖 (𝑥 − 𝑦) uniformly on ℝ 𝑛 as ℎ → 0 , and thus 𝑢 𝑥 𝑖 (𝑥) = ∫ ℝ 𝑛 Φ(𝑦)𝑓 𝑥 𝑖 (𝑥 − 𝑦) 𝑑𝑦 (𝑖 = 1, . . . , 𝑛). Similarly (10) 𝑢 𝑥 𝑖 𝑥 𝑗 (𝑥) = ∫ ℝ 𝑛 Φ(𝑦)𝑓 𝑥 𝑖 𝑥 𝑗 (𝑥 − 𝑦) 𝑑𝑦 (𝑖, 𝑗 = 1, . . . , 𝑛). As the expression on the right-hand side of (10) is continuous in the variable 𝑥 , we see 𝑢 ∈ 𝐶 2 (ℝ 𝑛 ) . 2. Since Φ blows up at 0 , we will need for subsequent calculations to isolate this singularity inside a small ball. So fix 𝜀 > 0 . Then (11) Δ𝑢(𝑥) = ∫ 𝐵(0,𝜀) Φ(𝑦)Δ 𝑥 𝑓(𝑥 − 𝑦) 𝑑𝑦 + ∫ ℝ 𝑛 −𝐵(0,𝜀) Φ(𝑦)Δ 𝑥 𝑓(𝑥 − 𝑦) 𝑑𝑦 ≕ 𝐼 𝜀 + 𝐽 𝜀 . Now (12) |𝐼 𝜀 | ≤ 𝐶‖𝐷 2 𝑓‖ 𝐿 ∞ (ℝ 𝑛 ) ∫ 𝐵(0,𝜀) |Φ(𝑦)| 𝑑𝑦 ≤ { 𝐶𝜀 2 | log 𝜀| (𝑛 = 2) 𝐶𝜀 2 (𝑛 ≥ 3). An integration by parts (see §C.2) yields (13) 𝐽 𝜀 = ∫ ℝ 𝑛 −𝐵(0,𝜀) Φ(𝑦)Δ 𝑦 𝑓(𝑥 − 𝑦) 𝑑𝑦 = − ∫ ℝ 𝑛 −𝐵(0,𝜀) 𝐷Φ(𝑦) ⋅ 𝐷 𝑦 𝑓(𝑥 − 𝑦) 𝑑𝑦 + ∫ 𝜕𝐵(0,𝜀) Φ(𝑦) 𝜕𝑓 𝜕𝜈 (𝑥 − 𝑦) 𝑑𝑆(𝑦) ≕ 𝐾 𝜀 + 𝐿 𝜀 , link to page 672 link to page 40 link to page 41 link to page 39 link to page 710 link to page 39 24 2. Four Important Linear Partial Differential Equations 𝝂 denoting the inward pointing unit normal along 𝜕𝐵(0, 𝜀) . We readily check (14) |𝐿 𝜀 | ≤ ‖𝐷𝑓‖ 𝐿 ∞ (ℝ 𝑛 ) ∫ 𝜕𝐵(0,𝜀) |Φ(𝑦)| 𝑑𝑆(𝑦) ≤ { 𝐶𝜀| log 𝜀| (𝑛 = 2) 𝐶𝜀 (𝑛 ≥ 3). 3. We continue by integrating by parts once again in the term 𝐾 𝜀 , to dis- cover 𝐾 𝜀 = ∫ ℝ 𝑛 −𝐵(0,𝜀) ΔΦ(𝑦)𝑓(𝑥 − 𝑦) 𝑑𝑦 − ∫ 𝜕𝐵(0,𝜀) 𝜕Φ 𝜕𝜈 (𝑦)𝑓(𝑥 − 𝑦) 𝑑𝑆(𝑦) = − ∫ 𝜕𝐵(0,𝜀) 𝜕Φ 𝜕𝜈 (𝑦)𝑓(𝑥 − 𝑦) 𝑑𝑆(𝑦), since Φ is harmonic away from the origin. Now 𝐷Φ(𝑦) = −1 𝑛𝛼(𝑛) 𝑦 |𝑦| 𝑛 (𝑦 ≠ 0) and 𝝂 = −𝑦 |𝑦| = − 𝑦 𝜀 on 𝜕𝐵(0, 𝜀) . Consequently 𝜕Φ 𝜕𝜈 (𝑦) = 𝝂 ⋅ 𝐷Φ(𝑦) = 1 𝑛𝛼(𝑛)𝜀 𝑛−1 on 𝜕𝐵(0, 𝜀) . Since 𝑛𝛼(𝑛)𝜀 𝑛−1 is the surface area of the sphere 𝜕𝐵(0, 𝜀) , we have (15) 𝐾 𝜀 = − 1 𝑛𝛼(𝑛)𝜀 𝑛−1 ∫ 𝜕𝐵(0,𝜀) 𝑓(𝑥 − 𝑦) 𝑑𝑆(𝑦) = − ⨍ 𝜕𝐵(𝑥,𝜀) 𝑓(𝑦) 𝑑𝑆(𝑦) → −𝑓(𝑥) as 𝜀 → 0. (Remember from §A.3 that a slash through an integral sign denotes an average.) 4. Combining now (11)–(15) and letting 𝜀 → 0 , we find −Δ𝑢(𝑥) = 𝑓(𝑥) , as asserted. □ Theorem 1 is in fact valid under far less stringent smoothness requirements for 𝑓 : see Gilbarg–Trudinger [ G-T ]. Interpretation of fundamental solution. We sometimes write −ΔΦ = 𝛿 0 in ℝ 𝑛 , 𝛿 0 denoting the Dirac measure on ℝ 𝑛 giving unit mass to the point 0 . Adopting this notation, we may formally compute −Δ𝑢(𝑥) = ∫ ℝ 𝑛 −Δ 𝑥 Φ(𝑥 − 𝑦)𝑓(𝑦) 𝑑𝑦 = ∫ ℝ 𝑛 𝛿 𝑥 𝑓(𝑦) 𝑑𝑦 = 𝑓(𝑥) (𝑥 ∈ ℝ 𝑛 ), in accordance with Theorem 1. link to page 684 link to page 685 2.2. Laplace’s Equation 25 2.2.2. Mean-value formulas. Consider now an open set 𝑈 ⊂ ℝ 𝑛 and sup- pose 𝑢 is a harmonic function within 𝑈 . We next derive the important mean- value formulas , which declare that 𝑢(𝑥) equals both the average of 𝑢 over the sphere 𝜕𝐵(𝑥, 𝑟) and the average of 𝑢 over the entire ball 𝐵(𝑥, 𝑟) , provided 𝐵(𝑥, 𝑟) ⊂ 𝑈 . These implicit formulas involving 𝑢 generate a remarkable num- ber of consequences, as we will momentarily see. THEOREM 2 (Mean-value formulas for Laplace’s equation) . If 𝑢 ∈ 𝐶 2 (𝑈) is harmonic, then (16) 𝑢(𝑥) = ⨍ 𝜕𝐵(𝑥,𝑟) 𝑢 𝑑𝑆 = ⨍ 𝐵(𝑥,𝑟) 𝑢 𝑑𝑦 for each ball 𝐵(𝑥, 𝑟) ⊂ 𝑈 . Proof. 1. Set 𝜙(𝑟) ≔ ⨍ 𝜕𝐵(𝑥,𝑟) 𝑢(𝑦) 𝑑𝑆(𝑦) = ⨍ 𝜕𝐵(0,1) 𝑢(𝑥 + 𝑟𝑧) 𝑑𝑆(𝑧). Then 𝜙 ′ (𝑟) = ⨍ 𝜕𝐵(0,1) 𝐷𝑢(𝑥 + 𝑟𝑧) ⋅ 𝑧 𝑑𝑆(𝑧), and consequently, using Green’s formulas from §C.2, we compute 𝜙 ′ (𝑟) = ⨍ 𝜕𝐵(𝑥,𝑟) 𝐷𝑢(𝑦) ⋅ 𝑦 − 𝑥 𝑟 𝑑𝑆(𝑦) = ⨍ 𝜕𝐵(𝑥,𝑟) 𝜕𝑢 𝜕𝜈 𝑑𝑆(𝑦) = 𝑟 𝑛 ⨍ 𝐵(𝑥,𝑟) Δ𝑢(𝑦) 𝑑𝑦 = 0. Hence 𝜙 is constant, and so 𝜙(𝑟) = lim 𝑡→0 𝜙(𝑡) = lim 𝑡→0 ⨍ 𝜕𝐵(𝑥,𝑡) 𝑢(𝑦) 𝑑𝑆(𝑦) = 𝑢(𝑥). 2. Observe next that our employing polar coordinates, as in §C.3, gives ∫ 𝐵(𝑥,𝑟) 𝑢 𝑑𝑦 = ∫ 𝑟 0 (∫ 𝜕𝐵(𝑥,𝑠) 𝑢 𝑑𝑆) 𝑑𝑠 = 𝑢(𝑥) ∫ 𝑟 0 𝑛𝛼(𝑛)𝑠 𝑛−1 𝑑𝑠 = 𝛼(𝑛)𝑟 𝑛 𝑢(𝑥). □ 26 2. Four Important Linear Partial Differential Equations THEOREM 3 (Converse to mean-value property) . If 𝑢 ∈ 𝐶 2 (𝑈) satisfies 𝑢(𝑥) = ⨍ 𝜕𝐵(𝑥,𝑟) 𝑢 𝑑𝑆 for each ball 𝐵(𝑥, 𝑟) ⊂ 𝑈 , then 𝑢 is harmonic. Proof. If Δ𝑢 ≢ 0 , there exists some ball 𝐵(𝑥, 𝑟) ⊂ 𝑈 such that, say, Δ𝑢 > 0 within 𝐵(𝑥, 𝑟) . But then for 𝜙 as above, 0 = 𝜙 ′ (𝑟) = 𝑟 𝑛 ⨍ 𝐵(𝑥,𝑟) Δ𝑢(𝑦) 𝑑𝑦 > 0, a contradiction. □ 2.2.3. Properties of harmonic functions. We now present a sequence of interesting deductions about harmonic functions, all based upon the mean- value formulas. Assume for the following that 𝑈 ⊂ ℝ 𝑛 is open and bounded. a. Strong maximum principle, uniqueness. We begin with the assertion that a harmonic function must attain its maximum on the boundary and cannot attain its maximum in the interior of a connected region unless it is constant. THEOREM 4 (Strong maximum principle) . Suppose 𝑢 ∈ 𝐶 2 (𝑈) ∩ 𝐶( ̄ 𝑈) is harmonic within 𝑈 . (i) Then max ̄ 𝑈 𝑢 = max 𝜕𝑈 𝑢. (ii) Furthermore, if 𝑈 is connected and there exists a point 𝑥 0 ∈ 𝑈 such that 𝑢(𝑥 0 ) = max ̄ 𝑈 𝑢, then 𝑢 is constant within 𝑈. Assertion (i) is the maximum principle for Laplace’s equation and (ii) is the strong maximum principle . Replacing 𝑢 by −𝑢 , we recover also similar asser- tions with “ min ” replacing “ max ”. Proof. Suppose there exists a point 𝑥 0 ∈ 𝑈 with 𝑢(𝑥 0 ) = 𝑀 ≔ max ̄ 𝑈 𝑢 . Then for 0 < 𝑟 < dist (𝑥 0 , 𝜕𝑈) , the mean-value property asserts 𝑀 = 𝑢(𝑥 0 ) = ⨍ 𝐵(𝑥 0 ,𝑟) 𝑢 𝑑𝑦 ≤ 𝑀. As equality holds only if 𝑢 ≡ 𝑀 within 𝐵(𝑥 0 , 𝑟) , we see 𝑢(𝑦) = 𝑀 for all 𝑦 ∈ 𝐵(𝑥 0 , 𝑟) . Hence the set { 𝑥 ∈ 𝑈 ∣ 𝑢(𝑥) = 𝑀 } is both open and relatively closed link to page 44 link to page 43 link to page 42 link to page 686 link to page 686 2.2. Laplace’s Equation 27 in 𝑈 and thus equals 𝑈 if 𝑈 is connected. This proves assertion (ii), from which (i) follows. □ Positivity. The strong maximum principle asserts in particular that if 𝑈 is connected and 𝑢 ∈ 𝐶 2 (𝑈) ∩ 𝐶( ̄ 𝑈) satisfies { Δ𝑢 = 0 in 𝑈 𝑢 = 𝑔 on 𝜕𝑈, where 𝑔 ≥ 0 , then 𝑢 is positive everywhere in 𝑈 if 𝑔 is positive somewhere on 𝜕𝑈 . An important application of the maximum principle is establishing the uniqueness of solutions to certain boundary-value problems for Poisson’s equa- tion. THEOREM 5 (Uniqueness) . Let 𝑔 ∈ 𝐶(𝜕𝑈) , 𝑓 ∈ 𝐶(𝑈) . Then there exists at most one solution 𝑢 ∈ 𝐶 2 (𝑈) ∩ 𝐶( ̄ 𝑈) of the boundary-value problem (17) { −Δ𝑢 = 𝑓 in 𝑈 𝑢 = 𝑔 on 𝜕𝑈 . Proof. If 𝑢 and ̃ 𝑢 both satisfy (17), apply Theorem 4 to the harmonic functions 𝑤 ≔ ±(𝑢 − ̃ 𝑢) . □ b. Regularity. Next we prove that if 𝑢 ∈ 𝐶 2 is harmonic, then necessarily 𝑢 ∈ 𝐶 ∞ . Thus harmonic functions are automatically infinitely differentiable . This sort of assertion is called a regularity theorem. The interesting point is that the algebraic structure of Laplace’s equation Δ𝑢 = ∑ 𝑛 𝑖=1 𝑢 𝑥 𝑖 𝑥 𝑖 = 0 leads to the analytic deduction that all the partial derivatives of 𝑢 exist, even those which do not appear in the PDE. THEOREM 6 (Smoothness) . If 𝑢 ∈ 𝐶(𝑈) satisfies the mean-value property (16) for each ball 𝐵(𝑥, 𝑟) ⊂ 𝑈 , then 𝑢 ∈ 𝐶 ∞ (𝑈). Note carefully that 𝑢 may not be smooth, or even continuous, up to 𝜕𝑈 . Proof. Let 𝜂 be a standard mollifier, as described in §C.5, and recall that 𝜂 is a radial function. Set 𝑢 𝜀 ≔ 𝜂 𝜀 ∗ 𝑢 in 𝑈 𝜀 = { 𝑥 ∈ 𝑈 ∣ dist(𝑥, 𝜕𝑈) > 𝜀 } . As shown in §C.5, 𝑢 𝜀 ∈ 𝐶 ∞ (𝑈 𝜀 ) . link to page 42 link to page 45 link to page 45 link to page 42 28 2. Four Important Linear Partial Differential Equations We will prove 𝑢 is smooth by demonstrating that in fact 𝑢 ≡ 𝑢 𝜀 on 𝑈 𝜀 . Indeed if 𝑥 ∈ 𝑈 𝜀 , then 𝑢 𝜀 (𝑥) = ∫ 𝑈 𝜂 𝜀 (𝑥 − 𝑦)𝑢(𝑦) 𝑑𝑦 = 1 𝜀 𝑛 ∫ 𝐵(𝑥,𝜀) 𝜂 ( |𝑥 − 𝑦| 𝜀 ) 𝑢(𝑦) 𝑑𝑦 = 1 𝜀 𝑛 ∫ 𝜀 0 𝜂 ( 𝑟 𝜀 ) (∫ 𝜕𝐵(𝑥,𝑟) 𝑢 𝑑𝑆) 𝑑𝑟 = 1 𝜀 𝑛 𝑢(𝑥) ∫ 𝜀 0 𝜂 ( 𝑟 𝜀 ) 𝑛𝛼(𝑛)𝑟 𝑛−1 𝑑𝑟 by (16) = 𝑢(𝑥) ∫ 𝐵(0,𝜀) 𝜂 𝜀 𝑑𝑦 = 𝑢(𝑥). Thus 𝑢 𝜀 ≡ 𝑢 in 𝑈 𝜀 , and so 𝑢 ∈ 𝐶 ∞ (𝑈 𝜀 ) for each 𝜀 > 0 . □ c. Local estimates for harmonic functions. Now we employ the mean- value formulas to derive careful estimates on the various partial derivatives of a harmonic function. The precise structure of these estimates will be needed below, when we prove analyticity. THEOREM 7 (Estimates on derivatives) . Assume 𝑢 is harmonic in 𝑈 . Then (18) |𝐷 𝛼 𝑢(𝑥 0 )| ≤ 𝐶 𝑘 𝑟 𝑛+𝑘 ‖𝑢‖ 𝐿 1 (𝐵(𝑥 0 ,𝑟)) for each ball 𝐵(𝑥 0 , 𝑟) ⊂ 𝑈 and each multiindex 𝛼 of order |𝛼| = 𝑘 . Here (19) 𝐶 0 = 1 𝛼(𝑛) , 𝐶 𝑘 = (2 𝑛+1 𝑛𝑘) 𝑘 𝛼(𝑛) (𝑘 = 1, . . . ). Proof. 1. We establish (18), (19) by induction on 𝑘 , the case 𝑘 = 0 being immediate from the mean-value formula (16). For 𝑘 = 1 , we note upon differentiating Laplace’s equation that 𝑢 𝑥 𝑖 (𝑖 = 1, . . . , 𝑛) is harmonic. Consequently (20) |𝑢 𝑥 𝑖 (𝑥 0 )| = ||⨍ 𝐵(𝑥 0 ,𝑟/2) 𝑢 𝑥 𝑖 𝑑𝑥|| = || 2 𝑛 𝛼(𝑛)𝑟 𝑛 ∫ 𝜕𝐵(𝑥 0 ,𝑟/2) 𝑢𝜈 𝑖 𝑑𝑆|| ≤ 2𝑛 𝑟 ‖𝑢‖ 𝐿 ∞ (𝜕𝐵(𝑥 0 , 𝑟 2 )) . link to page 45 link to page 45 link to page 45 link to page 45 link to page 45 link to page 45 link to page 45 link to page 45 link to page 45 link to page 45 link to page 45 link to page 45 2.2. Laplace’s Equation 29 Now if 𝑥 ∈ 𝜕𝐵(𝑥 0 , 𝑟/2) , then 𝐵(𝑥, 𝑟/2) ⊂ 𝐵(𝑥 0 , 𝑟) ⊂ 𝑈 , and so |𝑢(𝑥)| ≤ 1 𝛼(𝑛) ( 2 𝑟 ) 𝑛 ‖𝑢‖ 𝐿 1 (𝐵(𝑥 0 ,𝑟)) by (18), (19) for 𝑘 = 0 . Combining the inequalities above, we deduce |𝐷 𝛼 𝑢(𝑥 0 )| ≤ 2 𝑛+1 𝑛 𝛼(𝑛) 1 𝑟 𝑛+1 ‖𝑢‖ 𝐿 1 (𝐵(𝑥 0 ,𝑟)) if |𝛼| = 1 . This verifies (18), (19) for 𝑘 = 1 . 2. Assume now 𝑘 ≥ 2 and (18), (19) are valid for all balls in 𝑈 and each multiindex of order less than or equal to 𝑘 − 1 . Fix 𝐵(𝑥 0 , 𝑟) ⊂ 𝑈 and let 𝛼 be a multiindex with |𝛼| = 𝑘 . Then 𝐷 𝛼 𝑢 = (𝐷 𝛽 𝑢) 𝑥 𝑖 for some 𝑖 ∈ {1, . . . , 𝑛} , |𝛽| = 𝑘 − 1 . By calculations similar to those in (20), we establish that |𝐷 𝛼 𝑢(𝑥 0 )| ≤ 𝑛𝑘 𝑟 ‖𝐷 𝛽 𝑢‖ 𝐿 ∞ (𝜕𝐵(𝑥 0 , 𝑟 𝑘 )) . If 𝑥 ∈ 𝜕𝐵(𝑥 0 , 𝑟 𝑘 ) , then 𝐵(𝑥, 𝑘−1 𝑘 𝑟) ⊂ 𝐵(𝑥 0 , 𝑟) ⊂ 𝑈 . Thus (18), (19) for 𝑘−1 imply |𝐷 𝛽 𝑢(𝑥)| ≤ (2 𝑛+1 𝑛(𝑘 − 1)) 𝑘−1 𝛼(𝑛) ( 𝑘−1 𝑘 𝑟) 𝑛+𝑘−1 ‖𝑢‖ 𝐿 1 (𝐵(𝑥 0 ,𝑟)) . Combining the two previous estimates yields the bound (21) |𝐷 𝛼 𝑢(𝑥 0 )| ≤ (2 𝑛+1 𝑛𝑘) 𝑘 𝛼(𝑛)𝑟 𝑛+𝑘 ‖𝑢‖ 𝐿 1 (𝐵(𝑥 0 ,𝑟)) . This confirms (18), (19) for |𝛼| = 𝑘 . □ d. Liouville’s Theorem. We assert now that there are no nontrivial bounded harmonic functions on all of ℝ 𝑛 . THEOREM 8 (Liouville’s Theorem) . Suppose 𝑢 ∶ ℝ 𝑛 → ℝ is harmonic and bounded. Then 𝑢 is constant. Proof. Fix 𝑥 0 ∈ ℝ 𝑛 , 𝑟 > 0 , and apply Theorem 7 on 𝐵(𝑥 0 , 𝑟) : |𝐷𝑢(𝑥 0 )| ≤ √𝑛𝐶 1 𝑟 𝑛+1 ‖𝑢‖ 𝐿 1 (𝐵(𝑥 0 ,𝑟)) ≤ √𝑛𝐶 1 𝛼(𝑛) 𝑟 ‖𝑢‖ 𝐿 ∞ (ℝ 𝑛 ) → 0, as 𝑟 → ∞ . Thus 𝐷𝑢 ≡ 0 , and so 𝑢 is constant. □ THEOREM 9 (Representation formula) . Let 𝑓 ∈ 𝐶 2 𝑐 (ℝ 𝑛 ) , 𝑛 ≥ 3 . Then any bounded solution of −Δ𝑢 = 𝑓 in ℝ 𝑛 link to page 44 link to page 45 link to page 28 30 2. Four Important Linear Partial Differential Equations has the form 𝑢(𝑥) = ∫ ℝ 𝑛 Φ(𝑥 − 𝑦)𝑓(𝑦) 𝑑𝑦 + 𝐶 (𝑥 ∈ ℝ 𝑛 ) for some constant 𝐶 . Proof. Since Φ(𝑥) → 0 as |𝑥| → ∞ for 𝑛 ≥ 3 , ̃ 𝑢(𝑥) ≔ ∫ ℝ 𝑛 Φ(𝑥 − 𝑦)𝑓(𝑦) 𝑑𝑦 is a bounded solution of −Δ𝑢 = 𝑓 in ℝ 𝑛 . If 𝑢 is another solution, 𝑤 ≔ 𝑢 − ̃ 𝑢 is constant, according to Liouville’s Theorem. □ Remark. If 𝑛 = 2 , Φ(𝑥) = − 1 2𝜋 log |𝑥| is unbounded as |𝑥| → ∞ and so may be ∫ ℝ 2 Φ(𝑥 − 𝑦)𝑓(𝑦) 𝑑𝑦 . e. Analyticity. We refine Theorem 6: THEOREM 10 (Analyticity) . Assume 𝑢 is harmonic in 𝑈 . Then 𝑢 is analytic in 𝑈 . Proof. 1. Fix any point 𝑥 0 ∈ 𝑈 . We must show 𝑢 can be represented by a conver- gent power series in some neighborhood of 𝑥 0 . Let 𝑟 ≔ 1 4 dist(𝑥 0 , 𝜕𝑈) . Then 𝑀 ≔ 1 𝛼(𝑛)𝑟 𝑛 ‖𝑢‖ 𝐿 1 (𝐵(𝑥 0 ,2𝑟)) < ∞ . 2. Since 𝐵(𝑥, 𝑟) ⊂ 𝐵(𝑥 0 , 2𝑟) ⊂ 𝑈 for each 𝑥 ∈ 𝐵(𝑥 0 , 𝑟) , Theorem 7 provides the bound ‖𝐷 𝛼 𝑢‖ 𝐿 ∞ (𝐵(𝑥 0 ,𝑟)) ≤ 𝑀 ( 2 𝑛+1 𝑛 𝑟 ) |𝛼| |𝛼| |𝛼| . Now 𝑘 𝑘 𝑘! < 𝑒 𝑘 for all positive integers 𝑘 , and hence |𝛼| |𝛼| ≤ 𝑒 |𝛼| |𝛼|! for all multiindices 𝛼 . Furthermore, the Multinomial Theorem (§1.5) implies 𝑛 𝑘 = (1 + ⋯ + 1) 𝑘 = ∑ |𝛼|=𝑘 |𝛼|! 𝛼! , whence |𝛼|! ≤ 𝑛 |𝛼| 𝛼! . Combining the previous inequalities yields the estimate (22) ‖𝐷 𝛼 𝑢‖ 𝐿 ∞ (𝐵(𝑥 0 ,𝑟)) ≤ 𝑀 ( 2 𝑛+1 𝑛 2 𝑒 𝑟 ) |𝛼| 𝛼! . 3. The Taylor series for 𝑢 at 𝑥 0 is ∑ 𝛼 𝐷 𝛼 𝑢(𝑥 0 ) 𝛼! (𝑥 − 𝑥 0 ) 𝛼 , link to page 47 link to page 48 link to page 241 link to page 671 2.2. Laplace’s Equation 31 the sum taken over all multiindices. We assert this power series converges, provided (23) |𝑥 − 𝑥 0 | < 𝑟 2 𝑛+2 𝑛 3 𝑒 . To verify this, let us compute for each 𝑁 the remainder term: 𝑅 𝑁 (𝑥) ≔ 𝑢(𝑥) − 𝑁−1 ∑ 𝑘=0 ∑ |𝛼|=𝑘 𝐷 𝛼 𝑢(𝑥 0 )(𝑥 − 𝑥 0 ) 𝛼 𝛼! = ∑ |𝛼|=𝑁 𝐷 𝛼 𝑢(𝑥 0 + 𝑡(𝑥 − 𝑥 0 ))(𝑥 − 𝑥 0 ) 𝛼 𝛼! for some 0 ≤ 𝑡 ≤ 1 , 𝑡 depending on 𝑥 . We establish this formula by writing out the first 𝑁 terms and the error in the Taylor expansion about 0 for the function of one variable 𝑔(𝑡) ≔ 𝑢(𝑥 0 + 𝑡(𝑥 − 𝑥 0 )) , at 𝑡 = 1 . Employing (22), (23), we can estimate |𝑅 𝑁 (𝑥)| ≤ 𝑀 ∑ |𝛼|=𝑁 ( 2 𝑛+1 𝑛 2 𝑒 𝑟 ) 𝑁 ( 𝑟 2 𝑛+2 𝑛 3 𝑒 ) 𝑁 ≤ 𝑀𝑛 𝑁 1 (2𝑛) 𝑁 = 𝑀 2 𝑁 → 0 as 𝑁 → ∞. □ See §4.6.2 for more on analytic functions and partial differential equations. f. Harnack’s inequality. Recall from §A.2 that we write 𝑉 ⊂⊂ 𝑈 to mean 𝑉 ⊂ ̄ 𝑉 ⊂ 𝑈 and ̄ 𝑉 is compact. THEOREM 11 (Harnack’s inequality) . For each connected open set 𝑉 ⊂⊂ 𝑈 , there exists a positive constant 𝐶 , depending only on 𝑉 , such that sup 𝑉 𝑢 ≤ 𝐶 inf 𝑉 𝑢 for all nonnegative harmonic functions 𝑢 in 𝑈 . Thus in particular 1 𝐶 𝑢(𝑦) ≤ 𝑢(𝑥) ≤ 𝐶𝑢(𝑦) for all points 𝑥 , 𝑦 ∈ 𝑉 . These inequalities assert that the values of a nonnegative harmonic function within 𝑉 are all comparable : 𝑢 cannot be very small (or very large) at any point of 𝑉 unless 𝑢 is very small (or very large) everywhere in 𝑉 . The intuitive idea is that since 𝑉 is a positive distance away from 𝜕𝑈 , there is “room for the averaging effects of Laplace’s equation to occur”. link to page 684 link to page 39 32 2. Four Important Linear Partial Differential Equations Proof. Let 𝑟 ≔ 1 4 dist(𝑉, 𝜕𝑈) . Choose 𝑥 , 𝑦 ∈ 𝑉 , |𝑥 − 𝑦| ≤ 𝑟 . Then 𝑢(𝑥) = ⨍ 𝐵(𝑥,2𝑟) 𝑢 𝑑𝑧 ≥ 1 𝛼(𝑛)2 𝑛 𝑟 𝑛 ∫ 𝐵(𝑦,𝑟) 𝑢 𝑑𝑧 = 1 2 𝑛 ⨍ 𝐵(𝑦,𝑟) 𝑢 𝑑𝑧 = 1 2 𝑛 𝑢(𝑦). Thus 2 𝑛 𝑢(𝑦) ≥ 𝑢(𝑥) ≥ 1 2 𝑛 𝑢(𝑦) if 𝑥 , 𝑦 ∈ 𝑉 , |𝑥 − 𝑦| ≤ 𝑟 . Since 𝑉 is connected and ̄ 𝑉 is compact, we can cover ̄ 𝑉 by a chain of finitely many balls {𝐵 𝑖 } 𝑁 𝑖=1 , each of which has radius 𝑟 2 and 𝐵 𝑖 ∩𝐵 𝑖−1 ≠ ∅ for 𝑖 = 2, . . . , 𝑁 . Then 𝑢(𝑥) ≥ 1 2 𝑛(𝑁+1) 𝑢(𝑦) for all 𝑥 , 𝑦 ∈ 𝑉 . □ 2.2.4. Green’s function. Assume now 𝑈 ⊂ ℝ 𝑛 is open, bounded, and 𝜕𝑈 is 𝐶 1 . We propose next to obtain a general representation formula for the solution of Poisson’s equation −Δ𝑢 = 𝑓 in 𝑈, subject to the prescribed boundary condition 𝑢 = 𝑔 on 𝜕𝑈. a. Derivation of Green’s function. Suppose 𝑢 ∈ 𝐶 2 ( ̄ 𝑈) is an arbitrary func- tion. Fix 𝑥 ∈ 𝑈 , choose 𝜀 > 0 so small that 𝐵(𝑥, 𝜀) ⊂ 𝑈 , and apply Green’s formula from §C.2 on the region 𝑉 𝜀 ≔ 𝑈 − 𝐵(𝑥, 𝜀) to 𝑢(𝑦) and Φ(𝑦 − 𝑥) . We thereby compute (24) ∫ 𝑉 𝜀 𝑢(𝑦)ΔΦ(𝑦 − 𝑥) − Φ(𝑦 − 𝑥)Δ𝑢(𝑦) 𝑑𝑦 = ∫ 𝜕𝑉 𝜀 𝑢(𝑦) 𝜕Φ 𝜕𝜈 (𝑦 − 𝑥) − Φ(𝑦 − 𝑥) 𝜕𝑢 𝜕𝜈 (𝑦) 𝑑𝑆(𝑦), 𝝂 denoting the outer unit normal vector on 𝜕𝑉 𝜀 . Recall next ΔΦ(𝑥 − 𝑦) = 0 for 𝑥 ≠ 𝑦 . We observe also ||∫ 𝜕𝐵(𝑥,𝜀) Φ(𝑦 − 𝑥) 𝜕𝑢 𝜕𝜈 (𝑦) 𝑑𝑆(𝑦)|| ≤ 𝐶𝜀 𝑛−1 max 𝜕𝐵(0,𝜀) |Φ| = 𝑜(1) as 𝜀 → 0 . Furthermore the calculations in the proof of Theorem 1 show ∫ 𝜕𝐵(𝑥,𝜀) 𝑢(𝑦) 𝜕Φ 𝜕𝜈 (𝑦 − 𝑥) 𝑑𝑆(𝑦) = ⨍ 𝜕𝐵(𝑥,𝜀) 𝑢(𝑦) 𝑑𝑆(𝑦) → 𝑢(𝑥) link to page 49 link to page 50 link to page 50 link to page 50 link to page 50 link to page 50 2.2. Laplace’s Equation 33 as 𝜀 → 0 . Hence our sending 𝜀 → 0 in (24) yields the formula (25) 𝑢(𝑥) = ∫ 𝜕𝑈 Φ(𝑦 − 𝑥) 𝜕𝑢 𝜕𝜈 (𝑦) − 𝑢(𝑦) 𝜕Φ 𝜕𝜈 (𝑦 − 𝑥) 𝑑𝑆(𝑦) − ∫ 𝑈 Φ(𝑦 − 𝑥)Δ𝑢(𝑦) 𝑑𝑦. This identity is valid for any point 𝑥 ∈ 𝑈 and any function 𝑢 ∈ 𝐶 2 ( ̄ 𝑈) . Now formula (25) would permit us to solve for 𝑢(𝑥) if we knew the values of Δ𝑢 within 𝑈 and the values of 𝑢 , 𝜕𝑢/𝜕𝜈 along 𝜕𝑈 . However, for our application to Poisson’s equation with prescribed boundary values for 𝑢 , the normal deriv- ative 𝜕𝑢/𝜕𝜈 along 𝜕𝑈 is unknown to us. We must therefore somehow modify (25) to remove this term. The idea is now to introduce for fixed 𝑥 a corrector function 𝜙 𝑥 = 𝜙 𝑥 (𝑦) , solving the boundary-value problem (26) { Δ𝜙 𝑥 = 0 in 𝑈 𝜙 𝑥 = Φ(𝑦 − 𝑥) on 𝜕𝑈. Let us apply Green’s formula once more, to compute (27) − ∫ 𝑈 𝜙 𝑥 (𝑦)Δ𝑢(𝑦) 𝑑𝑦 = ∫ 𝜕𝑈 𝑢(𝑦) 𝜕𝜙 𝑥 𝜕𝜈 (𝑦) − 𝜙 𝑥 (𝑦) 𝜕𝑢 𝜕𝜈 (𝑦) 𝑑𝑆(𝑦) = ∫ 𝜕𝑈 𝑢(𝑦) 𝜕𝜙 𝑥 𝜕𝜈 (𝑦) − Φ(𝑦 − 𝑥) 𝜕𝑢 𝜕𝜈 (𝑦) 𝑑𝑆(𝑦). We introduce next this DEFINITION. Green’s function for the region 𝑈 is 𝐺(𝑥, 𝑦) ≔ Φ(𝑦 − 𝑥) − 𝜙 𝑥 (𝑦) (𝑥, 𝑦 ∈ 𝑈, 𝑥 ≠ 𝑦). Adopting this terminology and adding (27) to (25), we find (28) 𝑢(𝑥) = − ∫ 𝜕𝑈 𝑢(𝑦) 𝜕𝐺 𝜕𝜈 (𝑥, 𝑦) 𝑑𝑆(𝑦) − ∫ 𝑈 𝐺(𝑥, 𝑦)Δ𝑢(𝑦) 𝑑𝑦 (𝑥 ∈ 𝑈), where 𝜕𝐺 𝜕𝜈 (𝑥, 𝑦) = 𝐷 𝑦 𝐺(𝑥, 𝑦) ⋅ 𝝂(𝑦) is the outer normal derivative of 𝐺 with respect to the variable 𝑦 . Observe that the term 𝜕𝑢/𝜕𝜈 does not appear in equation (28): we introduced the corrector 𝜙 𝑥 precisely to achieve this. Suppose now 𝑢 ∈ 𝐶 2 ( ̄ 𝑈) solves the boundary-value problem (29) { −Δ𝑢 = 𝑓 in 𝑈 𝑢 = 𝑔 on 𝜕𝑈 , link to page 50 link to page 50 link to page 50 34 2. Four Important Linear Partial Differential Equations for given continuous functions 𝑓 , 𝑔 . Plugging into (28), we obtain THEOREM 12 (Representation formula using Green’s function) . If 𝑢 ∈ 𝐶 2 ( ̄ 𝑈) solves problem (29) , then (30) 𝑢(𝑥) = − ∫ 𝜕𝑈 𝑔(𝑦) 𝜕𝐺 𝜕𝜈 (𝑥, 𝑦) 𝑑𝑆(𝑦) + ∫ 𝑈 𝑓(𝑦)𝐺(𝑥, 𝑦) 𝑑𝑦 (𝑥 ∈ 𝑈). Here we have a formula for the solution of the boundary-value problem (29), provided we can construct Green’s function 𝐺 for the given domain 𝑈 . This is in general a difficult matter and can be done only when 𝑈 has simple geometry. Subsequent subsections identify some special cases for which an explicit calculation of 𝐺 is possible. Interpreting Green’s function. Fix 𝑥 ∈ 𝑈 . Then regarding 𝐺 as a function of 𝑦 , we may symbolically write { −Δ𝐺 = 𝛿 𝑥 in 𝑈 𝐺 = 0 on 𝜕𝑈 , 𝛿 𝑥 denoting the Dirac measure giving unit mass to the point 𝑥 . Before moving on to specific examples, let us record the general assertion that 𝐺 is symmetric in the variables 𝑥 and 𝑦 : THEOREM 13 (Symmetry of Green’s function) . For all 𝑥 , 𝑦 ∈ 𝑈 , 𝑥 ≠ 𝑦 , we have 𝐺(𝑦, 𝑥) = 𝐺(𝑥, 𝑦). Proof. Fix 𝑥 , 𝑦 ∈ 𝑈 , 𝑥 ≠ 𝑦 . Write 𝑣(𝑧) ≔ 𝐺(𝑥, 𝑧), 𝑤(𝑧) ≔ 𝐺(𝑦, 𝑧) (𝑧 ∈ 𝑈). Then Δ𝑣(𝑧) = 0 (𝑧 ≠ 𝑥) , Δ𝑤(𝑧) = 0 (𝑧 ≠ 𝑦) and 𝑤 = 𝑣 = 0 on 𝜕𝑈 . Thus our applying Green’s identity on 𝑉 ≔ 𝑈 − [𝐵(𝑥, 𝜀) ∪ 𝐵(𝑦, 𝜀)] for sufficiently small 𝜀 > 0 yields (31) ∫ 𝜕𝐵(𝑥,𝜀) 𝜕𝑣 𝜕𝜈 𝑤 − 𝜕𝑤 𝜕𝜈 𝑣 𝑑𝑆(𝑧) = ∫ 𝜕𝐵(𝑦,𝜀) 𝜕𝑤 𝜕𝜈 𝑣 − 𝜕𝑣 𝜕𝜈 𝑤 𝑑𝑆(𝑧), 𝝂 denoting the inward pointing unit vector field on 𝜕𝐵(𝑥, 𝜀) ∪ 𝜕𝐵(𝑦, 𝜀) . Now 𝑤 is smooth near 𝑥 , whence ||∫ 𝜕𝐵(𝑥,𝜀) 𝜕𝑤 𝜕𝜈 𝑣 𝑑𝑆|| ≤ 𝐶𝜀 𝑛−1 sup 𝜕𝐵(𝑥,𝜀) |𝑣| = 𝑜(1) as 𝜀 → 0. On the other hand, 𝑣(𝑧) = Φ(𝑧 − 𝑥) − 𝜙 𝑥 (𝑧) , where 𝜙 𝑥 is smooth in 𝑈 . Thus lim 𝜀→0 ∫ 𝜕𝐵(𝑥,𝜀) 𝜕𝑣 𝜕𝜈 𝑤 𝑑𝑆 = lim 𝜀→0 ∫ 𝜕𝐵(𝑥,𝜀) 𝜕Φ 𝜕𝜈 (𝑥 − 𝑧)𝑤(𝑧) 𝑑𝑆 = 𝑤(𝑥), link to page 39 link to page 51 link to page 50 link to page 50 2.2. Laplace’s Equation 35 by calculations as in the proof of Theorem 1. Thus the left-hand side of (31) converges to 𝑤(𝑥) as 𝜀 → 0 . Likewise the right-hand side converges to 𝑣(𝑦) . Consequently 𝐺(𝑦, 𝑥) = 𝑤(𝑥) = 𝑣(𝑦) = 𝐺(𝑥, 𝑦). □ b. Green’s function for a half-space. In this and the next subsection we will build Green’s functions for two regions with simple geometry, namely the half-space ℝ 𝑛 + and the unit ball 𝐵(0, 1) . Everything depends upon our explicitly solving the corrector problem (26) in these regions, and this in turn depends upon some clever geometric reflection tricks. First let us consider the half-space ℝ 𝑛 + = { 𝑥 = (𝑥 1 , . . . , 𝑥 𝑛 ) ∈ ℝ 𝑛 ∣ 𝑥 𝑛 > 0 }. Although this region is unbounded, and so the calculations in the previous sec- tion do not directly apply, we will attempt nevertheless to build Green’s func- tion using the ideas developed earlier. Later of course we must check directly that the corresponding representation formula is valid. DEFINITION. If 𝑥 = (𝑥 1 , . . . , 𝑥 𝑛−1 , 𝑥 𝑛 ) ∈ ℝ 𝑛 + , its reflection in the plane 𝜕ℝ 𝑛 + is the point ̃ 𝑥 = (𝑥 1 , . . . , 𝑥 𝑛−1 , −𝑥 𝑛 ). We will solve problem (26) for the half-space by setting 𝜙 𝑥 (𝑦) ≔ Φ(𝑦 − ̃ 𝑥) = Φ(𝑦 1 − 𝑥 1 , . . . , 𝑦 𝑛−1 − 𝑥 𝑛−1 , 𝑦 𝑛 + 𝑥 𝑛 ) (𝑥, 𝑦 ∈ ℝ 𝑛 + ). The idea is that the corrector 𝜙 𝑥 is built from Φ by “reflecting the singularity” from 𝑥 ∈ ℝ 𝑛 + to ̃ 𝑥 ∉ ℝ 𝑛 + . We note 𝜙 𝑥 (𝑦) = Φ(𝑦 − 𝑥) if 𝑦 ∈ 𝜕ℝ 𝑛 + , and thus { Δ𝜙 𝑥 = 0 in ℝ 𝑛 + 𝜙 𝑥 = Φ(𝑦 − 𝑥) on 𝜕ℝ 𝑛 + , as required. DEFINITION. Green’s function for the half-space ℝ 𝑛 + is 𝐺(𝑥, 𝑦) ≔ Φ(𝑦 − 𝑥) − Φ(𝑦 − ̃ 𝑥) (𝑥, 𝑦 ∈ ℝ 𝑛 + , 𝑥 ≠ 𝑦). Then 𝐺 𝑦 𝑛 (𝑥, 𝑦) = Φ 𝑦 𝑛 (𝑦 − 𝑥) − Φ 𝑦 𝑛 (𝑦 − ̃ 𝑥) = −1 𝑛𝛼(𝑛) [ 𝑦 𝑛 − 𝑥 𝑛 |𝑦 − 𝑥| 𝑛 − 𝑦 𝑛 + 𝑥 𝑛 |𝑦 − ̃ 𝑥| 𝑛 ] . link to page 51 link to page 53 link to page 53 link to page 53 link to page 53 link to page 53 36 2. Four Important Linear Partial Differential Equations Consequently if 𝑦 ∈ 𝜕ℝ 𝑛 + , 𝜕𝐺 𝜕𝜈 (𝑥, 𝑦) = −𝐺 𝑦 𝑛 (𝑥, 𝑦) = − 2𝑥 𝑛 𝑛𝛼(𝑛) 1 |𝑥 − 𝑦| 𝑛 . Suppose now 𝑢 solves the boundary-value problem (32) { Δ𝑢 = 0 in ℝ 𝑛 + 𝑢 = 𝑔 on 𝜕ℝ 𝑛 + . Then from (30) we expect (33) 𝑢(𝑥) = 2𝑥 𝑛 𝑛𝛼(𝑛) ∫ 𝜕ℝ 𝑛 + 𝑔(𝑦) |𝑥 − 𝑦| 𝑛 𝑑𝑦 (𝑥 ∈ ℝ 𝑛 + ) to be a representation formula for our solution. The function 𝐾(𝑥, 𝑦) ≔ 2𝑥 𝑛 𝑛𝛼(𝑛) 1 |𝑥 − 𝑦| 𝑛 (𝑥 ∈ ℝ 𝑛 + , 𝑦 ∈ 𝜕ℝ 𝑛 + ) is Poisson’s kernel for ℝ 𝑛 + , and (33) is Poisson’s formula . We must now check directly that formula (33) does indeed provide us with a solution of the boundary-value problem (32). THEOREM 14 (Poisson’s formula for half-space) . Assume 𝑔 ∈ 𝐶(ℝ 𝑛−1 ) ∩ 𝐿 ∞ (ℝ 𝑛−1 ) , and define 𝑢 by (33) . Then (i) 𝑢 ∈ 𝐶 ∞ (ℝ 𝑛 + ) ∩ 𝐿 ∞ (ℝ 𝑛 + ) , (ii) Δ𝑢 = 0 in ℝ 𝑛 + , and (iii) lim 𝑥→𝑥 0 𝑥∈ℝ 𝑛 + 𝑢(𝑥) = 𝑔(𝑥 0 ) for each point 𝑥 0 ∈ 𝜕ℝ 𝑛 + . Proof. 1. For each fixed 𝑥 , the mapping 𝑦 ↦ 𝐺(𝑥, 𝑦) is harmonic, except for 𝑦 = 𝑥 . As 𝐺(𝑥, 𝑦) = 𝐺(𝑦, 𝑥) , 𝑥 ↦ 𝐺(𝑥, 𝑦) is harmonic, except for 𝑥 = 𝑦 . Thus 𝑥 ↦ − 𝜕𝐺 𝜕𝑦 𝑛 (𝑥, 𝑦) = 𝐾(𝑥, 𝑦) is harmonic for 𝑥 ∈ ℝ 𝑛 + , 𝑦 ∈ 𝜕ℝ 𝑛 + . 2. A direct calculation, the details of which we omit, verifies (34) 1 = ∫ 𝜕ℝ 𝑛 + 𝐾(𝑥, 𝑦) 𝑑𝑦 for each 𝑥 ∈ ℝ 𝑛 + . As 𝑔 is bounded, 𝑢 defined by (33) is likewise bounded. Since 𝑥 ↦ 𝐾(𝑥, 𝑦) is smooth for 𝑥 ≠ 𝑦 , we easily verify as well 𝑢 ∈ 𝐶 ∞ (ℝ 𝑛 + ) , with Δ𝑢(𝑥) = ∫ 𝜕ℝ 𝑛 + Δ 𝑥 𝐾(𝑥, 𝑦)𝑔(𝑦) 𝑑𝑦 = 0 (𝑥 ∈ ℝ 𝑛 + ). link to page 53 link to page 54 link to page 54 2.2. Laplace’s Equation 37 3. Now fix 𝑥 0 ∈ 𝜕ℝ 𝑛 + , 𝜀 > 0 . Choose 𝛿 > 0 so small that (35) |𝑔(𝑦) − 𝑔(𝑥 0 )| < 𝜀 if |𝑦 − 𝑥 0 | < 𝛿, 𝑦 ∈ 𝜕ℝ 𝑛 + . Then if |𝑥 − 𝑥 0 | < 𝛿 2 , 𝑥 ∈ ℝ 𝑛 + , (36) |𝑢(𝑥) − 𝑔(𝑥 0 )| = | | | ∫ 𝜕ℝ 𝑛 + 𝐾(𝑥, 𝑦)[𝑔(𝑦) − 𝑔(𝑥 0 )] 𝑑𝑦 | | | ≤ ∫ 𝜕ℝ 𝑛 + ∩𝐵(𝑥 0 ,𝛿) 𝐾(𝑥, 𝑦)|𝑔(𝑦) − 𝑔(𝑥 0 )| 𝑑𝑦 + ∫ 𝜕ℝ 𝑛 + −𝐵(𝑥 0 ,𝛿) 𝐾(𝑥, 𝑦)|𝑔(𝑦) − 𝑔(𝑥 0 )| 𝑑𝑦 ≕ 𝐼 + 𝐽. Now (34), (35) imply 𝐼 ≤ 𝜀 ∫ 𝜕ℝ 𝑛 + 𝐾(𝑥, 𝑦) 𝑑𝑦 = 𝜀. Furthermore if |𝑥 − 𝑥 0 | ≤ 𝛿 2 and |𝑦 − 𝑥 0 | ≥ 𝛿 , we have |𝑦 − 𝑥 0 | ≤ |𝑦 − 𝑥| + 𝛿 2 ≤ |𝑦 − 𝑥| + 1 2 |𝑦 − 𝑥 0 |; and so |𝑦 − 𝑥| ≥ 1 2 |𝑦 − 𝑥 0 | . Thus 𝐽 ≤ 2‖𝑔‖ 𝐿 ∞ ∫ 𝜕ℝ 𝑛 + −𝐵(𝑥 0 ,𝛿) 𝐾(𝑥, 𝑦) 𝑑𝑦 ≤ 2 𝑛+2 ‖𝑔‖ 𝐿 ∞ 𝑥 𝑛 𝑛𝛼(𝑛) ∫ 𝜕ℝ 𝑛 + −𝐵(𝑥 0 ,𝛿) |𝑦 − 𝑥 0 | −𝑛 𝑑𝑦 → 0 as 𝑥 𝑛 → 0 + . Combining this calculation with estimate (36), we deduce |𝑢(𝑥) − 𝑔(𝑥 0 )| ≤ 2𝜀 , provided |𝑥 − 𝑥 0 | is sufficiently small. □ c. Green’s function for a ball. To construct Green’s function for the unit ball 𝐵(0, 1) , we will again employ a kind of reflection, this time through the sphere 𝜕𝐵(0, 1) . DEFINITION. If 𝑥 ∈ ℝ 𝑛 − {0} , the point ̃ 𝑥 = 𝑥 |𝑥| 2 is called the point dual to 𝑥 with respect to 𝜕𝐵(0, 1) . The mapping 𝑥 ↦ ̃ 𝑥 is inversion through the unit sphere 𝜕𝐵(0, 1) . link to page 51 link to page 55 38 2. Four Important Linear Partial Differential Equations We now employ inversion through the sphere to compute Green’s function for the unit ball 𝑈 = 𝐵 0 (0, 1) . Fix 𝑥 ∈ 𝐵 0 (0, 1) . Remember that we must find a corrector function 𝜙 𝑥 = 𝜙 𝑥 (𝑦) solving (37) { Δ𝜙 𝑥 = 0 in 𝐵 0 (0, 1) 𝜙 𝑥 = Φ(𝑦 − 𝑥) on 𝜕𝐵(0, 1); then Green’s function will be (38) 𝐺(𝑥, 𝑦) = Φ(𝑦 − 𝑥) − 𝜙 𝑥 (𝑦). The idea now is to “invert the singularity” from 𝑥 ∈ 𝐵 0 (0, 1) to ̃ 𝑥 ∉ 𝐵(0, 1) . Assume for the moment 𝑛 ≥ 3 . Now the mapping 𝑦 ↦ Φ(𝑦 − ̃ 𝑥) is harmonic for 𝑦 ≠ ̃ 𝑥 . Thus 𝑦 ↦ |𝑥| 2−𝑛 Φ(𝑦 − ̃ 𝑥) is harmonic for 𝑦 ≠ ̃ 𝑥 , and so (39) 𝜙 𝑥 (𝑦) ≔ Φ(|𝑥|(𝑦 − ̃ 𝑥)) is harmonic in 𝑈 . Furthermore, if 𝑦 ∈ 𝜕𝐵(0, 1) and 𝑥 ≠ 0 , |𝑥| 2 |𝑦 − ̃ 𝑥| 2 = |𝑥| 2 (|𝑦| 2 − 2𝑦 ⋅ 𝑥 |𝑥| 2 + 1 |𝑥| 2 ) = |𝑥| 2 − 2𝑦 ⋅ 𝑥 + 1 = |𝑥 − 𝑦| 2 . Thus (|𝑥||𝑦 − ̃ 𝑥|) −(𝑛−2) = |𝑥 − 𝑦| −(𝑛−2) . Consequently (40) 𝜙 𝑥 (𝑦) = Φ(𝑦 − 𝑥) (𝑦 ∈ 𝜕𝐵(0, 1)), as required. DEFINITION. Green’s function for the unit ball is (41) 𝐺(𝑥, 𝑦) ≔ Φ(𝑦 − 𝑥) − Φ(|𝑥|(𝑦 − ̃ 𝑥)) (𝑥, 𝑦 ∈ 𝐵(0, 1), 𝑥 ≠ 𝑦). The same formula is valid for 𝑛 = 2 as well. Assume now 𝑢 solves the boundary-value problem (42) { Δ𝑢 = 0 in 𝐵 0 (0, 1) 𝑢 = 𝑔 on 𝜕𝐵(0, 1) . Then using (30), we see (43) 𝑢(𝑥) = − ∫ 𝜕𝐵(0,1) 𝑔(𝑦) 𝜕𝐺 𝜕𝜈 (𝑥, 𝑦) 𝑑𝑆(𝑦). According to formula (41), 𝐺 𝑦 𝑖 (𝑥, 𝑦) = Φ 𝑦 𝑖 (𝑦 − 𝑥) − Φ(|𝑥|(𝑦 − ̃ 𝑥)) 𝑦 𝑖 . But Φ 𝑦 𝑖 (𝑦 − 𝑥) = 1 𝑛𝛼(𝑛) 𝑥 𝑖 − 𝑦 𝑖 |𝑥 − 𝑦| 𝑛 , link to page 55 link to page 55 link to page 55 link to page 56 link to page 56 link to page 56 link to page 53 2.2. Laplace’s Equation 39 and furthermore Φ(|𝑥|(𝑦 − ̃ 𝑥)) 𝑦 𝑖 . = −1 𝑛𝛼(𝑛) 𝑦 𝑖 |𝑥| 2 − 𝑥 𝑖 (|𝑥||𝑦 − ̃ 𝑥|) 𝑛 = − 1 𝑛𝛼(𝑛) 𝑦 𝑖 |𝑥| 2 − 𝑥 𝑖 |𝑥 − 𝑦| 𝑛 if 𝑦 ∈ 𝜕𝐵(0, 1) . Accordingly 𝜕𝐺 𝜕𝜈 (𝑥, 𝑦) = 𝑛 ∑ 𝑖=1 𝑦 𝑖 𝐺 𝑦 𝑖 (𝑥, 𝑦) = −1 𝑛𝛼(𝑛) 1 |𝑥 − 𝑦| 𝑛 𝑛 ∑ 𝑖=1 𝑦 𝑖 ((𝑦 𝑖 − 𝑥 𝑖 ) − 𝑦 𝑖 |𝑥| 2 + 𝑥 𝑖 ) = −1 𝑛𝛼(𝑛) 1 − |𝑥| 2 |𝑥 − 𝑦| 𝑛 . Hence formula (43) yields the representation formula 𝑢(𝑥) = 1 − |𝑥| 2 𝑛𝛼(𝑛) ∫ 𝜕𝐵(0,1) 𝑔(𝑦) |𝑥 − 𝑦| 𝑛 𝑑𝑆(𝑦). Suppose now instead of (42) 𝑢 solves the boundary-value problem (44) { Δ𝑢 = 0 in 𝐵 0 (0, 𝑟) 𝑢 = 𝑔 on 𝜕𝐵(0, 𝑟) for 𝑟 > 0 . Then ̃ 𝑢(𝑥) = 𝑢(𝑟𝑥) solves (42), with ̃ 𝑔(𝑥) = 𝑔(𝑟𝑥) replacing 𝑔 . We change variables to obtain Poisson’s formula (45) 𝑢(𝑥) = 𝑟 2 − |𝑥| 2 𝑛𝛼(𝑛)𝑟 ∫ 𝜕𝐵(0,𝑟) 𝑔(𝑦) |𝑥 − 𝑦| 𝑛 𝑑𝑆(𝑦) (𝑥 ∈ 𝐵 0 (0, 𝑟)). The function 𝐾(𝑥, 𝑦) ≔ 𝑟 2 − |𝑥| 2 𝑛𝛼(𝑛)𝑟 1 |𝑥 − 𝑦| 𝑛 (𝑥 ∈ 𝐵 0 (0, 𝑟), 𝑦 ∈ 𝜕𝐵(0, 𝑟)) is Poisson’s kernel for the ball 𝐵(0, 𝑟) . We have established (45) under the assumption that a smooth solution of (44) exists. We next assert that this formula in fact gives a solution: THEOREM 15 (Poisson’s formula for ball) . Assume 𝑔 ∈ 𝐶(𝜕𝐵(0, 𝑟)) and define 𝑢 by (45) . Then (i) 𝑢 ∈ 𝐶 ∞ (𝐵 0 (0, 𝑟)) , (ii) Δ𝑢 = 0 in 𝐵 0 (0, 𝑟) , and (iii) lim 𝑥→𝑥 0 𝑥∈𝐵 0 (0,𝑟) 𝑢(𝑥) = 𝑔(𝑥 0 ) for each point 𝑥 0 ∈ 𝜕𝐵(0, 𝑟) . The proof is similar to that for Theorem 14 and is left as an exercise. link to page 43 link to page 57 link to page 57 link to page 57 link to page 57 link to page 57 40 2. Four Important Linear Partial Differential Equations 2.2.5. Energy methods. Most of our analysis of harmonic functions thus far has depended upon fairly explicit representation formulas entailing the fun- damental solution, Green’s functions, etc. In this concluding subsection we illustrate some “energy” methods, which is to say techniques involving the 𝐿 2 - norms of various expressions. These ideas foreshadow later theoretical devel- opments in Parts II and III. a. Uniqueness. Consider first the boundary-value problem (46) { −Δ𝑢 = 𝑓 in 𝑈 𝑢 = 𝑔 on 𝜕𝑈 . We have already employed the maximum principle in §2.2.3 to show unique- ness, but now we set forth a simple alternative proof. Assume 𝑈 is open, bounded, and 𝜕𝑈 is 𝐶 1 . THEOREM 16 (Uniqueness) . There exists at most one solution 𝑢 ∈ 𝐶 2 ( ̄ 𝑈) of (46) . Proof. Assume ̃ 𝑢 is another solution and set 𝑤 ≔ 𝑢 − ̃ 𝑢 . Then Δ𝑤 = 0 in 𝑈 , and so an integration by parts shows 0 = − ∫ 𝑈 𝑤Δ𝑤 𝑑𝑥 = ∫ 𝑈 |𝐷𝑤| 2 𝑑𝑥. Thus 𝐷𝑤 ≡ 0 in 𝑈 , and, since 𝑤 = 0 on 𝜕𝑈 , we deduce 𝑤 = 𝑢 − ̃ 𝑢 ≡ 0 in 𝑈 . □ b. Dirichlet’s principle. Next let us demonstrate that a solution of the boundary-value problem (46) for Poisson’s equation can be characterized as the minimizer of an appropriate functional. For this, we define the energy func- tional 𝐼[𝑤] ≔ ∫ 𝑈 1 2 |𝐷𝑤| 2 − 𝑤𝑓 𝑑𝑥, 𝑤 belonging to the admissible set 𝒜 ≔ { 𝑤 ∈ 𝐶 2 ( ̄ 𝑈) ∣ 𝑤 = 𝑔 on 𝜕𝑈 }. THEOREM 17 (Dirichlet’s principle) . Assume 𝑢 ∈ 𝐶 2 ( ̄ 𝑈) solves (46) . Then (47) 𝐼[𝑢] = min 𝑤∈𝒜 𝐼[𝑤]. Conversely, if 𝑢 ∈ 𝒜 satisfies (47) , then 𝑢 solves the boundary-value problem (46) . In other words if 𝑢 ∈ 𝒜 , the PDE −Δ𝑢 = 𝑓 is equivalent to the statement that 𝑢 minimizes the energy 𝐼[ ⋅ ] among functions in 𝒜 . link to page 57 link to page 678 link to page 57 link to page 58 link to page 57 2.2. Laplace’s Equation 41 Proof. 1. Choose 𝑤 ∈ 𝒜 . Then (46) implies 0 = ∫ 𝑈 (−Δ𝑢 − 𝑓)(𝑢 − 𝑤) 𝑑𝑥. An integration by parts yields 0 = ∫ 𝑈 𝐷𝑢 ⋅ 𝐷(𝑢 − 𝑤) − 𝑓(𝑢 − 𝑤) 𝑑𝑥, and there is no boundary term since 𝑢 − 𝑤 = 𝑔 − 𝑔 ≡ 0 on 𝜕𝑈 . Hence ∫ 𝑈 |𝐷𝑢| 2 − 𝑢𝑓 𝑑𝑥 = ∫ 𝑈 𝐷𝑢 ⋅ 𝐷𝑤 − 𝑤𝑓 𝑑𝑥 ≤ ∫ 𝑈 1 2 |𝐷𝑢| 2 𝑑𝑥 + ∫ 𝑈 1 2 |𝐷𝑤| 2 − 𝑤𝑓 𝑑𝑥, where we employed the estimates |𝐷𝑢 ⋅ 𝐷𝑤| ≤ |𝐷𝑢||𝐷𝑤| ≤ 1 2 |𝐷𝑢| 2 + 1 2 |𝐷𝑤| 2 , following from the Cauchy–Schwarz and Cauchy inequalities (§B.2). Rearrang- ing, we conclude (48) 𝐼[𝑢] ≤ 𝐼[𝑤] (𝑤 ∈ 𝒜). Since 𝑢 ∈ 𝒜 , (47) follows from (48). 2. Now, conversely, suppose (47) holds. Fix any 𝑣 ∈ 𝐶 ∞ 𝑐 (𝑈) and write 𝑖(𝜏) ≔ 𝐼[𝑢 + 𝜏𝑣] (𝜏 ∈ ℝ). Since 𝑢 + 𝜏𝑣 ∈ 𝒜 for each 𝜏 , the scalar function 𝑖(⋅) has a minimum at zero, and thus 𝑖 ′ (0) = 0 ( ′ = 𝑑 𝑑𝜏 ) , provided this derivative exists. But 𝑖(𝜏) = ∫ 𝑈 1 2 |𝐷𝑢 + 𝜏𝐷𝑣| 2 − (𝑢 + 𝜏𝑣)𝑓 𝑑𝑥 = ∫ 𝑈 1 2 |𝐷𝑢| 2 + 𝜏𝐷𝑢 ⋅ 𝐷𝑣 + 𝜏 2 2 |𝐷𝑣| 2 − (𝑢 + 𝜏𝑣)𝑓 𝑑𝑥. Consequently 0 = 𝑖 ′ (0) = ∫ 𝑈 𝐷𝑢 ⋅ 𝐷𝑣 − 𝑣𝑓 𝑑𝑥 = ∫ 𝑈 (−Δ𝑢 − 𝑓)𝑣 𝑑𝑥. This identity is valid for each function 𝑣 ∈ 𝐶 ∞ 𝑐 (𝑈) and so −Δ𝑢 = 𝑓 in 𝑈 . □ link to page 444 link to page 59 link to page 59 42 2. Four Important Linear Partial Differential Equations Dirichlet’s principle is an instance of the calculus of variations applied to Laplace’s equation. See Chapter 8 for more. 2.3. HEAT EQUATION Next we study the heat equation (1) 𝑢 𝑡 − Δ𝑢 = 0 and the nonhomogeneous heat equation (2) 𝑢 𝑡 − Δ𝑢 = 𝑓, subject to appropriate initial and boundary conditions. Here 𝑡 > 0 and 𝑥 ∈ 𝑈 , where 𝑈 ⊆ ℝ 𝑛 is open. The unknown is 𝑢 ∶ ̄ 𝑈 × [0, ∞) → ℝ , 𝑢 = 𝑢(𝑥, 𝑡) , and the Laplacian Δ is taken with respect to the spatial variables 𝑥 = (𝑥 1 , . . . , 𝑥 𝑛 ) : Δ𝑢 = Δ 𝑥 𝑢 = ∑ 𝑛 𝑖=1 𝑢 𝑥 𝑖 𝑥 𝑖 . In (2) the function 𝑓 ∶ 𝑈 × [0, ∞) → ℝ is given. A guiding principle is that any assertion about harmonic functions yields an analogous (but more complicated) statement about solutions of the heat equation. Accordingly our development will largely parallel the corresponding theory for Laplace’s equation. Physical interpretation. The heat equation, also known as the diffusion equation , describes in typical applications the evolution in time of the density 𝑢 of some quantity such as heat, chemical concentration, etc. If 𝑉 ⊂ 𝑈 is any smooth subregion, the rate of change of the total quantity within 𝑉 equals the negative of the net flux through 𝜕𝑉 : 𝑑 𝑑𝑡 ∫ 𝑉 𝑢 𝑑𝑥 = − ∫ 𝜕𝑉 F ⋅ 𝝂 𝑑𝑆, F being the flux density. Thus (3) 𝑢 𝑡 = − div F , as 𝑉 was arbitrary. In many situations F is proportional to the gradient of 𝑢 but points in the opposite direction (since the flow is from regions of higher to lower concentration): F = −𝑎𝐷𝑢 (𝑎 > 0). Substituting into (3), we obtain the PDE 𝑢 𝑡 = 𝑎 div(𝐷𝑢) = 𝑎Δ𝑢, which for 𝑎 = 1 is the heat equation. The heat equation appears as well in the study of Brownian motion. link to page 38 link to page 59 link to page 59 link to page 100 link to page 60 link to page 60 link to page 60 link to page 59 link to page 60 link to page 60 link to page 60 2.3. Heat Equation 43 2.3.1. Fundamental solution. a. Derivation of the fundamental solution. As noted in §2.2.1 an impor- tant first step in studying any PDE is often to come up with some specific solu- tions. We observe that the heat equation involves one derivative with respect to the time variable 𝑡 , but two derivatives with respect to the space variables 𝑥 𝑖 (𝑖 = 1, . . . , 𝑛) . Consequently we see that if 𝑢 solves (1), then so does 𝑢(𝜆𝑥, 𝜆 2 𝑡) for 𝜆 ∈ ℝ . This scaling indicates the ratio 𝑟 2 𝑡 (𝑟 = |𝑥|) is important for the heat equation and suggests that we search for a solution of (1) having the form 𝑢(𝑥, 𝑡) = 𝑣( 𝑟 2 𝑡 ) = 𝑣( |𝑥| 2 𝑡 ) (𝑡 > 0, 𝑥 ∈ ℝ 𝑛 ) , for some function 𝑣 as yet undeter- mined. Although this approach eventually leads to what we want (see Problem 13), it is quicker to seek a solution 𝑢 having the special structure (4) 𝑢(𝑥, 𝑡) = 1 𝑡 𝛼 𝑣 ( 𝑥 𝑡 𝛽 ) (𝑥 ∈ ℝ 𝑛 , 𝑡 > 0), where the constants 𝛼 , 𝛽 and the function 𝑣 ∶ ℝ 𝑛 → ℝ must be found. We come to (4) if we look for a solution 𝑢 of the heat equation invariant under the dilation scaling 𝑢(𝑥, 𝑡) ↦ 𝜆 𝛼 𝑢(𝜆 𝛽 𝑥, 𝜆𝑡). That is, we ask that 𝑢(𝑥, 𝑡) = 𝜆 𝛼 𝑢(𝜆 𝛽 𝑥, 𝜆𝑡) for all 𝜆 > 0 , 𝑥 ∈ ℝ 𝑛 , 𝑡 > 0 . Setting 𝜆 = 𝑡 −1 , we derive (4) for 𝑣(𝑦) ≔ 𝑢(𝑦, 1) . Let us insert (4) into (1) and thereafter compute (5) 𝛼𝑡 −(𝛼+1) 𝑣(𝑦) + 𝛽𝑡 −(𝛼+1) 𝑦 ⋅ 𝐷𝑣(𝑦) + 𝑡 −(𝛼+2𝛽) Δ𝑣(𝑦) = 0 for 𝑦 ≔ 𝑡 −𝛽 𝑥 . In order to transform (5) into an expression involving the variable 𝑦 alone, we take 𝛽 = 1 2 . Then the terms with 𝑡 are identical, and so (5) reduces to (6) 𝛼𝑣 + 1 2 𝑦 ⋅ 𝐷𝑣 + Δ𝑣 = 0. We simplify further by guessing 𝑣 to be radial; that is, 𝑣(𝑦) = 𝑤(|𝑦|) for some 𝑤 ∶ ℝ → ℝ . Thereupon (6) becomes 𝛼𝑤 + 1 2 𝑟𝑤 ′ + 𝑤 ″ + 𝑛 − 1 𝑟 𝑤 ′ = 0, for 𝑟 = |𝑦| , ′ = 𝑑 𝑑𝑟 . Now if we set 𝛼 = 𝑛 2 , this simplifies to read (𝑟 𝑛−1 𝑤 ′ ) ′ + 1 2 (𝑟 𝑛 𝑤) ′ = 0. link to page 60 link to page 61 link to page 59 link to page 194 44 2. Four Important Linear Partial Differential Equations Thus 𝑟 𝑛−1 𝑤 ′ + 1 2 𝑟 𝑛 𝑤 = 𝑎 for some constant 𝑎 . Assuming 𝑤, 𝑤 ′ → 0 fast enough as 𝑟 → ∞ , we conclude 𝑎 = 0 ; whence 𝑤 ′ = − 1 2 𝑟𝑤. But then for some constant 𝑏 (7) 𝑤 = 𝑏𝑒 − 𝑟2 4 . Combining (4), (7) and our choices for 𝛼 , 𝛽 , we conclude that 𝑏 𝑡 𝑛/2 𝑒 − |𝑥|2 4𝑡 solves the heat equation (1). This computation motivates the following DEFINITION. The function Φ(𝑥, 𝑡) ≔ { 1 (4𝜋𝑡) 𝑛/2 𝑒 − |𝑥|2 4𝑡 (𝑥 ∈ ℝ 𝑛 , 𝑡 > 0) 0 (𝑥 ∈ ℝ 𝑛 , 𝑡 < 0) is called the fundamental solution of the heat equation . Notice that Φ is singular at the point (0, 0) . We will sometimes write Φ(𝑥, 𝑡) = Φ(|𝑥|, 𝑡) to emphasize that the fundamental solution is radial in the variable 𝑥 . The choice of the normalizing constant (4𝜋) −𝑛/2 is dictated by the following LEMMA (Integral of fundamental solution) . For each time 𝑡 > 0 , ∫ ℝ 𝑛 Φ(𝑥, 𝑡) 𝑑𝑥 = 1. Proof. We calculate ∫ ℝ 𝑛 Φ(𝑥, 𝑡) 𝑑𝑥 = 1 (4𝜋𝑡) 𝑛/2 ∫ ℝ 𝑛 𝑒 − |𝑥|2 4𝑡 𝑑𝑥 = 1 𝜋 𝑛/2 ∫ ℝ 𝑛 𝑒 −|𝑧| 2 𝑑𝑧 = 1 𝜋 𝑛/2 𝑛 ∏ 𝑖=1 ∫ ∞ −∞ 𝑒 −𝑧 2 𝑖 𝑑𝑧 𝑖 = 1. □ A different derivation of the fundamental solution of the heat equation ap- pears in §4.3.1. link to page 62 2.3. Heat Equation 45 b. Initial-value problem. We now employ Φ to fashion a solution to the initial-value (or Cauchy ) problem (8) { 𝑢 𝑡 − Δ𝑢 = 0 in ℝ 𝑛 × (0, ∞) 𝑢 = 𝑔 on ℝ 𝑛 × {𝑡 = 0} . Let us note that the function (𝑥, 𝑡) ↦ Φ(𝑥, 𝑡) solves the heat equation away from the singularity at (0, 0) , and thus so does (𝑥, 𝑡) ↦ Φ(𝑥 − 𝑦, 𝑡) for each fixed 𝑦 ∈ ℝ 𝑛 . Consequently the convolution (9) 𝑢(𝑥, 𝑡) = ∫ ℝ 𝑛 Φ(𝑥 − 𝑦, 𝑡)𝑔(𝑦) 𝑑𝑦 = 1 (4𝜋𝑡) 𝑛/2 ∫ ℝ 𝑛 𝑒 − |𝑥−𝑦|2 4𝑡 𝑔(𝑦) 𝑑𝑦 (𝑥 ∈ ℝ 𝑛 , 𝑡 > 0) should also be a solution. THEOREM 1 (Solution of initial-value problem) . Assume 𝑔 ∈ 𝐶(ℝ 𝑛 )∩𝐿 ∞ (ℝ 𝑛 ) , and define 𝑢 by (9) . Then (i) 𝑢 ∈ 𝐶 ∞ (ℝ 𝑛 × (0, ∞)) , (ii) 𝑢 𝑡 (𝑥, 𝑡) − Δ𝑢(𝑥, 𝑡) = 0 ( 𝑥 ∈ ℝ 𝑛 , 𝑡 > 0 ), and (iii) lim (𝑥,𝑡)→(𝑥 0 ,0) 𝑥∈ℝ 𝑛 , 𝑡>0 𝑢(𝑥, 𝑡) = 𝑔(𝑥 0 ) for each point 𝑥 0 ∈ ℝ 𝑛 . Proof. 1. Since the function 1 𝑡 𝑛/2 𝑒 − |𝑥|2 4𝑡 is infinitely differentiable, with uniformly bounded derivatives of all orders, on ℝ 𝑛 × [𝛿, ∞) for each 𝛿 > 0 , we see that 𝑢 ∈ 𝐶 ∞ (ℝ 𝑛 × (0, ∞)) . Furthermore (10) 𝑢 𝑡 (𝑥, 𝑡) − Δ𝑢(𝑥, 𝑡) = ∫ ℝ 𝑛 [(Φ 𝑡 − Δ 𝑥 Φ)(𝑥 − 𝑦, 𝑡)]𝑔(𝑦) 𝑑𝑦 = 0 (𝑥 ∈ ℝ 𝑛 , 𝑡 > 0), since Φ itself solves the heat equation. 2. Fix 𝑥 0 ∈ ℝ 𝑛 , 𝜀 > 0 . Choose 𝛿 > 0 such that (11) |𝑔(𝑦) − 𝑔(𝑥 0 )| < 𝜀 if |𝑦 − 𝑥 0 | < 𝛿, 𝑦 ∈ ℝ 𝑛 . link to page 62 link to page 62 46 2. Four Important Linear Partial Differential Equations Then if |𝑥 − 𝑥 0 | < 𝛿 2 , we have, according to the Lemma, |𝑢(𝑥, 𝑡) − 𝑔(𝑥 0 )| = ||∫ ℝ 𝑛 Φ(𝑥 − 𝑦, 𝑡)[𝑔(𝑦) − 𝑔(𝑥 0 )] 𝑑𝑦|| ≤ ∫ 𝐵(𝑥 0 ,𝛿) Φ(𝑥 − 𝑦, 𝑡)|𝑔(𝑦) − 𝑔(𝑥 0 )| 𝑑𝑦 + ∫ ℝ 𝑛 −𝐵(𝑥 0 ,𝛿) Φ(𝑥 − 𝑦, 𝑡)|𝑔(𝑦) − 𝑔(𝑥 0 )| 𝑑𝑦 ≕ 𝐼 + 𝐽. Now 𝐼 ≤ 𝜀 ∫ ℝ 𝑛 Φ(𝑥 − 𝑦, 𝑡) 𝑑𝑦 = 𝜀, owing to (11) and the Lemma. Furthermore, if |𝑥 − 𝑥 0 | ≤ 𝛿 2 and |𝑦 − 𝑥 0 | ≥ 𝛿 , then |𝑦 − 𝑥 0 | ≤ |𝑦 − 𝑥| + 𝛿 2 ≤ |𝑦 − 𝑥| + 1 2 |𝑦 − 𝑥 0 |. Thus |𝑦 − 𝑥| ≥ 1 2 |𝑦 − 𝑥 0 | . Consequently 𝐽 ≤ 2‖𝑔‖ 𝐿 ∞ ∫ ℝ 𝑛 −𝐵(𝑥 0 ,𝛿) Φ(𝑥 − 𝑦, 𝑡) 𝑑𝑦 ≤ 𝐶 𝑡 𝑛/2 ∫ ℝ 𝑛 −𝐵(𝑥 0 ,𝛿) 𝑒 − |𝑥−𝑦|2 4𝑡 𝑑𝑦 ≤ 𝐶 𝑡 𝑛/2 ∫ ℝ 𝑛 −𝐵(𝑥 0 ,𝛿) 𝑒 − |𝑦−𝑥0|2 16𝑡 𝑑𝑦 = 𝐶 ∫ ℝ 𝑛 −𝐵(0,𝛿/√𝑡) 𝑒 − |𝑧|2 16 𝑑𝑧 → 0 as 𝑡 → 0 + . Hence if |𝑥 − 𝑥 0 | < 𝛿 2 and 𝑡 > 0 is small enough, |𝑢(𝑥, 𝑡) − 𝑔(𝑥 0 )| < 2𝜀 . □ Interpretation of fundamental solution. In view of Theorem 1 we some- times write { Φ 𝑡 − ΔΦ = 0 in ℝ 𝑛 × (0, ∞) Φ = 𝛿 0 on ℝ 𝑛 × {𝑡 = 0}, 𝛿 0 denoting the Dirac measure on ℝ 𝑛 giving unit mass to the point 0 . Infinite propagation speed. Notice that if 𝑔 is bounded, continuous, 𝑔 ≥ 0 , 𝑔 ≢ 0 , then 𝑢(𝑥, 𝑡) = 1 (4𝜋𝑡) 𝑛/2 ∫ ℝ 𝑛 𝑒 − |𝑥−𝑦|2 4𝑡 𝑔(𝑦) 𝑑𝑦 is in fact positive for all points 𝑥 ∈ ℝ 𝑛 and times 𝑡 > 0 . We interpret this observation by saying the heat equation forces infinite propagation speed for link to page 96 link to page 62 link to page 62 link to page 64 link to page 64 link to page 64 link to page 64 link to page 64 link to page 36 link to page 94 2.3. Heat Equation 47 disturbances. If the initial temperature is nonnegative and is positive some- where, the temperature at any later time (no matter how small) is everywhere positive. (We will learn in §2.4.3 that the wave equation in contrast supports finite propagation speed for disturbances.) c. Nonhomogeneous problem. Now let us turn our attention to the nonho- mogeneous initial-value problem (12) { 𝑢 𝑡 − Δ𝑢 = 𝑓 in ℝ 𝑛 × (0, ∞) 𝑢 = 0 on ℝ 𝑛 × {𝑡 = 0} . How can we produce a formula for the solution? If we recall the motivation leading up to (9), we should note further that the mapping (𝑥, 𝑡) ↦ Φ(𝑥−𝑦, 𝑡−𝑠) is a solution of the heat equation (for given 𝑦 ∈ ℝ 𝑛 , 0 < 𝑠 < 𝑡 ). Now for fixed 𝑠 , the function 𝑢 = 𝑢(𝑥, 𝑡; 𝑠) = ∫ ℝ 𝑛 Φ(𝑥 − 𝑦, 𝑡 − 𝑠)𝑓(𝑦, 𝑠) 𝑑𝑦 solves ( 12 𝑠 ) { 𝑢 𝑡 (⋅; 𝑠) − Δ𝑢(⋅; 𝑠) = 0 in ℝ 𝑛 × (𝑠, ∞) 𝑢(⋅; 𝑠) = 𝑓(⋅, 𝑠) on ℝ 𝑛 × {𝑡 = 𝑠}, which is just an initial-value problem of the form (8), with the starting time 𝑡 = 0 replaced by 𝑡 = 𝑠 and 𝑔 replaced by 𝑓(⋅, 𝑠) . Thus 𝑢(⋅; 𝑠) is certainly not a solution of (12). However Duhamel’s principle ∗ asserts that we can build a solution of (12) out of the solutions of ( 12 𝑠 ), by integrating with respect to 𝑠 . The idea is to consider 𝑢(𝑥, 𝑡) = ∫ 𝑡 0 𝑢(𝑥, 𝑡; 𝑠) 𝑑𝑠 (𝑥 ∈ ℝ 𝑛 , 𝑡 ≥ 0). Rewriting, we have (13) 𝑢(𝑥, 𝑡) = ∫ 𝑡 0 ∫ ℝ 𝑛 Φ(𝑥 − 𝑦, 𝑡 − 𝑠)𝑓(𝑦, 𝑠) 𝑑𝑦𝑑𝑠 = ∫ 𝑡 0 1 (4𝜋(𝑡 − 𝑠)) 𝑛/2 ∫ ℝ 𝑛 𝑒 − |𝑥−𝑦|2 4(𝑡−𝑠) 𝑓(𝑦, 𝑠) 𝑑𝑦𝑑𝑠, for 𝑥 ∈ ℝ 𝑛 , 𝑡 > 0 . To confirm that formula (13) works, let us for simplicity assume 𝑓 ∈ 𝐶 2 1 (ℝ 𝑛 × [0, ∞)) and 𝑓 has compact support in the variable 𝑥 . ∗ Duhamel’s principle has wide applicability to linear ODE and PDE and does not depend on the spe- cific structure of the heat equation. It yields, for example, the solution of the nonhomogeneous transport equation, obtained by different means in §2.1.2. We will invoke Duhamel’s principle for the wave equation in §2.4.2. link to page 64 link to page 39 link to page 38 48 2. Four Important Linear Partial Differential Equations THEOREM 2 (Solution of nonhomogeneous problem) . Define 𝑢 by (13) . Then (i) 𝑢 ∈ 𝐶 2 1 (ℝ 𝑛 × (0, ∞)) , (ii) 𝑢 𝑡 (𝑥, 𝑡) − Δ𝑢(𝑥, 𝑡) = 𝑓(𝑥, 𝑡) (𝑥 ∈ ℝ 𝑛 , 𝑡 > 0) , and (iii) lim (𝑥,𝑡)→(𝑥 0 ,0) 𝑥∈ℝ 𝑛 , 𝑡>0 𝑢(𝑥, 𝑡) = 0 for each point 𝑥 0 ∈ ℝ 𝑛 . Proof. 1. Since Φ has a singularity at (0, 0) , we cannot directly justify differenti- ating under the integral sign. We instead proceed somewhat as in the proof of Theorem 1 in §2.2.1. First we change variables, to write 𝑢(𝑥, 𝑡) = ∫ 𝑡 0 ∫ ℝ 𝑛 Φ(𝑦, 𝑠)𝑓(𝑥 − 𝑦, 𝑡 − 𝑠) 𝑑𝑦𝑑𝑠. As 𝑓 ∈ 𝐶 2 1 (ℝ 𝑛 × [0, ∞)) has compact support and Φ = Φ(𝑦, 𝑠) is smooth near 𝑠 = 𝑡 > 0 , we compute 𝑢 𝑡 (𝑥, 𝑡) = ∫ 𝑡 0 ∫ ℝ 𝑛 Φ(𝑦, 𝑠)𝑓 𝑡 (𝑥 − 𝑦, 𝑡 − 𝑠) 𝑑𝑦𝑑𝑠 + ∫ ℝ 𝑛 Φ(𝑦, 𝑡)𝑓(𝑥 − 𝑦, 0) 𝑑𝑦 and 𝑢 𝑥 𝑖 𝑥 𝑗 (𝑥, 𝑡) = ∫ 𝑡 0 ∫ ℝ 𝑛 Φ(𝑦, 𝑠)𝑓 𝑥 𝑖 𝑥 𝑗 (𝑥 − 𝑦, 𝑡 − 𝑠) 𝑑𝑦𝑑𝑠 (𝑖, 𝑗 = 1, . . . , 𝑛). Thus 𝑢 𝑡 , 𝐷 2 𝑥 𝑢 , and likewise 𝑢 , 𝐷 𝑥 𝑢 , belong to 𝐶(ℝ 𝑛 × (0, ∞)) . 2. We then calculate (14) 𝑢 𝑡 (𝑥, 𝑡)−Δ𝑢(𝑥, 𝑡) = ∫ 𝑡 0 ∫ ℝ 𝑛 Φ(𝑦, 𝑠)[( 𝜕 𝜕𝑡 −Δ 𝑥 )𝑓(𝑥−𝑦, 𝑡−𝑠)] 𝑑𝑦𝑑𝑠 + ∫ ℝ 𝑛 Φ(𝑦, 𝑡)𝑓(𝑥−𝑦, 0) 𝑑𝑦 = ∫ 𝑡 𝜀 ∫ ℝ 𝑛 Φ(𝑦, 𝑠)[(− 𝜕 𝜕𝑠 −Δ 𝑦 )𝑓(𝑥−𝑦, 𝑡−𝑠)] 𝑑𝑦𝑑𝑠 + ∫ 𝜀 0 ∫ ℝ 𝑛 Φ(𝑦, 𝑠)[(− 𝜕 𝜕𝑠 −Δ 𝑦 )𝑓(𝑥−𝑦, 𝑡−𝑠)] 𝑑𝑦𝑑𝑠 + ∫ ℝ 𝑛 Φ(𝑦, 𝑡)𝑓(𝑥−𝑦, 0) 𝑑𝑦. ≕ 𝐼 𝜀 + 𝐽 𝜀 + 𝐾. link to page 65 link to page 66 link to page 62 link to page 62 link to page 65 link to page 671 2.3. Heat Equation 49 Now (15) |𝐽 𝜀 | ≤ (‖𝑓 𝑡 ‖ 𝐿 ∞ + ‖𝐷 2 𝑓‖ 𝐿 ∞ ) ∫ 𝜀 0 ∫ ℝ 𝑛 Φ(𝑦, 𝑠) 𝑑𝑦𝑑𝑠 ≤ 𝜀𝐶, by the Lemma. Integrating by parts, we also find (16) 𝐼 𝜀 = ∫ 𝑡 𝜀 ∫ ℝ 𝑛 [( 𝜕 𝜕𝑠 − Δ 𝑦 )Φ(𝑦, 𝑠)]𝑓(𝑥 − 𝑦, 𝑡 − 𝑠) 𝑑𝑦𝑑𝑠 + ∫ ℝ 𝑛 Φ(𝑦, 𝜀)𝑓(𝑥 − 𝑦, 𝑡 − 𝜀) 𝑑𝑦 − ∫ ℝ 𝑛 Φ(𝑦, 𝑡)𝑓(𝑥 − 𝑦, 0) 𝑑𝑦 = ∫ ℝ 𝑛 Φ(𝑦, 𝜀)𝑓(𝑥 − 𝑦, 𝑡 − 𝜀) 𝑑𝑦 − 𝐾, since Φ solves the heat equation. Combining (14)–(16), we ascertain 𝑢 𝑡 (𝑥, 𝑡) − Δ𝑢(𝑥, 𝑡) = lim 𝜀→0 ∫ ℝ 𝑛 Φ(𝑦, 𝜀)𝑓(𝑥 − 𝑦, 𝑡 − 𝜀) 𝑑𝑦 = 𝑓(𝑥, 𝑡) (𝑥 ∈ ℝ 𝑛 , 𝑡 > 0), the limit as 𝜀 → 0 being computed as in the proof of Theorem 1. Finally note ‖𝑢(⋅, 𝑡)‖ 𝐿 ∞ ≤ 𝑡‖𝑓‖ 𝐿 ∞ → 0 . □ Solution of nonhomogeneous problem with general initial data. We can of course combine Theorems 1 and 2 to discover that (17) 𝑢(𝑥, 𝑡) = ∫ ℝ 𝑛 Φ(𝑥 − 𝑦, 𝑡)𝑔(𝑦) 𝑑𝑦 + ∫ 𝑡 0 ∫ ℝ 𝑛 Φ(𝑥 − 𝑦, 𝑡 − 𝑠)𝑓(𝑦, 𝑠) 𝑑𝑦𝑑𝑠 is, under the hypotheses on 𝑔 and 𝑓 as above, a solution of (18) { 𝑢 𝑡 − Δ𝑢 = 𝑓 in ℝ 𝑛 × (0, ∞) 𝑢 = 𝑔 on ℝ 𝑛 × {𝑡 = 0} . 2.3.2. Mean-value formula. First we recall some useful notation from §A.2. Assume 𝑈 ⊂ ℝ 𝑛 is open and bounded, and fix a time 𝑇 > 0 . DEFINITIONS. (i) We define the parabolic cylinder 𝑈 𝑇 ≔ 𝑈 × (0, 𝑇]. (ii) The parabolic boundary of 𝑈 𝑇 is Γ 𝑇 ≔ ̄ 𝑈 𝑇 − 𝑈 𝑇 . link to page 42 50 2. Four Important Linear Partial Differential Equations We interpret 𝑈 𝑇 as being the parabolic interior of ̄ 𝑈 × [0, 𝑇] : note carefully that 𝑈 𝑇 includes the top 𝑈 × {𝑡 = 𝑇} . The parabolic boundary Γ 𝑇 comprises the bottom and vertical sides of 𝑈 × [0, 𝑇] , but not the top. The region U 𝑇 We want next to derive a kind of analogue to the mean-value property for harmonic functions, as discussed in §2.2.2. There is no such simple formula. However let us observe that for fixed 𝑥 the spheres 𝜕𝐵(𝑥, 𝑟) are level sets of the fundamental solution Φ(𝑥 − 𝑦) for Laplace’s equation. This suggests that perhaps for fixed (𝑥, 𝑡) the level sets of fundamental solution Φ(𝑥 − 𝑦, 𝑡 − 𝑠) for the heat equation may be relevant. DEFINITION. For fixed 𝑥 ∈ ℝ 𝑛 , 𝑡 ∈ ℝ , 𝑟 > 0 , we define 𝐸(𝑥, 𝑡; 𝑟) ≔ { (𝑦, 𝑠) ∈ ℝ 𝑛+1 ∣ 𝑠 ≤ 𝑡, Φ(𝑥 − 𝑦, 𝑡 − 𝑠) ≥ 1 𝑟 𝑛 } . This is a region in space-time, the boundary of which is a level set of Φ(𝑥 − 𝑦, 𝑡 − 𝑠) . Note that the point (𝑥, 𝑡) is at the center of the top. 𝐸(𝑥, 𝑡; 𝑟) is sometimes called a “heat ball”. THEOREM 3 (A mean-value property for the heat equation) . Let 𝑢 ∈ 𝐶 2 1 (𝑈 𝑇 ) solve the heat equation. Then (19) 𝑢(𝑥, 𝑡) = 1 4𝑟 𝑛 ∬ 𝐸(𝑥,𝑡; 𝑟) 𝑢(𝑦, 𝑠) |𝑥 − 𝑦| 2 (𝑡 − 𝑠) 2 𝑑𝑦𝑑𝑠 for each 𝐸(𝑥, 𝑡; 𝑟) ⊂ 𝑈 𝑇 . link to page 67 link to page 68 2.3. Heat Equation 51 A “heat ball” Formula (19) is a sort of analogue for the heat equation of the mean-value formulas for Laplace’s equation. Observe that the right-hand side involves only 𝑢(𝑦, 𝑠) for times 𝑠 ≤ 𝑡 . This is reasonable, as the value 𝑢(𝑥, 𝑡) should not depend upon future times. Proof. Shift the space and time coordinates so that 𝑥 = 0 and 𝑡 = 0 . Upon mollifying if necessary, we may assume 𝑢 is smooth. Write 𝐸(𝑟) = 𝐸(0, 0; 𝑟) and set (20) 𝜙(𝑟) ≔ 1 𝑟 𝑛 ∬ 𝐸(𝑟) 𝑢(𝑦, 𝑠) |𝑦| 2 𝑠 2 𝑑𝑦𝑑𝑠 = ∬ 𝐸(1) 𝑢(𝑟𝑦, 𝑟 2 𝑠) |𝑦| 2 𝑠 2 𝑑𝑦𝑑𝑠. We compute 𝜙 ′ (𝑟) = ∬ 𝐸(1) 𝑛 ∑ 𝑖=1 𝑢 𝑦 𝑖 𝑦 𝑖 |𝑦| 2 𝑠 2 + 2𝑟𝑢 𝑠 |𝑦| 2 𝑠 𝑑𝑦𝑑𝑠 = 1 𝑟 𝑛+1 ∬ 𝐸(𝑟) 𝑛 ∑ 𝑖=1 𝑢 𝑦 𝑖 𝑦 𝑖 |𝑦| 2 𝑠 2 + 2𝑢 𝑠 |𝑦| 2 𝑠 𝑑𝑦𝑑𝑠 ≕ 𝐴 + 𝐵. Also, let us introduce the useful function (21) 𝜓 ≔ − 𝑛 2 log(−4𝜋𝑠) + |𝑦| 2 4𝑠 + 𝑛 log 𝑟 and observe 𝜓 = 0 on 𝜕𝐸(𝑟) − {(0, 0)} , since Φ(𝑦, −𝑠) = 𝑟 −𝑛 on 𝜕𝐸(𝑟) . We utilize (21) to write 𝐵 = 1 𝑟 𝑛+1 ∬ 𝐸(𝑟) 4𝑢 𝑠 𝑛 ∑ 𝑖=1 𝑦 𝑖 𝜓 𝑦 𝑖 𝑑𝑦𝑑𝑠 = − 1 𝑟 𝑛+1 ∬ 𝐸(𝑟) 4𝑛𝑢 𝑠 𝜓 + 4 𝑛 ∑ 𝑖=1 𝑢 𝑠𝑦 𝑖 𝑦 𝑖 𝜓 𝑑𝑦𝑑𝑠; link to page 68 52 2. Four Important Linear Partial Differential Equations there is no boundary term since 𝜓 = 0 on 𝜕𝐸(𝑟) − {(0, 0)} . Integrating by parts with respect to 𝑠 , we discover 𝐵 = 1 𝑟 𝑛+1 ∬ 𝐸(𝑟) −4𝑛𝑢 𝑠 𝜓 + 4 𝑛 ∑ 𝑖=1 𝑢 𝑦 𝑖 𝑦 𝑖 𝜓 𝑠 𝑑𝑦𝑑𝑠 = 1 𝑟 𝑛+1 ∬ 𝐸(𝑟) −4𝑛𝑢 𝑠 𝜓 + 4 𝑛 ∑ 𝑖=1 𝑢 𝑦 𝑖 𝑦 𝑖 (− 𝑛 2𝑠 − |𝑦| 2 4𝑠 2 ) 𝑑𝑦𝑑𝑠 = 1 𝑟 𝑛+1 ∬ 𝐸(𝑟) −4𝑛𝑢 𝑠 𝜓 − 2𝑛 𝑠 𝑛 ∑ 𝑖=1 𝑢 𝑦 𝑖 𝑦 𝑖 𝑑𝑦𝑑𝑠 − 𝐴. Consequently, since 𝑢 solves the heat equation, 𝜙 ′ (𝑟) = 𝐴 + 𝐵 = 1 𝑟 𝑛+1 ∬ 𝐸(𝑟) −4𝑛Δ𝑢𝜓 − 2𝑛 𝑠 𝑛 ∑ 𝑖=1 𝑢 𝑦 𝑖 𝑦 𝑖 𝑑𝑦𝑑𝑠 = 𝑛 ∑ 𝑖=1 1 𝑟 𝑛+1 ∬ 𝐸(𝑟) 4𝑛𝑢 𝑦 𝑖 𝜓 𝑦 𝑖 − 2𝑛 𝑠 𝑢 𝑦 𝑖 𝑦 𝑖 𝑑𝑦𝑑𝑠 = 0 , according to (21). Thus 𝜙 is constant, and therefore 𝜙(𝑟) = lim 𝑡→0 𝜙(𝑡) = 𝑢(0, 0)(lim 𝑡→0 1 𝑡 𝑛 ∬ 𝐸(𝑡) |𝑦| 2 𝑠 2 𝑑𝑦𝑑𝑠) = 4𝑢(0, 0), as 1 𝑡 𝑛 ∬ 𝐸(𝑡) |𝑦| 2 𝑠 2 𝑑𝑦𝑑𝑠 = ∬ 𝐸(1) |𝑦| 2 𝑠 2 𝑑𝑦𝑑𝑠 = 4. We omit the details of this last computation. □ 2.3.3. Properties of solutions. a. Strong maximum principle, uniqueness. First we employ the mean- value property to give a quick proof of the strong maximum principle. THEOREM 4 (Strong maximum principle for the heat equation) . Assume 𝑢 ∈ 𝐶 2 1 (𝑈 𝑇 ) ∩ 𝐶( ̄ 𝑈 𝑇 ) solves the heat equation in 𝑈 𝑇 . (i) Then max ̄ 𝑈 𝑇 𝑢 = max Γ 𝑇 𝑢. (ii) Furthermore, if 𝑈 is connected and there exists a point (𝑥 0 , 𝑡 0 ) ∈ 𝑈 𝑇 such that 𝑢(𝑥 0 , 𝑡 0 ) = max ̄ 𝑈 𝑇 𝑢, 2.3. Heat Equation 53 Strong maximum principle for the heat equation then 𝑢 is constant in ̄ 𝑈 𝑡 0 . Assertion (i) is the maximum principle for the heat equation and (ii) is the strong maximum principle . Similar assertions are valid with “min” replacing “max”. Interpretation. So if 𝑢 attains its maximum (or minimum) at an interior point, then 𝑢 is constant at all earlier times . This accords with our strong intuitive understanding of the variable 𝑡 as denoting time: the solution will be constant on the time interval [0, 𝑡 0 ] provided the initial and boundary conditions are constant. However, the solution may change at times 𝑡 > 𝑡 0 , provided the boundary conditions alter after 𝑡 0 . The solution will however not respond to changes in boundary conditions until these changes happen. Take note that whereas all this is obvious on intuitive, physical grounds, such insights do not constitute a proof. The task is to deduce such behavior from the PDE. Proof. 1. Suppose there exists a point (𝑥 0 , 𝑡 0 ) ∈ 𝑈 𝑇 with 𝑢(𝑥 0 , 𝑡 0 ) = 𝑀 ≔ max ̄ 𝑈 𝑇 𝑢 . Then for all sufficiently small 𝑟 > 0 , 𝐸(𝑥 0 , 𝑡 0 ; 𝑟) ⊂ 𝑈 𝑇 ; and we employ 54 2. Four Important Linear Partial Differential Equations the mean-value property to deduce 𝑀 = 𝑢(𝑥 0 , 𝑡 0 ) = 1 4𝑟 𝑛 ∬ 𝐸(𝑥 0 ,𝑡 0 ; 𝑟) 𝑢(𝑦, 𝑠) |𝑥 0 − 𝑦| 2 (𝑡 0 − 𝑠) 2 𝑑𝑦𝑑𝑠 ≤ 𝑀, since 1 = 1 4𝑟 𝑛 ∬ 𝐸(𝑥 0 ,𝑡 0 ;𝑟) |𝑥 0 − 𝑦| 2 (𝑡 0 − 𝑠) 2 𝑑𝑦𝑑𝑠. Equality holds only if 𝑢 is identically equal to 𝑀 within 𝐸(𝑥 0 , 𝑡 0 ; 𝑟) . Conse- quently 𝑢(𝑦, 𝑠) = 𝑀 for all (𝑦, 𝑠) ∈ 𝐸(𝑥 0 , 𝑡 0 ; 𝑟). Draw any line segment 𝐿 in 𝑈 𝑇 connecting (𝑥 0 , 𝑡 0 ) with some other point (𝑦 0 , 𝑠 0 ) ∈ 𝑈 𝑇 , with 𝑠 0 < 𝑡 0 . Consider 𝑟 0 ≔ min{ 𝑠 ≥ 𝑠 0 ∣ 𝑢(𝑥, 𝑡) = 𝑀 for all points (𝑥, 𝑡) ∈ 𝐿 , 𝑠 ≤ 𝑡 ≤ 𝑡 0 }. Since 𝑢 is continuous, the minimum is attained. Assume 𝑟 0 > 𝑠 0 . Then 𝑢(𝑧 0 , 𝑟 0 ) = 𝑀 for some point (𝑧 0 , 𝑟 0 ) on 𝐿 ∩ 𝑈 𝑇 and so 𝑢 ≡ 𝑀 on 𝐸(𝑧 0 , 𝑟 0 ; 𝑟) for all sufficiently small 𝑟 > 0 . Since 𝐸(𝑧 0 , 𝑟 0 ; 𝑟) contains 𝐿 ∩ {𝑟 0 − 𝜎 ≤ 𝑡 ≤ 𝑟 0 } for some small 𝜎 > 0 , we have a contradiction. Thus 𝑟 0 = 𝑠 0 , and hence 𝑢 ≡ 𝑀 on 𝐿 . 2. Now fix any point 𝑥 ∈ 𝑈 and any time 0 ≤ 𝑡 < 𝑡 0 . There exist points {𝑥 0 , 𝑥 1 , . . . , 𝑥 𝑚 = 𝑥} such that the line segments in ℝ 𝑛 connecting 𝑥 𝑖−1 to 𝑥 𝑖 lie in 𝑈 for 𝑖 = 1 , . . . , 𝑚 . (This follows since the set of points in 𝑈 which can be so connected to 𝑥 0 by a polygonal path is nonempty, open and relatively closed in 𝑈 .) Select times 𝑡 0 > 𝑡 1 > ⋯ > 𝑡 𝑚 = 𝑡 . Then the line segments in ℝ 𝑛+1 connecting (𝑥 𝑖−1 , 𝑡 𝑖−1 ) to (𝑥 𝑖 , 𝑡 𝑖 ) (𝑖 = 1, . . . , 𝑚) lie in 𝑈 𝑇 . According to step 1, 𝑢 ≡ 𝑀 on each such segment and so 𝑢(𝑥, 𝑡) = 𝑀 . □ Infinite propagation speed again. The strong maximum principle implies that if 𝑈 is connected and 𝑢 ∈ 𝐶 2 1 (𝑈 𝑇 ) ∩ 𝐶( ̄ 𝑈 𝑇 ) satisfies ⎧ ⎨ ⎩ 𝑢 𝑡 − Δ𝑢 = 0 in 𝑈 𝑇 𝑢 = 0 on 𝜕𝑈 × [0, 𝑇] 𝑢 = 𝑔 on 𝑈 × {𝑡 = 0} where 𝑔 ≥ 0 , then 𝑢 is positive everywhere within 𝑈 𝑇 if 𝑔 is positive somewhere on 𝑈 . This is another illustration of infinite propagation speed for disturbances. An important application of the maximum principle is the following uniqueness assertion. link to page 72 link to page 69 link to page 60 link to page 69 2.3. Heat Equation 55 THEOREM 5 (Uniqueness on bounded domains) . Let 𝑔 ∈ 𝐶(Γ 𝑇 ) , 𝑓 ∈ 𝐶(𝑈 𝑇 ) . Then there exists at most one solution 𝑢 ∈ 𝐶 2 1 (𝑈 𝑇 )∩𝐶( ̄ 𝑈 𝑇 ) of the initial/boundary- value problem (22) { 𝑢 𝑡 − Δ𝑢 = 𝑓 in 𝑈 𝑇 𝑢 = 𝑔 on Γ 𝑇 . Proof. If 𝑢 and ̃ 𝑢 are two solutions of (22), apply Theorem 4 to 𝑤 ≔ ±(𝑢 − ̃ 𝑢) . □ We next extend our uniqueness assertion to the Cauchy problem , that is, the initial-value problem for 𝑈 = ℝ 𝑛 . As we are no longer on a bounded region, we must introduce some control on the behavior of solutions for large |𝑥| . THEOREM 6 (Maximum principle for the Cauchy problem) . Suppose 𝑢 ∈ 𝐶 2 1 (ℝ 𝑛 × (0, 𝑇]) ∩ 𝐶(ℝ 𝑛 × [0, 𝑇]) solves (23) { 𝑢 𝑡 − Δ𝑢 = 0 in ℝ 𝑛 × (0, 𝑇) 𝑢 = 𝑔 on ℝ 𝑛 × {𝑡 = 0} and satisfies the growth estimate (24) 𝑢(𝑥, 𝑡) ≤ 𝐴𝑒 𝑎|𝑥| 2 (𝑥 ∈ ℝ 𝑛 , 0 ≤ 𝑡 ≤ 𝑇) for constants 𝐴 , 𝑎 > 0 . Then sup ℝ 𝑛 ×[0,𝑇] 𝑢 = sup ℝ 𝑛 𝑔. Proof. 1. First assume (25) 4𝑎𝑇 < 1, in which case (26) 4𝑎(𝑇 + 𝜀) < 1 for some 𝜀 > 0 . Fix 𝑦 ∈ ℝ 𝑛 , 𝜇 > 0 , and define 𝑣(𝑥, 𝑡) ≔ 𝑢(𝑥, 𝑡) − 𝜇 (𝑇 + 𝜀 − 𝑡) 𝑛/2 𝑒 |𝑥−𝑦|2 4(𝑇+𝜀−𝑡) (𝑥 ∈ ℝ 𝑛 , 𝑡 > 0). A direct calculation (cf. §2.3.1) shows 𝑣 𝑡 − Δ𝑣 = 0 in ℝ 𝑛 × (0, 𝑇]. Fix 𝑟 > 0 and set 𝑈 ≔ 𝐵 0 (𝑦, 𝑟) , 𝑈 𝑇 = 𝐵 0 (𝑦, 𝑟) × (0, 𝑇] . Then according to Theorem 4, (27) max ̄ 𝑈 𝑇 𝑣 = max Γ 𝑇 𝑣 link to page 72 link to page 72 link to page 72 link to page 73 link to page 72 link to page 72 link to page 73 link to page 73 link to page 72 56 2. Four Important Linear Partial Differential Equations where Γ 𝑇 is the parabolic boundary of 𝑈 𝑇 . 2. Now if 𝑥 ∈ ℝ 𝑛 , (28) 𝑣(𝑥, 0) = 𝑢(𝑥, 0) − 𝜇 (𝑇 + 𝜀) 𝑛/2 𝑒 |𝑥−𝑦|2 4(𝑇+𝜀) ≤ 𝑢(𝑥, 0) = 𝑔(𝑥); and if |𝑥 − 𝑦| = 𝑟 , 0 ≤ 𝑡 ≤ 𝑇 , then 𝑣(𝑥, 𝑡) = 𝑢(𝑥, 𝑡) − 𝜇 (𝑇 + 𝜀 − 𝑡) 𝑛/2 𝑒 𝑟2 4(𝑇+𝜀−𝑡) ≤ 𝐴𝑒 𝑎|𝑥| 2 − 𝜇 (𝑇 + 𝜀 − 𝑡) 𝑛/2 𝑒 𝑟2 4(𝑇+𝜀−𝑡) by (24) ≤ 𝐴𝑒 𝑎(|𝑦|+𝑟) 2 − 𝜇 (𝑇 + 𝜀) 𝑛/2 𝑒 𝑟2 4(𝑇+𝜀) . Now according to (26), 1 4(𝑇+𝜀) = 𝑎 + 𝛾 for some 𝛾 > 0 . Thus we may continue the calculation above to find (29) 𝑣(𝑥, 𝑡) ≤ 𝐴𝑒 𝑎(|𝑦|+𝑟) 2 − 𝜇(4(𝑎 + 𝛾)) 𝑛/2 𝑒 (𝑎+𝛾)𝑟 2 ≤ sup ℝ 𝑛 𝑔, for 𝑟 selected sufficiently large. Thus (27)–(29) imply 𝑣(𝑦, 𝑡) ≤ sup ℝ 𝑛 𝑔 for all 𝑦 ∈ ℝ 𝑛 , 0 ≤ 𝑡 ≤ 𝑇 , provided (25) is valid. Let 𝜇 → 0 . 3. In the general case that (25) fails, we repeatedly apply the result above on the time intervals [0, 𝑇 1 ] , [𝑇 1 , 2𝑇 1 ] , etc., for 𝑇 1 = 1 8𝑎 . □ THEOREM 7 (Uniqueness for Cauchy problem) . Let 𝑔 ∈ 𝐶(ℝ 𝑛 ) , 𝑓 ∈ 𝐶(ℝ 𝑛 × [0, 𝑇]) . Then there exists at most one solution 𝑢 ∈ 𝐶 2 1 (ℝ 𝑛 ×(0, 𝑇])∩𝐶(ℝ 𝑛 ×[0, 𝑇]) of the initial-value problem (30) { 𝑢 𝑡 − Δ𝑢 = 𝑓 in ℝ 𝑛 × (0, 𝑇) 𝑢 = 𝑔 on ℝ 𝑛 × {𝑡 = 0} satisfying the growth estimate (31) |𝑢(𝑥, 𝑡)| ≤ 𝐴𝑒 𝑎|𝑥| 2 (𝑥 ∈ ℝ 𝑛 , 0 ≤ 𝑡 ≤ 𝑇) for constants 𝐴 , 𝑎 > 0 . Proof. If 𝑢 and ̃ 𝑢 both satisfy (30), (31), we apply Theorem 6 to 𝑤 ≔ ±(𝑢 − ̃ 𝑢) . □ link to page 711 link to page 74 link to page 73 link to page 106 link to page 560 link to page 588 link to page 671 2.3. Heat Equation 57 Nonphysical solutions. There are in fact infinitely many solutions of (32) { 𝑢 𝑡 − Δ𝑢 = 0 in ℝ 𝑛 × (0, 𝑇) 𝑢 = 0 on ℝ 𝑛 × {𝑡 = 0} ; see for instance John [ J2 , Chapter 7]. Each of these solutions besides 𝑢 ≡ 0 grows very rapidly as |𝑥| → ∞ . There is an interesting point here: although 𝑢 ≡ 0 is certainly the “phys- ically correct” solution of (32), this initial-value problem in fact admits other, “nonphysical”, solutions. Theorem 7 provides a criterion which excludes the “wrong” solutions. We will encounter somewhat analogous situations in our study of Hamilton–Jacobi equations and conservation laws, in Chapters 3, 10 and 11. b. Regularity. We next demonstrate that solutions of the heat equation are automatically smooth. THEOREM 8 (Smoothness) . Suppose 𝑢 ∈ 𝐶 2 1 (𝑈 𝑇 ) solves the heat equation in 𝑈 𝑇 . Then 𝑢 ∈ 𝐶 ∞ (𝑈 𝑇 ). This regularity assertion is valid even if 𝑢 attains nonsmooth boundary val- ues on Γ 𝑇 . Proof. 1. Recall from §A.2 that we write 𝐶(𝑥, 𝑡; 𝑟) = { (𝑦, 𝑠) ∣ |𝑥 − 𝑦| ≤ 𝑟, 𝑡 − 𝑟 2 ≤ 𝑠 ≤ 𝑡 } to denote the closed circular cylinder of radius 𝑟 , height 𝑟 2 , and top center point (𝑥, 𝑡) . Fix (𝑥 0 , 𝑡 0 ) ∈ 𝑈 𝑇 and choose 𝑟 > 0 so small that 𝐶 ≔ 𝐶(𝑥 0 , 𝑡 0 ; 𝑟) ⊂ 𝑈 𝑇 . Define also the smaller cylinders 𝐶 ′ ≔ 𝐶(𝑥 0 , 𝑡 0 ; 3 4 𝑟) , 𝐶 ″ ≔ 𝐶(𝑥 0 , 𝑡 0 ; 1 2 𝑟) , which have the same top center point (𝑥 0 , 𝑡 0 ) . Choose a smooth cutoff function 𝜁 = 𝜁(𝑥, 𝑡) such that { 0 ≤ 𝜁 ≤ 1, 𝜁 ≡ 1 on 𝐶 ′ , 𝜁 ≡ 0 near the parabolic boundary of 𝐶 . Extend 𝜁 ≡ 0 in (ℝ 𝑛 × [0, 𝑡 0 ]) − 𝐶 . 2. Assume temporarily that 𝑢 ∈ 𝐶 ∞ (𝑈 𝑇 ) and set 𝑣(𝑥, 𝑡) ≔ 𝜁(𝑥, 𝑡)𝑢(𝑥, 𝑡) (𝑥 ∈ ℝ 𝑛 , 0 ≤ 𝑡 ≤ 𝑡 0 ). Then 𝑣 𝑡 = 𝜁𝑢 𝑡 + 𝜁 𝑡 𝑢, Δ𝑣 = 𝜁Δ𝑢 + 2𝐷𝜁 ⋅ 𝐷𝑢 + 𝑢Δ𝜁. link to page 65 link to page 73 link to page 75 link to page 75 58 2. Four Important Linear Partial Differential Equations Consequently (33) 𝑣 = 0 on ℝ 𝑛 × {𝑡 = 0}, and (34) 𝑣 𝑡 − Δ𝑣 = 𝜁 𝑡 𝑢 − 2𝐷𝜁 ⋅ 𝐷𝑢 − 𝑢Δ𝜁 ≕ ̃ 𝑓 in ℝ 𝑛 × (0, 𝑡 0 ) . Now set ̃ 𝑣(𝑥, 𝑡) ≔ ∫ 𝑡 0 ∫ ℝ 𝑛 Φ(𝑥 − 𝑦, 𝑡 − 𝑠) ̃ 𝑓(𝑦, 𝑠) 𝑑𝑦𝑑𝑠. According to Theorem 2 (35) { ̃ 𝑣 𝑡 − Δ ̃ 𝑣 = ̃ 𝑓 in ℝ 𝑛 × (0, 𝑡 0 ) ̃ 𝑣 = 0 on ℝ 𝑛 × {𝑡 = 0} . Since |𝑣|, | ̃ 𝑣| ≤ 𝐴 for some constant 𝐴 , Theorem 7 implies 𝑣 ≡ ̃ 𝑣 ; that is, (36) 𝑣(𝑥, 𝑡) = ∫ 𝑡 0 ∫ ℝ 𝑛 Φ(𝑥 − 𝑦, 𝑡 − 𝑠) ̃ 𝑓(𝑦, 𝑠) 𝑑𝑦𝑑𝑠. Now suppose (𝑥, 𝑡) ∈ 𝐶 ″ . As 𝜁 ≡ 0 off the cylinder 𝐶 , (34) and (36) imply 𝑢(𝑥, 𝑡) = ∬ 𝐶 Φ(𝑥 − 𝑦, 𝑡 − 𝑠)[(𝜁 𝑠 (𝑦, 𝑠) − Δ𝜁(𝑦, 𝑠))𝑢(𝑦, 𝑠) − 2𝐷𝜁(𝑦, 𝑠) ⋅ 𝐷𝑢(𝑦, 𝑠)] 𝑑𝑦𝑑𝑠. Note in this equation that the expression in the square brackets vanishes in some region near the singularity of Φ . Integrate the last term by parts: (37) 𝑢(𝑥, 𝑡) = ∬ 𝐶 [Φ(𝑥 − 𝑦, 𝑡 − 𝑠)(𝜁 𝑠 (𝑦, 𝑠) + Δ𝜁(𝑦, 𝑠)) + 2𝐷 𝑦 Φ(𝑥 − 𝑦, 𝑡 − 𝑠) ⋅ 𝐷𝜁(𝑦, 𝑠)]𝑢(𝑦, 𝑠) 𝑑𝑦𝑑𝑠. link to page 75 link to page 75 link to page 76 link to page 74 2.3. Heat Equation 59 We have proved this formula assuming 𝑢 ∈ 𝐶 ∞ . If 𝑢 satisfies only the hypothe- ses of the theorem, we derive (37) with 𝑢 𝜀 = 𝜂 𝜀 ∗ 𝑢 replacing 𝑢 , 𝜂 𝜀 being the standard mollifier in the variables 𝑥 and 𝑡 , and let 𝜀 → 0 . 3. Formula (37) has the form (38) 𝑢(𝑥, 𝑡) = ∬ 𝐶 𝐾(𝑥, 𝑡, 𝑦, 𝑠)𝑢(𝑦, 𝑠) 𝑑𝑦𝑑𝑠 ((𝑥, 𝑡) ∈ 𝐶 ″ ), where 𝐾(𝑥, 𝑡, 𝑦, 𝑠) = 0 for all points (𝑦, 𝑠) ∈ 𝐶 ′ , since 𝜁 ≡ 1 on 𝐶 ′ . Note also 𝐾 is smooth on 𝐶 − 𝐶 ′ . In view of expression (38), we see 𝑢 is 𝐶 ∞ within 𝐶 ″ = 𝐶(𝑥 0 , 𝑡 0 ; 1 2 𝑟) . □ c. Local estimates for solutions of the heat equation. Let us now record some estimates on the derivatives of solutions to the heat equation, paying at- tention to the differences between derivatives with respect to 𝑥 𝑖 ( 𝑖 = 1 , . . . , 𝑛 ) and with respect to 𝑡 . THEOREM 9 (Estimates on derivatives) . There exists for each pair of integers 𝑘 , 𝑙 = 0 , 1 , . . . a constant 𝐶 𝑘,𝑙 such that max 𝐶(𝑥,𝑡; 𝑟/2) |𝐷 𝑘 𝑥 𝐷 𝑙 𝑡 𝑢| ≤ 𝐶 𝑘𝑙 𝑟 𝑘+2𝑙+𝑛+2 ‖𝑢‖ 𝐿 1 (𝐶(𝑥,𝑡;𝑟)) for all cylinders 𝐶(𝑥, 𝑡; 𝑟/2) ⊂ 𝐶(𝑥, 𝑡; 𝑟) ⊂ 𝑈 𝑇 and all solutions 𝑢 of the heat equation in 𝑈 𝑇 . Proof. 1. Fix some point in 𝑈 𝑇 . Upon shifting the coordinates, we may as well assume the point is (0, 0) . Suppose first that the cylinder 𝐶(1) ≔ 𝐶(0, 0; 1) lies in 𝑈 𝑇 . Let 𝐶 ( 1 2 ) ≔ 𝐶 (0, 0; 1 2 ) . Then, as in the proof of Theorem 8, 𝑢(𝑥, 𝑡) = ∬ 𝐶(1) 𝐾(𝑥, 𝑡, 𝑦, 𝑠)𝑢(𝑦, 𝑠) 𝑑𝑦𝑑𝑠 ((𝑥, 𝑡) ∈ 𝐶( 1 2 )) for some smooth function 𝐾 . Consequently (39) |𝐷 𝑘 𝑥 𝐷 𝑙 𝑡 𝑢(𝑥, 𝑡)| ≤ ∬ 𝐶(1) |𝐷 𝑙 𝑡 𝐷 𝑘 𝑥 𝐾(𝑥, 𝑡, 𝑦, 𝑠)||𝑢(𝑦, 𝑠)| 𝑑𝑦𝑑𝑠 ≤ 𝐶 𝑘𝑙 ‖𝑢‖ 𝐿 1 (𝐶(1)) for some constant 𝐶 𝑘𝑙 . 2. Now suppose the cylinder 𝐶(𝑟) ≔ 𝐶(0, 0; 𝑟) lies in 𝑈 𝑇 . Let 𝐶(𝑟/2) = 𝐶(0, 0; 𝑟/2) . We rescale by defining 𝑣(𝑥, 𝑡) ≔ 𝑢(𝑟𝑥, 𝑟 2 𝑡). link to page 76 link to page 714 link to page 57 link to page 77 60 2. Four Important Linear Partial Differential Equations Then 𝑣 𝑡 − Δ𝑣 = 0 in the cylinder 𝐶(1) . According to (39), |𝐷 𝑘 𝑥 𝐷 𝑙 𝑡 𝑣(𝑥, 𝑡)| ≤ 𝐶 𝑘𝑙 ‖𝑣‖ 𝐿 1 (𝐶(1)) ((𝑥, 𝑡) ∈ 𝐶( 1 2 )). But 𝐷 𝑘 𝑥 𝐷 𝑙 𝑡 𝑣(𝑥, 𝑡) = 𝑟 2𝑙+𝑘 𝐷 𝑘 𝑥 𝐷 𝑙 𝑡 𝑢(𝑟𝑥, 𝑟 2 𝑡) and ‖𝑣‖ 𝐿 1 (𝐶(1)) = 1 𝑟 𝑛+2 ‖𝑢‖ 𝐿 1 (𝐶(𝑟)) . There- fore max 𝐶(𝑟/2) |𝐷 𝑘 𝑥 𝐷 𝑙 𝑡 𝑢| ≤ 𝐶 𝑘𝑙 𝑟 2𝑙+𝑘+𝑛+2 ‖𝑢‖ 𝐿 1 (𝐶(𝑟)) . □ If 𝑢 solves the heat equation within 𝑈 𝑇 , then for each time 0 < 𝑡 ≤ 𝑇 , the mapping 𝑥 ↦ 𝑢(𝑥, 𝑡) is analytic. (See Mikhailov [ M ].) However the mapping 𝑡 ↦ 𝑢(𝑥, 𝑡) is not in general analytic. 2.3.4. Energy methods. a. Uniqueness. We investigate again the initial/boundary-value problem (40) { 𝑢 𝑡 − Δ𝑢 = 𝑓 in 𝑈 𝑇 𝑢 = 𝑔 on Γ 𝑇 . We earlier invoked the maximum principle to show uniqueness and now—by analogy with §2.2.5—provide an alternative argument based upon integration by parts. We assume as usual that 𝑈 ⊂ ℝ 𝑛 is open and bounded and that 𝜕𝑈 is 𝐶 1 . The terminal time 𝑇 > 0 is given. THEOREM 10 (Uniqueness) . There exists only one solution 𝑢 ∈ 𝐶 2 1 ( ̄ 𝑈 𝑇 ) of the initial/ boundary-value problem (40) . Proof. 1. If ̃ 𝑢 is another solution, 𝑤 ≔ 𝑢 − ̃ 𝑢 solves (41) { 𝑤 𝑡 − Δ𝑤 = 0 in 𝑈 𝑇 𝑤 = 0 on Γ 𝑇 . 2. Set 𝑒(𝑡) ≔ ∫ 𝑈 𝑤 2 (𝑥, 𝑡) 𝑑𝑥 (0 ≤ 𝑡 ≤ 𝑇). Then ̇ 𝑒(𝑡) = 2 ∫ 𝑈 𝑤𝑤 𝑡 𝑑𝑥 ( ̇ = 𝑑 𝑑𝑡 ) = 2 ∫ 𝑈 𝑤Δ𝑤 𝑑𝑥 = −2 ∫ 𝑈 |𝐷𝑤| 2 𝑑𝑥 ≤ 0, and so 𝑒(𝑡) ≤ 𝑒(0) = 0 (0 ≤ 𝑡 ≤ 𝑇). link to page 57 link to page 57 link to page 78 link to page 78 link to page 77 2.3. Heat Equation 61 Consequently 𝑤 = 𝑢 − ̃ 𝑢 ≡ 0 in 𝑈 𝑇 . □ Observe that the foregoing is a time-dependent variant of the proof of The- orem 16 in §2.2.5. b. Backwards uniqueness. A rather more subtle question asks about uniqueness backwards in time for the heat equation. For this, suppose 𝑢 and ̃ 𝑢 are both smooth solutions of the heat equation in 𝑈 𝑇 , with the same boundary conditions on 𝜕𝑈 : (42) { 𝑢 𝑡 − Δ𝑢 = 0 in 𝑈 𝑇 𝑢 = 𝑔 on 𝜕𝑈 × [0, 𝑇] , (43) { ̃ 𝑢 𝑡 − Δ ̃ 𝑢 = 0 in 𝑈 𝑇 ̃ 𝑢 = 𝑔 on 𝜕𝑈 × [0, 𝑇] , for some function 𝑔 . Note carefully that we are not supposing 𝑢 = ̃ 𝑢 at time 𝑡 = 0 . THEOREM 11 (Backwards uniqueness) . Suppose 𝑢 , ̃ 𝑢 ∈ 𝐶 2 ( ̄ 𝑈 𝑇 ) solve (42) , (43) . If 𝑢(𝑥, 𝑇) = ̃ 𝑢(𝑥, 𝑇) (𝑥 ∈ 𝑈), then 𝑢 ≡ ̃ 𝑢 within 𝑈 𝑇 . In other words, if two temperature distributions on 𝑈 agree at some time 𝑇 > 0 and have had the same boundary values for times 0 ≤ 𝑡 ≤ 𝑇 , then these temperatures must have been identically equal within 𝑈 at all earlier times. This is not at all obvious. Proof. 1. Write 𝑤 ≔ 𝑢 − ̃ 𝑢 and, as in the proof of Theorem 10, set 𝑒(𝑡) ≔ ∫ 𝑈 𝑤 2 (𝑥, 𝑡) 𝑑𝑥 (0 ≤ 𝑡 ≤ 𝑇). As before (44) ̇ 𝑒(𝑡) = −2 ∫ 𝑈 |𝐷𝑤| 2 𝑑𝑥 ( ̇ = 𝑑 𝑑𝑡 ) . link to page 77 link to page 78 link to page 79 link to page 79 link to page 79 62 2. Four Important Linear Partial Differential Equations Furthermore (45) ̈ 𝑒(𝑡) = −4 ∫ 𝑈 𝐷𝑤 ⋅ 𝐷𝑤 𝑡 𝑑𝑥 = 4 ∫ 𝑈 Δ𝑤𝑤 𝑡 𝑑𝑥 = 4 ∫ 𝑈 (Δ𝑤) 2 𝑑𝑥 by (41). Now since 𝑤 = 0 on 𝜕𝑈 , ∫ 𝑈 |𝐷𝑤| 2 𝑑𝑥 = − ∫ 𝑈 𝑤Δ𝑤 𝑑𝑥 ≤ (∫ 𝑈 𝑤 2 𝑑𝑥) 1/2 (∫ 𝑈 (Δ𝑤) 2 𝑑𝑥) 1/2 . Thus (44) and (45) imply ( ̇ 𝑒(𝑡)) 2 = 4 (∫ 𝑈 |𝐷𝑤| 2 𝑑𝑥) 2 ≤ (∫ 𝑈 𝑤 2 𝑑𝑥) (4 ∫ 𝑈 (Δ𝑤) 2 𝑑𝑥) = 𝑒(𝑡) ̈ 𝑒(𝑡). Hence (46) ̈ 𝑒(𝑡)𝑒(𝑡) ≥ ( ̇ 𝑒(𝑡)) 2 (0 ≤ 𝑡 ≤ 𝑇). 2. Now if 𝑒(𝑡) = 0 for all 0 ≤ 𝑡 ≤ 𝑇 , we are done. Otherwise there exists an interval [𝑡 1 , 𝑡 2 ] ⊂ [0, 𝑇] , with (47) 𝑒(𝑡) > 0 for 𝑡 1 ≤ 𝑡 < 𝑡 2 , 𝑒(𝑡 2 ) = 0. 3. Write (48) 𝑓(𝑡) ≔ log 𝑒(𝑡) (𝑡 1 ≤ 𝑡 < 𝑡 2 ). Then ̈ 𝑓(𝑡) = ̈ 𝑒(𝑡) 𝑒(𝑡) − ̇ 𝑒(𝑡) 2 𝑒(𝑡) 2 ≥ 0 by (46), and so 𝑓 is convex on the interval (𝑡 1 , 𝑡 2 ) . Consequently if 0 < 𝜏 < 1 , 𝑡 1 < 𝑡 < 𝑡 2 , we have 𝑓((1 − 𝜏)𝑡 1 + 𝜏𝑡) ≤ (1 − 𝜏)𝑓(𝑡 1 ) + 𝜏𝑓(𝑡). Recalling (48), we deduce 𝑒((1 − 𝜏)𝑡 1 + 𝜏𝑡) ≤ 𝑒(𝑡 1 ) 1−𝜏 𝑒(𝑡) 𝜏 , and so 0 ≤ 𝑒((1 − 𝜏)𝑡 1 + 𝜏𝑡 2 ) ≤ 𝑒(𝑡 1 ) 1−𝜏 𝑒(𝑡 2 ) 𝜏 (0 < 𝜏 < 1). link to page 79 link to page 80 2.4. Wave Equation 63 But in view of (47) this inequality implies 𝑒(𝑡) = 0 for all times 𝑡 1 ≤ 𝑡 ≤ 𝑡 2 , a contradiction. □ 2.4. WAVE EQUATION In this section we investigate the wave equation (1) 𝑢 𝑡𝑡 − Δ𝑢 = 0 and the nonhomogeneous wave equation (2) 𝑢 𝑡𝑡 − Δ𝑢 = 𝑓, subject to appropriate initial and boundary conditions. Here 𝑡 > 0 and 𝑥 ∈ 𝑈 , where 𝑈 ⊂ ℝ 𝑛 is open. The unknown is 𝑢 ∶ ̄ 𝑈 × [0, ∞) → ℝ , 𝑢 = 𝑢(𝑥, 𝑡) , and the Laplacian Δ is taken with respect to the spatial variables 𝑥 = (𝑥 1 , . . . , 𝑥 𝑛 ) . In (2) the function 𝑓 ∶ 𝑈 × [0, ∞) → ℝ is given. A common abbreviation is to write □𝑢 ≔ 𝑢 𝑡𝑡 − Δ𝑢. We shall discover that solutions of the wave equation behave quite differ- ently than solutions of Laplace’s equation or the heat equation. For example, these solutions are generally not 𝐶 ∞ , exhibit finite speed of propagation, etc. Physical interpretation. The wave equation is a simplified model for a vi- brating string (𝑛 = 1) , membrane (𝑛 = 2) , or elastic solid (𝑛 = 3) . In these physical interpretations 𝑢(𝑥, 𝑡) represents the displacement in some direction of the point 𝑥 at time 𝑡 ≥ 0 . Let 𝑉 represent any smooth subregion of 𝑈 . The acceleration within 𝑉 is then 𝑑 2 𝑑𝑡 2 ∫ 𝑉 𝑢 𝑑𝑥 = ∫ 𝑉 𝑢 𝑡𝑡 𝑑𝑥 and the net contact force is − ∫ 𝜕𝑉 F ⋅ 𝝂 𝑑𝑆, where F denotes the force acting on 𝑉 through 𝜕𝑉 and the mass density is taken to be unity. Newton’s law asserts that the mass times the acceleration equals the net force: ∫ 𝑉 𝑢 𝑡𝑡 𝑑𝑥 = − ∫ 𝜕𝑉 F ⋅ 𝝂 𝑑𝑆. This identity obtains for each subregion 𝑉 and so 𝑢 𝑡𝑡 = − div F . link to page 38 link to page 60 link to page 80 link to page 81 link to page 81 link to page 81 link to page 35 link to page 81 link to page 81 64 2. Four Important Linear Partial Differential Equations For elastic bodies, F is a function of the displacement gradient 𝐷𝑢 , whence 𝑢 𝑡𝑡 + div F (𝐷𝑢) = 0. For small 𝐷𝑢 , the linearization F (𝐷𝑢) ≈ −𝑎𝐷𝑢 is often appropriate; and so 𝑢 𝑡𝑡 − 𝑎Δ𝑢 = 0. This is the wave equation if 𝑎 = 1 . This physical interpretation strongly suggests it will be mathematically ap- propriate to specify two initial conditions, on the displacement 𝑢 and the velocity 𝑢 𝑡 , at time 𝑡 = 0 . 2.4.1. Solution by spherical means. We began §§2.2.1 and 2.3.1 by search- ing for certain scaling invariant solutions of Laplace’s equation and the heat equation. For the wave equation however we will instead present the (reason- ably) elegant method of solving (1) first for 𝑛 = 1 directly and then for 𝑛 ≥ 2 by the method of spherical means. a. Solution for n = 1, d’Alembert’s formula. We first focus our attention on the initial-value problem for the one-dimensional wave equation in all of ℝ : (3) { 𝑢 𝑡𝑡 − 𝑢 𝑥𝑥 = 0 in ℝ × (0, ∞) 𝑢 = 𝑔, 𝑢 𝑡 = ℎ on ℝ × {𝑡 = 0}, where 𝑔 , ℎ are given. We desire to derive a formula for 𝑢 in terms of 𝑔 and ℎ . Let us first note that the PDE in (3) can be “factored”, to read (4) ( 𝜕 𝜕𝑡 + 𝜕 𝜕𝑥 ) ( 𝜕 𝜕𝑡 − 𝜕 𝜕𝑥 ) 𝑢 = 𝑢 𝑡𝑡 − 𝑢 𝑥𝑥 = 0. Write (5) 𝑣(𝑥, 𝑡) ≔ ( 𝜕 𝜕𝑡 − 𝜕 𝜕𝑥 ) 𝑢(𝑥, 𝑡). Then (4) says 𝑣 𝑡 (𝑥, 𝑡) + 𝑣 𝑥 (𝑥, 𝑡) = 0 (𝑥 ∈ ℝ, 𝑡 > 0). This is a transport equation with constant coefficients. Applying formula (3) from §2.1.1 (with 𝑛 = 1 , 𝑏 = 1 ), we find (6) 𝑣(𝑥, 𝑡) = 𝑎(𝑥 − 𝑡) for 𝑎(𝑥) ≔ 𝑣(𝑥, 0) . Combining now (4)–(6), we obtain 𝑢 𝑡 (𝑥, 𝑡) − 𝑢 𝑥 (𝑥, 𝑡) = 𝑎(𝑥 − 𝑡) in ℝ × (0, ∞). link to page 81 link to page 36 link to page 81 link to page 81 link to page 81 link to page 82 link to page 82 link to page 81 link to page 82 link to page 82 2.4. Wave Equation 65 This is a nonhomogeneous transport equation; and so formula (5) from §2.1.2 (with 𝑛 = 1 , 𝑏 = −1 , 𝑓(𝑥, 𝑡) = 𝑎(𝑥 − 𝑡) ) implies for 𝑏(𝑥) ≔ 𝑢(𝑥, 0) that (7) 𝑢(𝑥, 𝑡) = ∫ 𝑡 0 𝑎(𝑥 + (𝑡 − 𝑠) − 𝑠) 𝑑𝑠 + 𝑏(𝑥 + 𝑡) = 1 2 ∫ 𝑥+𝑡 𝑥−𝑡 𝑎(𝑦) 𝑑𝑦 + 𝑏(𝑥 + 𝑡). We lastly invoke the initial conditions in (3) to compute 𝑎 and 𝑏 . The first initial condition in (3) gives 𝑏(𝑥) = 𝑔(𝑥) (𝑥 ∈ ℝ), whereas the second initial condition and (5) imply 𝑎(𝑥) = 𝑣(𝑥, 0) = 𝑢 𝑡 (𝑥, 0) − 𝑢 𝑥 (𝑥, 0) = ℎ(𝑥) − 𝑔 ′ (𝑥) (𝑥 ∈ ℝ). Our substituting into (7) now yields 𝑢(𝑥, 𝑡) = 1 2 ∫ 𝑥+𝑡 𝑥−𝑡 ℎ(𝑦) − 𝑔 ′ (𝑦) 𝑑𝑦 + 𝑔(𝑥 + 𝑡). Hence (8) 𝑢(𝑥, 𝑡) = 1 2 [𝑔(𝑥 + 𝑡) + 𝑔(𝑥 − 𝑡)] + 1 2 ∫ 𝑥+𝑡 𝑥−𝑡 ℎ(𝑦) 𝑑𝑦 (𝑥 ∈ ℝ, 𝑡 ≥ 0). This is d’Alembert’s formula . We have derived formula (8) assuming 𝑢 is a (sufficiently smooth) solution of (3). We need to check that this really is a solution. THEOREM 1 (Solution of wave equation, 𝑛 = 1 ) . Assume 𝑔 ∈ 𝐶 2 (ℝ) , ℎ ∈ 𝐶 1 (ℝ) , and define 𝑢 by d’Alembert’s formula (8) . Then (i) 𝑢 ∈ 𝐶 2 (ℝ × [0, ∞)) , (ii) 𝑢 𝑡𝑡 − 𝑢 𝑥𝑥 = 0 in ℝ × (0, ∞) , (iii) lim (𝑥,𝑡)→(𝑥 0 ,0) 𝑡>0 𝑢(𝑥, 𝑡) = 𝑔(𝑥 0 ) , lim (𝑥,𝑡)→(𝑥 0 ,0) 𝑡>0 𝑢 𝑡 (𝑥, 𝑡) = ℎ(𝑥 0 ) for each point 𝑥 0 ∈ ℝ . The proof is a straightforward calculation. Remarks. (i) In view of (8), our solution 𝑢 has the form 𝑢(𝑥, 𝑡) = 𝐹(𝑥 + 𝑡) + 𝐺(𝑥 − 𝑡) link to page 81 link to page 102 link to page 82 link to page 83 link to page 81 link to page 83 link to page 82 link to page 83 66 2. Four Important Linear Partial Differential Equations for appropriate functions 𝐹 and 𝐺 . Conversely any function of this form solves 𝑢 𝑡𝑡 − 𝑢 𝑥𝑥 = 0 . Hence the general solution of the one-dimensional wave equa- tion is a sum of the general solution of 𝑢 𝑡 − 𝑢 𝑥 = 0 and the general solution of 𝑢 𝑡 + 𝑢 𝑥 = 0 . This is a consequence of the factorization (4). See Problem 19. (ii) We see from (8) that if 𝑔 ∈ 𝐶 𝑘 and ℎ ∈ 𝐶 𝑘−1 , then 𝑢 ∈ 𝐶 𝑘 but is not in general smoother. Thus the wave equation does not cause instantaneous smoothing of the initial data, as does the heat equation. A reflection method. To illustrate a further application of d’Alembert’s for- mula, let us next consider this initial/boundary-value problem on the half-line ℝ + = {𝑥 > 0} : (9) ⎧ ⎨ ⎩ 𝑢 𝑡𝑡 − 𝑢 𝑥𝑥 = 0 in ℝ + × (0, ∞) 𝑢 = 𝑔, 𝑢 𝑡 = ℎ on ℝ + × {𝑡 = 0} 𝑢 = 0 on {𝑥 = 0} × (0, ∞) , where 𝑔 , ℎ are given, with 𝑔(0) = ℎ(0) = 0 . We convert (9) into the form (3) by extending 𝑢 , 𝑔 , ℎ to all of ℝ by odd reflection . That is, we set ̃ 𝑢(𝑥, 𝑡) ≔ { 𝑢(𝑥, 𝑡) (𝑥 ≥ 0, 𝑡 ≥ 0) −𝑢(−𝑥, 𝑡) (𝑥 ≤ 0, 𝑡 ≥ 0), ̃ 𝑔(𝑥) ≔ { 𝑔(𝑥) (𝑥 ≥ 0) −𝑔(−𝑥) (𝑥 ≤ 0), ̃ ℎ(𝑥) ≔ { ℎ(𝑥) (𝑥 ≥ 0) −ℎ(−𝑥) (𝑥 ≤ 0). Then (9) becomes { ̃ 𝑢 𝑡𝑡 = ̃ 𝑢 𝑥𝑥 in ℝ × (0, ∞) ̃ 𝑢 = ̃ 𝑔, ̃ 𝑢 𝑡 = ̃ ℎ on ℝ × {𝑡 = 0} . Hence d’Alembert’s formula (8) implies ̃ 𝑢(𝑥, 𝑡) = 1 2 [ ̃ 𝑔(𝑥 + 𝑡) + ̃ 𝑔(𝑥 − 𝑡)] + 1 2 ∫ 𝑥+𝑡 𝑥−𝑡 ̃ ℎ(𝑦) 𝑑𝑦. Recalling the definitions of ̃ 𝑢, ̃ 𝑔, ̃ ℎ above, we can transform this expression to read for 𝑥 ≥ 0 , 𝑡 ≥ 0 : (10) 𝑢(𝑥, 𝑡) = { 1 2 [𝑔(𝑥 + 𝑡) + 𝑔(𝑥 − 𝑡)] + 1 2 ∫ 𝑥+𝑡 𝑥−𝑡 ℎ(𝑦) 𝑑𝑦 if 𝑥 ≥ 𝑡 ≥ 0 1 2 [𝑔(𝑥 + 𝑡) − 𝑔(𝑡 − 𝑥)] + 1 2 ∫ 𝑥+𝑡 −𝑥+𝑡 ℎ(𝑦) 𝑑𝑦 if 0 ≤ 𝑥 ≤ 𝑡 . If ℎ ≡ 0 , we can understand formula (10) as saying that an initial displace- ment 𝑔 splits into two parts, one moving to the right with speed one and the link to page 83 link to page 84 link to page 84 2.4. Wave Equation 67 other to the left with speed one. The latter then reflects off the point 𝑥 = 0 , where the vibrating string is held fixed. Note that our solution does not belong to 𝐶 2 , unless 𝑔 ″ (0) = 0 . b. Spherical means. Now suppose 𝑛 ≥ 2 , 𝑚 ≥ 2 , and 𝑢 ∈ 𝐶 𝑚 (ℝ 𝑛 × [0, ∞)) solves the initial-value problem (11) { 𝑢 𝑡𝑡 − Δ𝑢 = 0 in ℝ 𝑛 × (0, ∞) 𝑢 = 𝑔, 𝑢 𝑡 = ℎ on ℝ 𝑛 × {𝑡 = 0} . We intend to derive an explicit formula for 𝑢 in terms of 𝑔 , ℎ . The plan will be to study first the average of 𝑢 over certain spheres. These averages, taken as functions of the time 𝑡 and the radius 𝑟 , turn out to solve the Euler–Poisson– Darboux equation, a PDE which we can for odd 𝑛 convert into the ordinary one- dimensional wave equation. Applying d’Alembert’s formula, or more precisely its variant (10), eventually leads us to a formula for the solution. NOTATION. (i) Let 𝑥 ∈ ℝ 𝑛 , 𝑡 > 0 , 𝑟 > 0 . Define (12) 𝑈(𝑥; 𝑟, 𝑡) ≔ ⨍ 𝜕𝐵(𝑥,𝑟) 𝑢(𝑦, 𝑡) 𝑑𝑆(𝑦), the average of 𝑢(⋅, 𝑡) over the sphere 𝜕𝐵(𝑥, 𝑟) . (ii) Similarly, (13) { 𝐺(𝑥; 𝑟) ≔ ⨍ 𝜕𝐵(𝑥,𝑟) 𝑔(𝑦) 𝑑𝑆(𝑦) 𝐻(𝑥; 𝑟) ≔ ⨍ 𝜕𝐵(𝑥,𝑟) ℎ(𝑦) 𝑑𝑆(𝑦). For fixed 𝑥 , we hereafter regard 𝑈 as a function of 𝑟 and 𝑡 and discover a partial differential equation that 𝑈 solves: LEMMA 1 (Euler–Poisson–Darboux equation) . Fix 𝑥 ∈ ℝ 𝑛 , and let 𝑢 satisfy (11) . Then 𝑈 ∈ 𝐶 𝑚 ( ̄ ℝ + × [0, ∞)) and (14) { 𝑈 𝑡𝑡 − 𝑈 𝑟𝑟 − 𝑛−1 𝑟 𝑈 𝑟 = 0 in ℝ + × (0, ∞) 𝑈 = 𝐺, 𝑈 𝑡 = 𝐻 on ℝ + × {𝑡 = 0}. The partial differential equation in (14) is the Euler–Poisson–Darboux equa- tion . (Note that the term 𝑈 𝑟𝑟 + 𝑛−1 𝑟 𝑈 𝑟 is the radial part of the Laplacian Δ in polar coordinates.) link to page 42 link to page 42 link to page 85 link to page 85 link to page 85 link to page 84 link to page 84 link to page 84 link to page 84 link to page 84 68 2. Four Important Linear Partial Differential Equations Proof. 1. As in the proof of Theorem 2 in §2.2.2 we compute for 𝑟 > 0 (15) 𝑈 𝑟 (𝑥; 𝑟, 𝑡) = 𝑟 𝑛 ⨍ 𝐵(𝑥,𝑟) Δ𝑢(𝑦, 𝑡) 𝑑𝑦. From this equality we deduce lim 𝑟→0 + 𝑈 𝑟 (𝑥; 𝑟, 𝑡) = 0 . We next differentiate (15), to discover after some computations that (16) 𝑈 𝑟𝑟 (𝑥; 𝑟, 𝑡) = ⨍ 𝜕𝐵(𝑥,𝑟) Δ𝑢 𝑑𝑆 + ( 1 𝑛 − 1) ⨍ 𝐵(𝑥,𝑟) Δ𝑢 𝑑𝑦. Thus lim 𝑟→0 + 𝑈 𝑟𝑟 (𝑥; 𝑟, 𝑡) = 1 𝑛 Δ𝑢(𝑥, 𝑡). Using formula (16), we can similarly compute 𝑈 𝑟𝑟𝑟 , etc., and so verify that 𝑈 ∈ 𝐶 𝑚 ( ̄ ℝ + × [0, ∞)) . 2. Continuing the calculation above, we see from (15) that 𝑈 𝑟 = 𝑟 𝑛 ⨍ 𝐵(𝑥,𝑟) 𝑢 𝑡𝑡 𝑑𝑦 by (11) = 1 𝑛𝛼(𝑛) 1 𝑟 𝑛−1 ∫ 𝐵(𝑥,𝑟) 𝑢 𝑡𝑡 𝑑𝑦. Thus 𝑟 𝑛−1 𝑈 𝑟 = 1 𝑛𝛼(𝑛) ∫ 𝐵(𝑥,𝑟) 𝑢 𝑡𝑡 𝑑𝑦, and so (𝑟 𝑛−1 𝑈 𝑟 ) 𝑟 = 1 𝑛𝛼(𝑛) ∫ 𝜕𝐵(𝑥,𝑟) 𝑢 𝑡𝑡 𝑑𝑆 = 𝑟 𝑛−1 ⨍ 𝜕𝐵(𝑥,𝑟) 𝑢 𝑡𝑡 𝑑𝑆 = 𝑟 𝑛−1 𝑈 𝑡𝑡 . □ c. Solution for n = 3, 2, Kirchhoff’s and Poisson’s formulas. The plan in the ensuing subsections will be to transform the Euler–Poisson–Darboux equa- tion (14) into the usual one-dimensional wave equation. As the full procedure is rather complicated, we pause here to handle the simpler cases 𝑛 = 3 , 2 , in that order. Solution for n = 3. Let us therefore hereafter take 𝑛 = 3 , and suppose 𝑢 ∈ 𝐶 2 (ℝ 3 × [0, ∞)) solves the initial-value problem (11). We recall the definitions (12), (13) of 𝑈 , 𝐺 , 𝐻 and then set (17) ̃ 𝑈 ≔ 𝑟𝑈, (18) ̃ 𝐺 ≔ 𝑟𝐺, ̃ 𝐻 ≔ 𝑟𝐻. link to page 84 link to page 83 link to page 86 link to page 84 link to page 85 link to page 85 link to page 86 link to page 84 link to page 86 link to page 84 2.4. Wave Equation 69 We now assert that ̃ 𝑈 solves (19) ⎧ ⎨ ⎩ ̃ 𝑈 𝑡𝑡 − ̃ 𝑈 𝑟𝑟 = 0 in ℝ + × (0, ∞) ̃ 𝑈 = ̃ 𝐺, ̃ 𝑈 𝑡 = ̃ 𝐻 on ℝ + × {𝑡 = 0} ̃ 𝑈 = 0 on {𝑟 = 0} × (0, ∞). Indeed ̃ 𝑈 𝑡𝑡 = 𝑟𝑈 𝑡𝑡 = 𝑟 [𝑈 𝑟𝑟 + 2 𝑟 𝑈 𝑟 ] by (14), with 𝑛 = 3 = 𝑟𝑈 𝑟𝑟 + 2𝑈 𝑟 = (𝑈 + 𝑟𝑈 𝑟 ) 𝑟 = ̃ 𝑈 𝑟𝑟 . Notice also that ̃ 𝐺 𝑟𝑟 (0) = 0 . Applying formula (10) to (19), we find for 0 ≤ 𝑟 ≤ 𝑡 (20) ̃ 𝑈(𝑥; 𝑟, 𝑡) = 1 2 [ ̃ 𝐺(𝑟 + 𝑡) − ̃ 𝐺(𝑡 − 𝑟)] + 1 2 ∫ 𝑟+𝑡 −𝑟+𝑡 ̃ 𝐻(𝑦) 𝑑𝑦. Since (12) implies 𝑢(𝑥, 𝑡) = lim 𝑟→0 + 𝑈(𝑥; 𝑟, 𝑡) , we conclude from (17), (18), (20) that 𝑢(𝑥, 𝑡) = lim 𝑟→0 + ̃ 𝑈(𝑥; 𝑟, 𝑡) 𝑟 = lim 𝑟→0 + [ ̃ 𝐺(𝑡 + 𝑟) − ̃ 𝐺(𝑡 − 𝑟) 2𝑟 + 1 2𝑟 ∫ 𝑡+𝑟 𝑡−𝑟 ̃ 𝐻(𝑦) 𝑑𝑦] = ̃ 𝐺 ′ (𝑡) + ̃ 𝐻(𝑡). Owing then to (13), we deduce (21) 𝑢(𝑥, 𝑡) = 𝜕 𝜕𝑡 (𝑡 ⨍ 𝜕𝐵(𝑥,𝑡) 𝑔 𝑑𝑆) + 𝑡 ⨍ 𝜕𝐵(𝑥,𝑡) ℎ 𝑑𝑆. But ⨍ 𝜕𝐵(𝑥,𝑡) 𝑔(𝑦) 𝑑𝑆(𝑦) = ⨍ 𝜕𝐵(0,1) 𝑔(𝑥 + 𝑡𝑧) 𝑑𝑆(𝑧); and so 𝜕 𝜕𝑡 (⨍ 𝜕𝐵(𝑥,𝑡) 𝑔 𝑑𝑆) = ⨍ 𝜕𝐵(0,1) 𝐷𝑔(𝑥 + 𝑡𝑧) ⋅ 𝑧 𝑑𝑆(𝑧) = ⨍ 𝜕𝐵(𝑥,𝑡) 𝐷𝑔(𝑦) ⋅ ( 𝑦 − 𝑥 𝑡 ) 𝑑𝑆(𝑦). Returning to (21), we therefore conclude (22) 𝑢(𝑥, 𝑡) = ⨍ 𝜕𝐵(𝑥,𝑡) 𝑡ℎ(𝑦) + 𝑔(𝑦) + 𝐷𝑔(𝑦) ⋅ (𝑦 − 𝑥) 𝑑𝑆(𝑦) (𝑥 ∈ ℝ 3 , 𝑡 > 0). This is Kirchhoff’s formula for the solution of the initial-value problem (11) in three dimensions. link to page 85 link to page 84 link to page 84 link to page 84 link to page 87 link to page 86 link to page 87 70 2. Four Important Linear Partial Differential Equations Solution for n = 2. No transformation like (17) works to convert the Euler– Poisson–Darboux equation into the one-dimensional wave equation when 𝑛 = 2 . Instead we will take the initial-value problem (11) for 𝑛 = 2 and simply regard it as a problem for 𝑛 = 3 , in which the third spatial variable 𝑥 3 does not appear. Indeed, assuming 𝑢 ∈ 𝐶 2 (ℝ 2 × [0, ∞)) solves (11) for 𝑛 = 2 , let us write (23) ̄ 𝑢(𝑥 1 , 𝑥 2 , 𝑥 3 , 𝑡) ≔ 𝑢(𝑥 1 , 𝑥 2 , 𝑡). Then (11) implies (24) { ̄ 𝑢 𝑡𝑡 − Δ ̄ 𝑢 = 0 in ℝ 3 × (0, ∞) ̄ 𝑢 = ̄ 𝑔, ̄ 𝑢 𝑡 = ̄ ℎ on ℝ 3 × {𝑡 = 0} , for ̄ 𝑔(𝑥 1 , 𝑥 2 , 𝑥 3 ) ≔ 𝑔(𝑥 1 , 𝑥 2 ), ̄ ℎ(𝑥 1 , 𝑥 2 , 𝑥 3 ) ≔ ℎ(𝑥 1 , 𝑥 2 ). If we write 𝑥 = (𝑥 1 , 𝑥 2 ) ∈ ℝ 2 and ̄ 𝑥 = (𝑥 1 , 𝑥 2 , 0) ∈ ℝ 3 , then (24) and Kirchhoff’s formula (in the form (21)) imply (25) 𝑢(𝑥, 𝑡) = ̄ 𝑢( ̄ 𝑥, 𝑡) = 𝜕 𝜕𝑡 (𝑡 ⨍ 𝜕 ̄ 𝐵( ̄ 𝑥,𝑡) ̄ 𝑔 𝑑 ̄ 𝑆) + 𝑡 ⨍ 𝜕 ̄ 𝐵( ̄ 𝑥,𝑡) ̄ ℎ 𝑑 ̄ 𝑆, where ̄ 𝐵( ̄ 𝑥, 𝑡) denotes the ball in ℝ 3 with center ̄ 𝑥 , radius 𝑡 > 0 and where 𝑑 ̄ 𝑆 denotes two-dimensional surface measure on 𝜕 ̄ 𝐵( ̄ 𝑥, 𝑡) . We simplify (25) by observing ⨍ 𝜕 ̄ 𝐵( ̄ 𝑥,𝑡) ̄ 𝑔 𝑑 ̄ 𝑆 = 1 4𝜋𝑡 2 ∫ 𝜕 ̄ 𝐵( ̄ 𝑥,𝑡) ̄ 𝑔 𝑑 ̄ 𝑆 = 2 4𝜋𝑡 2 ∫ 𝐵(𝑥,𝑡) 𝑔(𝑦)(1 + |𝐷𝛾(𝑦)| 2 ) 1/2 𝑑𝑦, where 𝛾(𝑦) = (𝑡 2 −|𝑦−𝑥| 2 ) 1/2 for 𝑦 ∈ 𝐵(𝑥, 𝑡) . The factor “ 2 ” enters since 𝜕 ̄ 𝐵( ̄ 𝑥, 𝑡) consists of two hemispheres. Observe that (1 + |𝐷𝛾| 2 ) 1/2 = 𝑡(𝑡 2 − |𝑦 − 𝑥| 2 ) −1/2 . Therefore ⨍ 𝜕 ̄ 𝐵( ̄ 𝑥,𝑡) ̄ 𝑔 𝑑 ̄ 𝑆 = 1 2𝜋𝑡 ∫ 𝐵(𝑥,𝑡) 𝑔(𝑦) (𝑡 2 − |𝑦 − 𝑥| 2 ) 1/2 𝑑𝑦 = 𝑡 2 ⨍ 𝐵(𝑥,𝑡) 𝑔(𝑦) (𝑡 2 − |𝑦 − 𝑥| 2 ) 1/2 𝑑𝑦. link to page 87 link to page 88 link to page 84 2.4. Wave Equation 71 Consequently formula (25) becomes (26) 𝑢(𝑥, 𝑡) = 1 2 𝜕 𝜕𝑡 (𝑡 2 ⨍ 𝐵(𝑥,𝑡) 𝑔(𝑦) (𝑡 2 − |𝑦 − 𝑥| 2 ) 1/2 𝑑𝑦) + 𝑡 2 2 ⨍ 𝐵(𝑥,𝑡) ℎ(𝑦) (𝑡 2 − |𝑦 − 𝑥| 2 ) 1/2 𝑑𝑦. But 𝑡 2 ⨍ 𝐵(𝑥,𝑡) 𝑔(𝑦) (𝑡 2 − |𝑦 − 𝑥| 2 ) 1/2 𝑑𝑦 = 𝑡 ⨍ 𝐵(0,1) 𝑔(𝑥 + 𝑡𝑧) (1 − |𝑧| 2 ) 1/2 𝑑𝑧, and so 𝜕 𝜕𝑡 (𝑡 2 ⨍ 𝐵(𝑥,𝑡) 𝑔(𝑦) (𝑡 2 − |𝑦 − 𝑥| 2 ) 1/2 𝑑𝑦) = ⨍ 𝐵(0,1) 𝑔(𝑥 + 𝑡𝑧) (1 − |𝑧| 2 ) 1/2 𝑑𝑧 + 𝑡 ⨍ 𝐵(0,1) 𝐷𝑔(𝑥 + 𝑡𝑧) ⋅ 𝑧 (1 − |𝑧| 2 ) 1/2 𝑑𝑧 = 𝑡 ⨍ 𝐵(𝑥,𝑡) 𝑔(𝑦) (𝑡 2 − |𝑦 − 𝑥| 2 ) 1/2 𝑑𝑦 + 𝑡 ⨍ 𝐵(𝑥,𝑡) 𝐷𝑔(𝑦) ⋅ (𝑦 − 𝑥) (𝑡 2 − |𝑦 − 𝑥| 2 ) 1/2 𝑑𝑦. Hence we can rewrite (26) and obtain the relation (27) 𝑢(𝑥, 𝑡) = 1 2 ⨍ 𝐵(𝑥,𝑡) 𝑡𝑔(𝑦) + 𝑡 2 ℎ(𝑦) + 𝑡𝐷𝑔(𝑦) ⋅ (𝑦 − 𝑥) (𝑡 2 − |𝑦 − 𝑥| 2 ) 1/2 𝑑𝑦 for 𝑥 ∈ ℝ 2 , 𝑡 > 0 . This is Poisson’s formula for the solution of the initial-value problem (11) in two dimensions. The trick of solving the problem for 𝑛 = 3 first and then dropping to 𝑛 = 2 is the method of descent . d. Solution for odd n. In this subsection we solve the Euler–Poisson– Dar- boux PDE for odd 𝑛 ≥ 3 . We first record some technical facts. LEMMA 2 (Some useful identities) . Let 𝜙 ∶ ℝ → ℝ be 𝐶 𝑘+1 . Then for 𝑘 = 1 , 2 , . . . (i) ( 𝑑 2 𝑑𝑟 2 ) ( 1 𝑟 𝑑 𝑑𝑟 ) 𝑘−1 (𝑟 2𝑘−1 𝜙(𝑟)) = ( 1 𝑟 𝑑 𝑑𝑟 ) 𝑘 (𝑟 2𝑘 𝑑𝜙 𝑑𝑟 (𝑟)) , (ii) ( 1 𝑟 𝑑 𝑑𝑟 ) 𝑘−1 (𝑟 2𝑘−1 𝜙(𝑟)) = ∑ 𝑘−1 𝑗=0 𝛽 𝑘 𝑗 𝑟 𝑗+1 𝑑 𝑗 𝜙 𝑑𝑟 𝑗 (𝑟) , where the constants 𝛽 𝑘 𝑗 ( 𝑗 = 0 , . . . , 𝑘 − 1 ) are independent of 𝜙 . Furthermore, (iii) 𝛽 𝑘 0 = 1 ⋅ 3 ⋅ 5 ⋯ (2𝑘 − 1) . The proof by induction is left as an exercise. Now assume 𝑛 ≥ 3 is an odd integer link to page 84 link to page 84 link to page 84 link to page 88 link to page 89 link to page 88 link to page 84 link to page 88 link to page 89 link to page 89 link to page 83 72 2. Four Important Linear Partial Differential Equations and set 𝑛 = 2𝑘 + 1 (𝑘 ≥ 1). Henceforth suppose 𝑢 ∈ 𝐶 𝑘+1 (ℝ 𝑛 × [0, ∞)) solves the initial-value problem (11). Then the function 𝑈 defined by (12) is 𝐶 𝑘+1 . NOTATION. We write (28) ⎧ ⎪ ⎨ ⎪ ⎩ ̃ 𝑈(𝑟, 𝑡) ≔ ( 1 𝑟 𝜕 𝜕𝑟 ) 𝑘−1 (𝑟 2𝑘−1 𝑈(𝑥; 𝑟, 𝑡)) ̃ 𝐺(𝑟) ≔ ( 1 𝑟 𝜕 𝜕𝑟 ) 𝑘−1 (𝑟 2𝑘−1 𝐺(𝑥; 𝑟)) ̃ 𝐻(𝑟) ≔ ( 1 𝑟 𝜕 𝜕𝑟 ) 𝑘−1 (𝑟 2𝑘−1 𝐻(𝑥; 𝑟)) (𝑟 > 0, 𝑡 ≥ 0). Then (29) ̃ 𝑈(𝑟, 0) = ̃ 𝐺(𝑟), ̃ 𝑈 𝑡 (𝑟, 0) = ̃ 𝐻(𝑟). Next we combine Lemma 1 and the identities provided by Lemma 2 to demonstrate that the transformation (28) of 𝑈 into ̃ 𝑈 in effect converts the Euler–Poisson–Darboux equation into the wave equation. LEMMA 3 ( ̃ 𝑈 solves the one-dimensional wave equation) . We have ⎧ ⎨ ⎩ ̃ 𝑈 𝑡𝑡 − ̃ 𝑈 𝑟𝑟 = 0 in ℝ + × (0, ∞) ̃ 𝑈 = ̃ 𝐺, ̃ 𝑈 𝑡 = ̃ 𝐻 on ℝ + × {𝑡 = 0} ̃ 𝑈 = 0 on {𝑟 = 0} × (0, ∞). Proof. If 𝑟 > 0 , ̃ 𝑈 𝑟𝑟 = ( 𝜕 2 𝜕𝑟 2 ) ( 1 𝑟 𝜕 𝜕𝑟 ) 𝑘−1 (𝑟 2𝑘−1 𝑈) = ( 1 𝑟 𝜕 𝜕𝑟 ) 𝑘 (𝑟 2𝑘 𝑈 𝑟 ) by Lemma 2(i) = ( 1 𝑟 𝜕 𝜕𝑟 ) 𝑘−1 [𝑟 2𝑘−1 𝑈 𝑟𝑟 + 2𝑘𝑟 2𝑘−2 𝑈 𝑟 ] = ( 1 𝑟 𝜕 𝜕𝑟 ) 𝑘−1 [𝑟 2𝑘−1 (𝑈 𝑟𝑟 + 𝑛 − 1 𝑟 𝑈 𝑟 )] (𝑛 = 2𝑘 + 1) = ( 1 𝑟 𝜕 𝜕𝑟 ) 𝑘−1 (𝑟 2𝑘−1 𝑈 𝑡𝑡 ) = ̃ 𝑈 𝑡𝑡 , the next-to-last equality holding according to (14). Using Lemma 2(ii) we con- clude as well that ̃ 𝑈 = 0 on {𝑟 = 0} . □ In view of Lemma 3, (29), and formula (10), we conclude for 0 ≤ 𝑟 ≤ 𝑡 that (30) ̃ 𝑈(𝑟, 𝑡) = 1 2 [ ̃ 𝐺(𝑟 + 𝑡) − ̃ 𝐺(𝑡 − 𝑟)] + 1 2 ∫ 𝑡+𝑟 𝑡−𝑟 ̃ 𝐻(𝑦) 𝑑𝑦. link to page 88 link to page 89 link to page 89 link to page 88 link to page 90 link to page 86 link to page 86 link to page 90 link to page 84 link to page 90 2.4. Wave Equation 73 But recall 𝑢(𝑥, 𝑡) = lim 𝑟→0 𝑈(𝑥; 𝑟, 𝑡) . Furthermore Lemma 2(ii) asserts ̃ 𝑈(𝑟, 𝑡) = ( 1 𝑟 𝜕 𝜕𝑟 ) 𝑘−1 (𝑟 2𝑘−1 𝑈(𝑥; 𝑟, 𝑡)) = 𝑘−1 ∑ 𝑗=0 𝛽 𝑘 𝑗 𝑟 𝑗+1 𝜕 𝑗 𝜕𝑟 𝑗 𝑈(𝑥; 𝑟, 𝑡), and so lim 𝑟→0 ̃ 𝑈(𝑟, 𝑡) 𝛽 𝑘 0 𝑟 = lim 𝑟→0 𝑈(𝑥; 𝑟, 𝑡) = 𝑢(𝑥, 𝑡). Thus (30) implies 𝑢(𝑥, 𝑡) = 1 𝛽 𝑘 0 lim 𝑟→0 [ ̃ 𝐺(𝑡 + 𝑟) − ̃ 𝐺(𝑡 − 𝑟) 2𝑟 + 1 2𝑟 ∫ 𝑡+𝑟 𝑡−𝑟 ̃ 𝐻(𝑦) 𝑑𝑦] = 1 𝛽 𝑘 0 [ ̃ 𝐺 ′ (𝑡) + ̃ 𝐻(𝑡)]. Finally then, since 𝑛 = 2𝑘 + 1 , (30) and Lemma 2(iii) yield this representa- tion formula: (31) ⎧ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎩ 𝑢(𝑥, 𝑡) = 1 𝛾 𝑛 [( 𝜕 𝜕𝑡 ) ( 1 𝑡 𝜕 𝜕𝑡 ) 𝑛−3 2 (𝑡 𝑛−2 ⨍ 𝜕𝐵(𝑥,𝑡) 𝑔 𝑑𝑆) + ( 1 𝑡 𝜕 𝜕𝑡 ) 𝑛−3 2 (𝑡 𝑛−2 ⨍ 𝜕𝐵(𝑥,𝑡) ℎ 𝑑𝑆)] where 𝑛 is odd and 𝛾 𝑛 = 1 ⋅ 3 ⋅ 5 ⋯ (𝑛 − 2) , for 𝑥 ∈ ℝ 𝑛 , 𝑡 > 0 . We note that 𝛾 3 = 1 , and so (31) agrees for 𝑛 = 3 with (21) and thus with Kirchhoff’s formula (22). It remains to check that formula (31) really provides a solution of (11). THEOREM 2 (Solution of wave equation in odd dimensions) . Assume 𝑛 is an odd integer, 𝑛 ≥ 3 , and suppose also 𝑔 ∈ 𝐶 𝑚+1 (ℝ 𝑛 ) , ℎ ∈ 𝐶 𝑚 (ℝ 𝑛 ) , for 𝑚 = 𝑛+1 2 . Define 𝑢 by (31) . Then (i) 𝑢 ∈ 𝐶 2 (ℝ 𝑛 × [0, ∞)) , (ii) 𝑢 𝑡𝑡 − Δ𝑢 = 0 in ℝ 𝑛 × (0, ∞) , and (iii) lim (𝑥,𝑡)→(𝑥 0 ,0) 𝑥∈ℝ 𝑛 , 𝑡>0 𝑢(𝑥, 𝑡) = 𝑔(𝑥 0 ) , lim (𝑥,𝑡)→(𝑥 0 ,0) 𝑥∈ℝ 𝑛 , 𝑡>0 𝑢 𝑡 (𝑥, 𝑡) = ℎ(𝑥 0 ) for each point 𝑥 0 ∈ ℝ 𝑛 . link to page 88 link to page 42 link to page 42 link to page 91 link to page 88 link to page 90 link to page 82 link to page 84 link to page 96 74 2. Four Important Linear Partial Differential Equations Proof. 1. Suppose first 𝑔 ≡ 0 , so that (32) 𝑢(𝑥, 𝑡) = 1 𝛾 𝑛 ( 1 𝑡 𝜕 𝜕𝑡 ) 𝑛−3 2 (𝑡 𝑛−2 𝐻(𝑥; 𝑡)) . Then Lemma 2(i) lets us compute 𝑢 𝑡𝑡 = 1 𝛾 𝑛 ( 1 𝑡 𝜕 𝜕𝑡 ) 𝑛−1 2 (𝑡 𝑛−1 𝐻 𝑡 ) . From the calculation in the proof of Theorem 2 in §2.2.2, we see as well that 𝐻 𝑡 = 𝑡 𝑛 ⨍ 𝐵(𝑥,𝑡) Δℎ 𝑑𝑦. Consequently 𝑢 𝑡𝑡 = 1 𝑛𝛼(𝑛)𝛾 𝑛 ( 1 𝑡 𝜕 𝜕𝑡 ) 𝑛−1 2 (∫ 𝐵(𝑥,𝑡) Δℎ 𝑑𝑦) = 1 𝑛𝛼(𝑛)𝛾 𝑛 ( 1 𝑡 𝜕 𝜕𝑡 ) 𝑛−3 2 ( 1 𝑡 ∫ 𝜕𝐵(𝑥,𝑡) Δℎ 𝑑𝑆) . On the other hand, Δ𝐻(𝑥; 𝑡) = Δ 𝑥 ⨍ 𝜕𝐵(0,𝑡) ℎ(𝑥 + 𝑦) 𝑑𝑆(𝑦) = ⨍ 𝜕𝐵(𝑥,𝑡) Δℎ 𝑑𝑆. Consequently (32) and the calculations above imply 𝑢 𝑡𝑡 = Δ𝑢 in ℝ 𝑛 × (0, ∞) . A similar computation works if ℎ ≡ 0 . 2. We leave it as an exercise to confirm, using Lemma 2(ii)–(iii), that 𝑢 takes on the correct initial conditions. □ Remarks. (i) Notice that to compute 𝑢(𝑥, 𝑡) we need only have information on 𝑔 , ℎ and their derivatives on the sphere 𝜕𝐵(𝑥, 𝑡) , and not on the entire ball 𝐵(𝑥, 𝑡) . (ii) Comparing formula (31) with d’Alembert’s formula (8) (𝑛 = 1) , we observe that the latter does not involve the derivatives of 𝑔 . This suggests that for 𝑛 > 1 , a solution of the wave equation (11) need not for times 𝑡 > 0 be as smooth as its initial value 𝑔 : irregularities in 𝑔 may focus at times 𝑡 > 0 , thereby causing 𝑢 to be less regular. (We will see later in §2.4.3 that the “energy norm” of 𝑢 does not deteriorate for 𝑡 > 0 .) (iii) Once again (as in the case 𝑛 = 1 ) we see the phenomenon of finite propagation speed of the initial disturbance. link to page 90 link to page 209 link to page 84 link to page 90 link to page 90 link to page 92 link to page 92 link to page 90 link to page 92 2.4. Wave Equation 75 (iv) A completely different derivation of formula (31) (using the heat equa- tion!) is in §4.3.3. e. Solution for even n. Assume now 𝑛 is an even integer. Suppose 𝑢 is a 𝐶 𝑚 solution of (11), 𝑚 = 𝑛+2 2 . We want to fashion a representa- tion formula like (31) for 𝑢 . The trick, as above for 𝑛 = 2 , is to note that (33) ̄ 𝑢(𝑥 1 , . . . , 𝑥 𝑛+1 , 𝑡) ≔ 𝑢(𝑥 1 , . . . , 𝑥 𝑛 , 𝑡) solves the wave equation in ℝ 𝑛+1 × (0, ∞) , with the initial conditions ̄ 𝑢 = ̄ 𝑔, ̄ 𝑢 𝑡 = ̄ ℎ on ℝ 𝑛+1 × {𝑡 = 0} , where (34) { ̄ 𝑔(𝑥 1 , . . . , 𝑥 𝑛+1 ) ≔ 𝑔(𝑥 1 , . . . , 𝑥 𝑛 ) ̄ ℎ(𝑥 1 , . . . , 𝑥 𝑛+1 ) ≔ ℎ(𝑥 1 , . . . , 𝑥 𝑛 ). As 𝑛 + 1 is odd, we may employ (31) (with 𝑛 + 1 replacing 𝑛 ) to secure a repre- sentation formula for ̄ 𝑢 in terms of ̄ 𝑔, ̄ ℎ . But then (33) and (34) yield at once a formula for 𝑢 in terms of 𝑔 , ℎ . This is again the method of descent. To carry out the details, let us fix 𝑥 ∈ ℝ 𝑛 , 𝑡 > 0 , and write ̄ 𝑥 = (𝑥 1 , . . . , 𝑥 𝑛 , 0) ∈ ℝ 𝑛+1 . Then (31), with 𝑛 + 1 replacing 𝑛 , gives (35) 𝑢(𝑥, 𝑡) = 1 𝛾 𝑛+1 [ 𝜕 𝜕𝑡 ( 1 𝑡 𝜕 𝜕𝑡 ) 𝑛−2 2 (𝑡 𝑛−1 ⨍ 𝜕 ̄ 𝐵( ̄ 𝑥,𝑡) ̄ 𝑔 𝑑 ̄ 𝑆) + ( 1 𝑡 𝜕 𝜕𝑡 ) 𝑛−2 2 (𝑡 𝑛−1 ⨍ 𝜕 ̄ 𝐵( ̄ 𝑥,𝑡) ̄ ℎ 𝑑 ̄ 𝑆)] , ̄ 𝐵( ̄ 𝑥, 𝑡) denoting the ball in ℝ 𝑛+1 with center ̄ 𝑥 and radius 𝑡 and 𝑑 ̄ 𝑆 denoting 𝑛 -dimensional surface measure on 𝜕 ̄ 𝐵( ̄ 𝑥, 𝑡) . Now (36) ⨍ 𝜕 ̄ 𝐵( ̄ 𝑥,𝑡) ̄ 𝑔 𝑑 ̄ 𝑆 = 1 (𝑛 + 1)𝛼(𝑛 + 1)𝑡 𝑛 ∫ 𝜕 ̄ 𝐵( ̄ 𝑥,𝑡) ̄ 𝑔 𝑑 ̄ 𝑆. Note that 𝜕 ̄ 𝐵(𝑥, 𝑡)∩{𝑦 𝑛+1 ≥ 0} is the graph of the function 𝛾(𝑦) ≔ (𝑡 2 −|𝑦−𝑥| 2 ) 1/2 for 𝑦 ∈ 𝐵(𝑥, 𝑡) ⊂ ℝ 𝑛 . Likewise 𝜕 ̄ 𝐵(𝑥, 𝑡) ∩ {𝑦 𝑛+1 ≤ 0} is the graph of −𝛾 . Thus (36) implies (37) ⨍ 𝜕 ̄ 𝐵( ̄ 𝑥,𝑡) ̄ 𝑔 𝑑 ̄ 𝑆 = 2 (𝑛 + 1)𝛼(𝑛 + 1)𝑡 𝑛 ∫ 𝐵(𝑥,𝑡) 𝑔(𝑦)(1 + |𝐷𝛾(𝑦)| 2 ) 1/2 𝑑𝑦, link to page 92 link to page 92 link to page 88 link to page 93 76 2. Four Important Linear Partial Differential Equations the factor “2” entering because 𝜕 ̄ 𝐵( ̄ 𝑥, 𝑡) comprises two hemispheres. Note that (1 + |𝐷𝛾(𝑦)| 2 ) 1/2 = 𝑡(𝑡 2 − |𝑦 − 𝑥| 2 ) −1/2 . Our substituting this into (37) yields ⨍ 𝜕 ̄ 𝐵( ̄ 𝑥,𝑡) ̄ 𝑔 𝑑 ̄ 𝑆 = 2 (𝑛 + 1)𝛼(𝑛 + 1)𝑡 𝑛−1 ∫ 𝐵(𝑥,𝑡) 𝑔(𝑦) (𝑡 2 − |𝑦 − 𝑥| 2 ) 1/2 𝑑𝑦 = 2𝑡𝛼(𝑛) (𝑛 + 1)𝛼(𝑛 + 1) ⨍ 𝐵(𝑥,𝑡) 𝑔(𝑦) (𝑡 2 − |𝑦 − 𝑥| 2 ) 1/2 𝑑𝑦. We insert this formula and the similar one with ℎ in place of 𝑔 into (35) and find 𝑢(𝑥, 𝑡) = 1 𝛾 𝑛+1 2𝛼(𝑛) (𝑛 + 1)𝛼(𝑛 + 1) [ 𝜕 𝜕𝑡 ( 1 𝑡 𝜕 𝜕𝑡 ) 𝑛−2 2 (𝑡 𝑛 ⨍ 𝐵(𝑥,𝑡) 𝑔(𝑦) (𝑡 2 − |𝑦 − 𝑥| 2 ) 1/2 𝑑𝑦) + ( 1 𝑡 𝜕 𝜕𝑡 ) 𝑛−2 2 (𝑡 𝑛 ⨍ 𝐵(𝑥,𝑡) ℎ(𝑦) (𝑡 2 − |𝑦 − 𝑥| 2 ) 1/2 𝑑𝑦)] . Since 𝛾 𝑛+1 = 1 ⋅ 3 ⋅ 5 ⋯ (𝑛 − 1) and 𝛼(𝑛) = 𝜋 𝑛/2 Γ( 𝑛+2 2 ) , we may compute 𝛾 𝑛 = 2 ⋅ 4 ⋯ (𝑛 − 2) ⋅ 𝑛 . Hence the resulting representation formula for even 𝑛 is (38) ⎧ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎩ 𝑢(𝑥, 𝑡) = 1 𝛾 𝑛 [( 𝜕 𝜕𝑡 ) ( 1 𝑡 𝜕 𝜕𝑡 ) 𝑛−2 2 (𝑡 𝑛 ⨍ 𝐵(𝑥,𝑡) 𝑔(𝑦) (𝑡 2 − |𝑦 − 𝑥| 2 ) 1/2 𝑑𝑦) + ( 1 𝑡 𝜕 𝜕𝑡 ) 𝑛−2 2 (𝑡 𝑛 ⨍ 𝐵(𝑥,𝑡) ℎ(𝑦) (𝑡 2 − |𝑦 − 𝑥| 2 ) 1/2 𝑑𝑦)] , where 𝑛 is even and 𝛾 𝑛 = 2 ⋅ 4 ⋯ (𝑛 − 2) ⋅ 𝑛 , for 𝑥 ∈ ℝ 𝑛 , 𝑡 > 0 . Since 𝛾 2 = 2 , this agrees with Poisson’s formula (27) if 𝑛 = 2 . THEOREM 3 (Solution of wave equation in even dimensions) . Assume 𝑛 is an even integer, 𝑛 ≥ 2 , and suppose also 𝑔 ∈ 𝐶 𝑚+1 (ℝ 𝑛 ) , ℎ ∈ 𝐶 𝑚 (ℝ 𝑛 ) , for 𝑚 = 𝑛+2 2 . Define 𝑢 by (38) . Then (i) 𝑢 ∈ 𝐶 2 (ℝ 𝑛 × [0, ∞)) , (ii) 𝑢 𝑡𝑡 − Δ𝑢 = 0 in ℝ 𝑛 × (0, ∞) , and (iii) lim (𝑥,𝑡)→(𝑥 0 ,0) 𝑥∈ℝ 𝑛 , 𝑡>0 𝑢(𝑥, 𝑡) = 𝑔(𝑥 0 ) , lim (𝑥,𝑡)→(𝑥 0 ,0) 𝑥∈ℝ 𝑛 , 𝑡>0 𝑢 𝑡 (𝑥, 𝑡) = ℎ(𝑥 0 ) for each point 𝑥 0 ∈ ℝ 𝑛 . link to page 90 link to page 90 link to page 90 link to page 93 link to page 60 link to page 94 link to page 90 link to page 93 2.4. Wave Equation 77 This follows from Theorem 2. Observe, in contrast to formula (31), that to compute 𝑢(𝑥, 𝑡) for even 𝑛 we need information on 𝑢 = 𝑔 , 𝑢 𝑡 = ℎ on all of 𝐵(𝑥, 𝑡) and not just on 𝜕𝐵(𝑥, 𝑡) . Huygens’ principle. Comparing (31) and (38), we observe that if 𝑛 is odd and 𝑛 ≥ 3 , the data 𝑔 and ℎ at a given point 𝑥 ∈ ℝ 𝑛 affect the solution 𝑢 only on the boundary { (𝑦, 𝑡) ∣ 𝑡 > 0 , |𝑥 − 𝑦| = 𝑡 } of the cone 𝐶 = { (𝑦, 𝑡) ∣ 𝑡 > 0 , |𝑥 − 𝑦| < 𝑡 } . On the other hand, if 𝑛 is even, the data 𝑔 and ℎ affect 𝑢 within all of 𝐶 . In other words, a “disturbance” originating at 𝑥 propagates along a sharp wavefront in odd dimensions, but in even dimensions it continues to have effects even after the leading edge of the wavefront passes . This is Huygens’ principle . 2.4.2. Nonhomogeneous problem. We next investigate the initial-value problem for the nonhomogeneous wave equation (39) { 𝑢 𝑡𝑡 − Δ𝑢 = 𝑓 in ℝ 𝑛 × (0, ∞) 𝑢 = 0, 𝑢 𝑡 = 0 on ℝ 𝑛 × {𝑡 = 0} . Motivated by Duhamel’s principle (introduced earlier in §2.3.1), we define 𝑢 = 𝑢(𝑥, 𝑡; 𝑠) to be the solution of ( 40 𝑠 ) { 𝑢 𝑡𝑡 (⋅; 𝑠) − Δ𝑢(⋅; 𝑠) = 0 in ℝ 𝑛 × (𝑠, ∞) 𝑢(⋅; 𝑠) = 0, 𝑢 𝑡 (⋅; 𝑠) = 𝑓(⋅, 𝑠) on ℝ 𝑛 × {𝑡 = 𝑠}. Now set (41) 𝑢(𝑥, 𝑡) ≔ ∫ 𝑡 0 𝑢(𝑥, 𝑡; 𝑠) 𝑑𝑠 (𝑥 ∈ ℝ 𝑛 , 𝑡 ≥ 0). Duhamel’s principle asserts this is a solution of (42) { 𝑢 𝑡𝑡 − Δ𝑢 = 𝑓 in ℝ 𝑛 × (0, ∞) 𝑢 = 0, 𝑢 𝑡 = 0 on ℝ 𝑛 × {𝑡 = 0} . THEOREM 4 (Solution of nonhomogeneous wave equation) . Assume that 𝑛 ≥ 2 and 𝑓 ∈ 𝐶 [𝑛/2]+1 (ℝ 𝑛 × [0, ∞)) . Define 𝑢 by (41) . Then (i) 𝑢 ∈ 𝐶 2 (ℝ 𝑛 × [0, ∞)) , (ii) 𝑢 𝑡𝑡 − Δ𝑢 = 𝑓 in ℝ 𝑛 × (0, ∞) , and (iii) lim (𝑥,𝑡)→(𝑥 0 ,0) 𝑥∈ℝ 𝑛 , 𝑡>0 𝑢(𝑥, 𝑡) = 0 , lim (𝑥,𝑡)→(𝑥 0 ,0) 𝑥∈ℝ 𝑛 , 𝑡>0 𝑢 𝑡 (𝑥, 𝑡) = 0 for each point 𝑥 0 ∈ ℝ 𝑛 . Proof. 1. If 𝑛 is odd, [ 𝑛 2 ] + 1 = 𝑛+1 2 . According to Theorem 2, we have 𝑢(⋅, ⋅; 𝑠) ∈ 𝐶 2 (ℝ 𝑛 × [𝑠, ∞)) for each 𝑠 ≥ 0 , and so 𝑢 ∈ 𝐶 2 (ℝ 𝑛 × [0, ∞)) . If 𝑛 is even, [ 𝑛 2 ] + 1 = 𝑛+2 2 . Hence 𝑢 ∈ 𝐶 2 (ℝ 𝑛 × [0, ∞)) , according to Theorem 3. link to page 84 link to page 82 link to page 90 link to page 93 link to page 94 link to page 94 link to page 94 link to page 82 link to page 86 78 2. Four Important Linear Partial Differential Equations 2. We then compute 𝑢 𝑡 (𝑥, 𝑡) = 𝑢(𝑥, 𝑡; 𝑡) + ∫ 𝑡 0 𝑢 𝑡 (𝑥, 𝑡; 𝑠) 𝑑𝑠 = ∫ 𝑡 0 𝑢 𝑡 (𝑥, 𝑡; 𝑠) 𝑑𝑠, 𝑢 𝑡𝑡 (𝑥, 𝑡) = 𝑢 𝑡 (𝑥, 𝑡; 𝑡) + ∫ 𝑡 0 𝑢 𝑡𝑡 (𝑥, 𝑡; 𝑠) 𝑑𝑠 = 𝑓(𝑥, 𝑡) + ∫ 𝑡 0 𝑢 𝑡𝑡 (𝑥, 𝑡; 𝑠) 𝑑𝑠. Furthermore Δ𝑢(𝑥, 𝑡) = ∫ 𝑡 0 Δ𝑢(𝑥, 𝑡; 𝑠) 𝑑𝑠 = ∫ 𝑡 0 𝑢 𝑡𝑡 (𝑥, 𝑡; 𝑠) 𝑑𝑠. Thus 𝑢 𝑡𝑡 (𝑥, 𝑡) − Δ𝑢(𝑥, 𝑡) = 𝑓(𝑥, 𝑡) (𝑥 ∈ ℝ 𝑛 , 𝑡 > 0), and clearly 𝑢(𝑥, 0) = 𝑢 𝑡 (𝑥, 0) = 0 for 𝑥 ∈ ℝ 𝑛 . □ The solution of the general nonhomogeneous problem is consequently the sum of the solution of (11) (given by formulas (8), (31) or (38)) and the solution of (42) (given by (41)). Examples. (i) Let us work out explicitly how to solve (42) for 𝑛 = 1 . In this case d’Alembert’s formula (8) gives 𝑢(𝑥, 𝑡; 𝑠) = 1 2 ∫ 𝑥+𝑡−𝑠 𝑥−𝑡+𝑠 𝑓(𝑦, 𝑠) 𝑑𝑦, 𝑢(𝑥, 𝑡) = 1 2 ∫ 𝑡 0 ∫ 𝑥+𝑡−𝑠 𝑥−𝑡+𝑠 𝑓(𝑦, 𝑠) 𝑑𝑦𝑑𝑠. That is, (43) 𝑢(𝑥, 𝑡) = 1 2 ∫ 𝑡 0 ∫ 𝑥+𝑠 𝑥−𝑠 𝑓(𝑦, 𝑡 − 𝑠) 𝑑𝑦𝑑𝑠 (𝑥 ∈ ℝ, 𝑡 ≥ 0). (ii) For 𝑛 = 3 , Kirchhoff’s formula (22) implies 𝑢(𝑥, 𝑡; 𝑠) = (𝑡 − 𝑠) ⨍ 𝜕𝐵(𝑥,𝑡−𝑠) 𝑓(𝑦, 𝑠) 𝑑𝑆, so that 𝑢(𝑥, 𝑡) = ∫ 𝑡 0 (𝑡 − 𝑠) (⨍ 𝜕𝐵(𝑥,𝑡−𝑠) 𝑓(𝑦, 𝑠) 𝑑𝑆) 𝑑𝑠 = 1 4𝜋 ∫ 𝑡 0 ∫ 𝜕𝐵(𝑥,𝑡−𝑠) 𝑓(𝑦, 𝑠) (𝑡 − 𝑠) 𝑑𝑆𝑑𝑠 = 1 4𝜋 ∫ 𝑡 0 ∫ 𝜕𝐵(𝑥,𝑟) 𝑓(𝑦, 𝑡 − 𝑟) 𝑟 𝑑𝑆𝑑𝑟. link to page 94 link to page 90 link to page 93 link to page 96 2.4. Wave Equation 79 Therefore (44) 𝑢(𝑥, 𝑡) = 1 4𝜋 ∫ 𝐵(𝑥,𝑡) 𝑓(𝑦, 𝑡 − |𝑦 − 𝑥|) |𝑦 − 𝑥| 𝑑𝑦 (𝑥 ∈ ℝ 3 , 𝑡 ≥ 0) solves (42) for 𝑛 = 3 . The integrand on the right is called a retarded potential . 2.4.3. Energy methods. The explicit formulas (31) and (38) demonstrate the necessity of making more and more smoothness assumptions upon the data 𝑔 and ℎ to ensure the existence of a 𝐶 2 solution of the wave equation for larger and larger 𝑛 . This suggests that perhaps some other way of measuring the size and smoothness of functions may be more appropriate. Indeed we will see in this subsection that the wave equation is nicely behaved (for all 𝑛 ) with respect to certain integral “energy” norms. a. Uniqueness. Let 𝑈 ⊂ ℝ 𝑛 be a bounded, open set with a smooth boundary 𝜕𝑈 , and as usual set 𝑈 𝑇 = 𝑈 × (0, 𝑇] , Γ 𝑇 = ̄ 𝑈 𝑇 − 𝑈 𝑇 , where 𝑇 > 0 . We are interested in the initial/boundary-value problem (45) ⎧ ⎨ ⎩ 𝑢 𝑡𝑡 − Δ𝑢 = 𝑓 in 𝑈 𝑇 𝑢 = 𝑔 on Γ 𝑇 𝑢 𝑡 = ℎ on 𝑈 × {𝑡 = 0} . THEOREM 5 (Uniqueness for wave equation) . There exists at most one func- tion 𝑢 ∈ 𝐶 2 ( ̄ 𝑈 𝑇 ) solving (45) . Proof. If ̃ 𝑢 is another such solution, then 𝑤 ≔ 𝑢 − ̃ 𝑢 solves ⎧ ⎨ ⎩ 𝑤 𝑡𝑡 − Δ𝑤 = 0 in 𝑈 𝑇 𝑤 = 0 on Γ 𝑇 𝑤 𝑡 = 0 on 𝑈 × {𝑡 = 0} . Define the “energy” 𝐸(𝑡) ≔ 1 2 ∫ 𝑈 𝑤 2 𝑡 (𝑥, 𝑡) + |𝐷𝑤(𝑥, 𝑡)| 2 𝑑𝑥 (0 ≤ 𝑡 ≤ 𝑇). We compute ̇ 𝐸(𝑡) = ∫ 𝑈 𝑤 𝑡 𝑤 𝑡𝑡 + 𝐷𝑤 ⋅ 𝐷𝑤 𝑡 𝑑𝑥 ( ̇ = 𝑑 𝑑𝑡 ) = ∫ 𝑈 𝑤 𝑡 (𝑤 𝑡𝑡 − Δ𝑤) 𝑑𝑥 = 0. There is no boundary term since 𝑤 = 0 , and hence 𝑤 𝑡 = 0 , on 𝜕𝑈 ×[0, 𝑇] . Thus for all 0 ≤ 𝑡 ≤ 𝑇 , 𝐸(𝑡) = 𝐸(0) = 0 , and so 𝑤 𝑡 , 𝐷𝑤 ≡ 0 within 𝑈 𝑇 . Since 𝑤 ≡ 0 on 𝑈 × {𝑡 = 0} , we conclude 𝑤 = 𝑢 − ̃ 𝑢 ≡ 0 in 𝑈 𝑇 . □ link to page 90 link to page 93 80 2. Four Important Linear Partial Differential Equations Cone of dependence b. Domain of dependence. As another illustration of energy methods, let us examine again the domain of dependence of solutions to the wave equation in all of space. For this, suppose 𝑢 ∈ 𝐶 2 solves 𝑢 𝑡𝑡 − Δ𝑢 = 0 in ℝ 𝑛 × (0, ∞). Fix 𝑥 0 ∈ ℝ 𝑛 , 𝑡 0 > 0 and consider the backwards wave cone with apex (𝑥 0 , 𝑡 0 ) 𝐾(𝑥 0 , 𝑡 0 ) ≔ { (𝑥, 𝑡) ∣ 0 ≤ 𝑡 ≤ 𝑡 0 , |𝑥 − 𝑥 0 | ≤ 𝑡 0 − 𝑡 }. THEOREM 6 (Finite propagation speed) . If 𝑢 ≡ 𝑢 𝑡 ≡ 0 on 𝐵(𝑥 0 , 𝑡 0 ) × {𝑡 = 0} , then 𝑢 ≡ 0 within the cone 𝐾(𝑥 0 , 𝑡 0 ) . In particular, we see that any “disturbance” originating outside 𝐵(𝑥 0 , 𝑡 0 ) has no effect on the solution within 𝐾(𝑥 0 , 𝑡 0 ) and consequently has finite prop- agation speed. We already know this from the representation formulas (31) and (38), at least assuming 𝑔 = 𝑢 and ℎ = 𝑢 𝑡 on ℝ 𝑛 × {𝑡 = 0} are sufficiently smooth. The point is that energy methods provide a much simpler proof. Proof. Define the local energy 𝑒(𝑡) ≔ 1 2 ∫ 𝐵(𝑥 0 ,𝑡 0 −𝑡) 𝑢 2 𝑡 (𝑥, 𝑡) + |𝐷𝑢(𝑥, 𝑡)| 2 𝑑𝑥 (0 ≤ 𝑡 ≤ 𝑡 0 ). link to page 678 link to page 98 link to page 98 link to page 407 link to page 634 2.5. Problems 81 Then (46) ̇ 𝑒(𝑡) = ∫ 𝐵(𝑥 0 ,𝑡 0 −𝑡) 𝑢 𝑡 𝑢 𝑡𝑡 + 𝐷𝑢 ⋅ 𝐷𝑢 𝑡 𝑑𝑥 − 1 2 ∫ 𝜕𝐵(𝑥 0 ,𝑡 0 −𝑡) 𝑢 2 𝑡 + |𝐷𝑢| 2 𝑑𝑆 = ∫ 𝐵(𝑥 0 ,𝑡 0 −𝑡) 𝑢 𝑡 (𝑢 𝑡𝑡 − Δ𝑢) 𝑑𝑥 + ∫ 𝜕𝐵(𝑥 0 ,𝑡 0 −𝑡) 𝜕𝑢 𝜕𝜈 𝑢 𝑡 𝑑𝑆 − 1 2 ∫ 𝜕𝐵(𝑥 0 ,𝑡 0 −𝑡) 𝑢 2 𝑡 + |𝐷𝑢| 2 𝑑𝑆 = ∫ 𝜕𝐵(𝑥 0 ,𝑡 0 −𝑡) 𝜕𝑢 𝜕𝜈 𝑢 𝑡 − 1 2 𝑢 2 𝑡 − 1 2 |𝐷𝑢| 2 𝑑𝑆. Now (47) | | | 𝜕𝑢 𝜕𝜈 𝑢 𝑡 | | | ≤ |𝑢 𝑡 ||𝐷𝑢| ≤ 1 2 𝑢 2 𝑡 + 1 2 |𝐷𝑢| 2 , by the Cauchy–Schwarz and Cauchy inequalities (§B.2). Inserting (47) into (46), we find ̇ 𝑒(𝑡) ≤ 0 ; and so 𝑒(𝑡) ≤ 𝑒(0) = 0 for all 0 ≤ 𝑡 ≤ 𝑡 0 . Thus 𝑢 𝑡 , 𝐷𝑢 ≡ 0 , and consequently 𝑢 ≡ 0 within the cone 𝐾(𝑥 0 , 𝑡 0 ) . □ A generalization of this proof to more complicated geometry appears later, in §7.2.4. See also §12.1 for a similar calculation for a nonlinear wave equation. 2.5. PROBLEMS In the following exercises, all given functions are assumed smooth, unless oth- erwise stated. 1. Write down an explicit formula for a function 𝑢 solving the initial-value problem { 𝑢 𝑡 + 𝑏 ⋅ 𝐷𝑢 + 𝑐𝑢 = 0 in ℝ 𝑛 × (0, ∞) 𝑢 = 𝑔 on ℝ 𝑛 × {𝑡 = 0} . Here 𝑐 ∈ ℝ and 𝑏 ∈ ℝ 𝑛 are constants. 2. Prove that Laplace’s equation Δ𝑢 = 0 is rotation invariant; that is, if 𝑂 is an orthogonal 𝑛 × 𝑛 matrix and we define 𝑣(𝑥) ≔ 𝑢(𝑂𝑥) (𝑥 ∈ ℝ 𝑛 ), then Δ𝑣 = 0 . 3. Modify the proof of the mean-value formulas to show for 𝑛 ≥ 3 that 𝑢(0) = ⨍ 𝜕𝐵(0,𝑟) 𝑔 𝑑𝑆 + 1 𝑛(𝑛 − 2)𝛼(𝑛) ∫ 𝐵(0,𝑟) ( 1 |𝑥| 𝑛−2 − 1 𝑟 𝑛−2 ) 𝑓 𝑑𝑥, link to page 56 link to page 49 link to page 96 82 2. Four Important Linear Partial Differential Equations provided { −Δ𝑢 = 𝑓 in 𝐵 0 (0, 𝑟) 𝑢 = 𝑔 on 𝜕𝐵(0, 𝑟). 4. Give a direct proof that if 𝑢 ∈ 𝐶 2 (𝑈) ∩ 𝐶( ̄ 𝑈) is harmonic within a bounded open set 𝑈 , then max ̄ 𝑈 𝑢 = max 𝜕𝑈 𝑢. (Hint: Define 𝑢 𝜀 ≔ 𝑢 + 𝜀|𝑥| 2 for 𝜀 > 0 , and show 𝑢 𝜀 cannot attain its maximum over ̄ 𝑈 at an interior point.) 5. We say 𝑣 ∈ 𝐶 2 ( ̄ 𝑈) is subharmonic if −Δ𝑣 ≤ 0 in 𝑈. (a) Prove for a subharmonic function 𝑣 that 𝑣(𝑥) ≤ ⨍ 𝐵(𝑥,𝑟) 𝑣 𝑑𝑦 for all 𝐵(𝑥, 𝑟) ⊂ 𝑈. (b) Prove that therefore max ̄ 𝑈 𝑣 = max 𝜕𝑈 𝑣 if 𝑈 is bounded. (c) Let 𝜙 ∶ ℝ → ℝ be smooth and convex. Assume 𝑢 is harmonic and 𝑣 ≔ 𝜙(𝑢) . Prove 𝑣 is subharmonic. (d) Prove 𝑣 ≔ |𝐷𝑢| 2 is subharmonic, whenever 𝑢 is harmonic. 6. Let 𝑈 be a bounded, open subset of ℝ 𝑛 . Prove that there exists a constant 𝐶 , depending only on 𝑈 , such that max ̄ 𝑈 |𝑢| ≤ 𝐶(max 𝜕𝑈 |𝑔| + max ̄ 𝑈 |𝑓|) whenever 𝑢 is a smooth solution of { −Δ𝑢 = 𝑓 in 𝑈 𝑢 = 𝑔 on 𝜕𝑈 . (Hint: −Δ(𝑢 + |𝑥| 2 2𝑛 𝜆) ≤ 0 , for 𝜆 ≔ max ̄ 𝑈 |𝑓| .) 7. Use Poisson’s formula for the ball to prove 𝑟 𝑛−2 𝑟 − |𝑥| (𝑟 + |𝑥|) 𝑛−1 𝑢(0) ≤ 𝑢(𝑥) ≤ 𝑟 𝑛−2 𝑟 + |𝑥| (𝑟 − |𝑥|) 𝑛−1 𝑢(0) whenever 𝑢 is positive and harmonic in 𝐵 0 (0, 𝑟) . This is an explicit form of Harnack’s inequality. 8. Prove Theorem 15 in §2.2.4. (Hint: Since 𝑢 ≡ 1 solves (44) for 𝑔 ≡ 1 , the theory automatically implies ∫ 𝜕𝐵(0,1) 𝐾(𝑥, 𝑦) 𝑑𝑆(𝑦) = 1 for each 𝑥 ∈ 𝐵 0 (0, 1) .) link to page 100 2.5. Problems 83 9. Let 𝑢 be the solution of { Δ𝑢 = 0 in ℝ 𝑛 + 𝑢 = 𝑔 on 𝜕ℝ 𝑛 + given by Poisson’s formula for the half-space. Assume 𝑔 is bounded and 𝑔(𝑥) = |𝑥| for 𝑥 ∈ 𝜕ℝ 𝑛 + , |𝑥| ≤ 1 . Show 𝐷𝑢 is not bounded near 𝑥 = 0 . (Hint: Estimate ᵆ(𝜆𝑒 𝑛 )−ᵆ(0) 𝜆 .) 10. (Reflection principle) (a) Let 𝑈 + denote the open half-ball { 𝑥 ∈ ℝ 𝑛 ∣ |𝑥| < 1, 𝑥 𝑛 > 0 } . Assume 𝑢 ∈ 𝐶 2 (𝑈 + ) is harmonic in 𝑈 + , with 𝑢 = 0 on 𝜕𝑈 + ∩ {𝑥 𝑛 = 0} . Set 𝑣(𝑥) ≔ { 𝑢(𝑥) if 𝑥 𝑛 ≥ 0 −𝑢(𝑥 1 , . . . , 𝑥 𝑛−1 , −𝑥 𝑛 ) if 𝑥 𝑛 < 0 for 𝑥 ∈ 𝑈 = 𝐵 0 (0, 1) . Prove 𝑣 ∈ 𝐶 2 (𝑈) and thus 𝑣 is harmonic within 𝑈 . (b) Now assume only that 𝑢 ∈ 𝐶 2 (𝑈 + ) ∩ 𝐶(𝑈 + ) . Show that 𝑣 is harmonic within 𝑈 . (Hint: Use Poisson’s formula for the ball.) 11. (Kelvin transform for Laplace’s equation) The Kelvin transform 𝒦𝑢 = ̄ 𝑢 of a function 𝑢 ∶ ℝ 𝑛 → ℝ is ̄ 𝑢(𝑥) ≔ 𝑢( ̄ 𝑥)| ̄ 𝑥| 𝑛−2 = 𝑢(𝑥/|𝑥| 2 )|𝑥| 2−𝑛 (𝑥 ≠ 0), where ̄ 𝑥 = 𝑥/|𝑥| 2 . Show that if 𝑢 is harmonic, then so is ̄ 𝑢 . (Hint: First show that 𝐷 𝑥 ̄ 𝑥(𝐷 𝑥 ̄ 𝑥) 𝑇 = | ̄ 𝑥| 4 𝐼 . The mapping 𝑥 → ̄ 𝑥 is conformal , meaning angle preserving.) 12. Suppose 𝑢 is smooth and solves 𝑢 𝑡 − Δ𝑢 = 0 in ℝ 𝑛 × (0, ∞) . (a) Show 𝑢 𝜆 (𝑥, 𝑡) ≔ 𝑢(𝜆𝑥, 𝜆 2 𝑡) also solves the heat equation for each 𝜆 ∈ ℝ . (b) Use (a) to show 𝑣(𝑥, 𝑡) ≔ 𝑥 ⋅ 𝐷𝑢(𝑥, 𝑡) + 2𝑡𝑢 𝑡 (𝑥, 𝑡) solves the heat equa- tion as well. 13. Assume 𝑛 = 1 and 𝑢(𝑥, 𝑡) = 𝑣( 𝑥 √𝑡 ) . (a) Show 𝑢 𝑡 = 𝑢 𝑥𝑥 if and only if ( ∗ ) 𝑣 ″ + 𝑧 2 𝑣 ′ = 0. Show that the general solution of ( ∗ ) is 𝑣(𝑧) = 𝑐 ∫ 𝑧 0 𝑒 −𝑠 2 /4 𝑑𝑠 + 𝑑. 84 2. Four Important Linear Partial Differential Equations (b) Differentiate 𝑢(𝑥, 𝑡) = 𝑣( 𝑥 √𝑡 ) with respect to 𝑥 and select the constant 𝑐 properly, to obtain the fundamental solution Φ for 𝑛 = 1 . Explain why this procedure produces the fundamental solution. (Hint: What is the initial condition for 𝑢 ?) 14. Write down an explicit formula for a solution of { 𝑢 𝑡 − Δ𝑢 + 𝑐𝑢 = 𝑓 in ℝ 𝑛 × (0, ∞) 𝑢 = 𝑔 on ℝ 𝑛 × {𝑡 = 0} , where 𝑐 ∈ ℝ . 15. Given 𝑔 ∶ [0, ∞) → ℝ , with 𝑔(0) = 0 , derive the formula 𝑢(𝑥, 𝑡) = 𝑥 √4𝜋 ∫ 𝑡 0 1 (𝑡 − 𝑠) 3/2 𝑒 −𝑥2 4(𝑡−𝑠) 𝑔(𝑠) 𝑑𝑠 for a solution of the initial/boundary-value problem ⎧ ⎨ ⎩ 𝑢 𝑡 − 𝑢 𝑥𝑥 = 0 in ℝ + × (0, ∞) 𝑢 = 0 on ℝ + × {𝑡 = 0} 𝑢 = 𝑔 on {𝑥 = 0} × [0, ∞). (Hint: Let 𝑣(𝑥, 𝑡) ≔ 𝑢(𝑥, 𝑡) − 𝑔(𝑡) and extend 𝑣 to {𝑥 < 0} by odd reflection.) 16. Give a direct proof that if 𝑈 is bounded and 𝑢 ∈ 𝐶 2 1 (𝑈 𝑇 ) ∩ 𝐶( ̄ 𝑈 𝑇 ) solves the heat equation, then max ̄ 𝑈 𝑇 𝑢 = max Γ 𝑇 𝑢. (Hint: Define 𝑢 𝜀 ≔ 𝑢−𝜀𝑡 for 𝜀 > 0 , and show 𝑢 𝜀 cannot attain its maximum over ̄ 𝑈 𝑇 at a point in 𝑈 𝑇 .) 17. We say 𝑣 ∈ 𝐶 2 1 (𝑈 𝑇 ) is a subsolution of the heat equation if 𝑣 𝑡 − Δ𝑣 ≤ 0 in 𝑈 𝑇 . (a) Prove for a subsolution 𝑣 that 𝑣(𝑥, 𝑡) ≤ 1 4𝑟 𝑛 ∬ 𝐸(𝑥,𝑡;𝑟) 𝑣(𝑦, 𝑠) |𝑥 − 𝑦| 2 (𝑡 − 𝑠) 2 𝑑𝑦𝑑𝑠 for all 𝐸(𝑥, 𝑡; 𝑟) ⊂ 𝑈 𝑇 . (b) Prove that therefore max ̄ 𝑈 𝑇 𝑣 = max Γ 𝑇 𝑣 . (c) Let 𝜙 ∶ ℝ → ℝ be smooth and convex. Assume 𝑢 solves the heat equation and 𝑣 ≔ 𝜙(𝑢) . Prove 𝑣 is a subsolution. (d) Prove 𝑣 ≔ |𝐷𝑢| 2 + 𝑢 2 𝑡 is a subsolution, whenever 𝑢 solves the heat equation. 2.5. Problems 85 18. Assume 𝑢 solves the initial-value problem { 𝑢 𝑡𝑡 − Δ𝑢 = 0 in ℝ 𝑛 × (0, ∞) 𝑢 = 0, 𝑢 𝑡 = ℎ on ℝ 𝑛 × {𝑡 = 0} . Show that 𝑣 ≔ 𝑢 𝑡 solves { 𝑣 𝑡𝑡 − Δ𝑣 = 0 in ℝ 𝑛 × (0, ∞) 𝑣 = ℎ, 𝑣 𝑡 = 0 on ℝ 𝑛 × {𝑡 = 0} . This is Stokes’ rule . 19. (a) Show the general solution of the PDE 𝑢 𝑥𝑦 = 0 is 𝑢(𝑥, 𝑦) = 𝐹(𝑥) + 𝐺(𝑦) for arbitrary functions 𝐹 , 𝐺 . (b) Using the change of variables 𝜉 = 𝑥 + 𝑡 , 𝜂 = 𝑥 − 𝑡 , show 𝑢 𝑡𝑡 − 𝑢 𝑥𝑥 = 0 if and only if 𝑢 𝜉𝜂 = 0 . (c) Use (a) and (b) to rederive d’Alembert’s formula. (d) Under what conditions on the initial data 𝑔 , ℎ is the solution 𝑢 a right- moving wave? A left-moving wave? 20. Assume that for some attenuation function 𝛼 = 𝛼(𝑟) and delay function 𝛽 = 𝛽(𝑟) ≥ 0 , there exist for all profiles 𝜙 solutions of the wave equation in (ℝ 𝑛 − {0}) × ℝ having the form 𝑢(𝑥, 𝑡) = 𝛼(𝑟)𝜙(𝑡 − 𝛽(𝑟)). Here 𝑟 = |𝑥| and we assume 𝛽(0) = 0 . Show that this is possible only if 𝑛 = 1 or 3 , and compute the form of the functions 𝛼 , 𝛽 . (T. Morley, SIAM Review 27 (1985), 69–71) 21. (a) Assume E = (𝐸 1 , 𝐸 2 , 𝐸 3 ) and B = (𝐵 1 , 𝐵 2 , 𝐵 3 ) solve Maxwell’s equa- tions ⎧ ⎨ ⎩ E 𝑡 = curl B , B 𝑡 = − curl E div B = div E = 0. Show E 𝑡𝑡 − Δ E = 0, B 𝑡𝑡 − Δ B = 0. (b) Assume that u = (𝑢 1 , 𝑢 2 , 𝑢 3 ) solves the evolution equations of linear elasticity u 𝑡𝑡 − 𝜇Δ u − (𝜆 + 𝜇)𝐷(div u ) = 0 in ℝ 3 × (0, ∞). Show 𝑤 ≔ div u and w ≔ curl u each solve wave equations, but with differing speeds of propagation. link to page 36 link to page 710 link to page 714 link to page 59 link to page 711 link to page 709 link to page 67 86 2. Four Important Linear Partial Differential Equations 22. Let 𝑢 denote the density of particles moving to the right with speed one along the real line and let 𝑣 denote the density of particles moving to the left with speed one. If at rate 𝑑 > 0 right-moving particles randomly become left-moving, and vice versa, we have the system of PDE { 𝑢 𝑡 + 𝑢 𝑥 = 𝑑(𝑣 − 𝑢) 𝑣 𝑡 − 𝑣 𝑥 = 𝑑(𝑢 − 𝑣). Show that both 𝑤 ≔ 𝑢 and 𝑤 ≔ 𝑣 solve the telegraph equation 𝑤 𝑡𝑡 + 2𝑑𝑤 𝑡 − 𝑤 𝑥𝑥 = 0. 23. Let 𝑆 denote the square lying in ℝ × (0, ∞) with corners at the points (0, 1) , (1, 2) , (0, 3) , (−1, 2) . Define 𝑓(𝑥, 𝑡) ≔ ⎧ ⎨ ⎩ −1 for (𝑥, 𝑡) ∈ 𝑆 ∩ {𝑡 > 𝑥 + 2} 1 for (𝑥, 𝑡) ∈ 𝑆 ∩ {𝑡 < 𝑥 + 2} 0 otherwise . Assume 𝑢 solves { 𝑢 𝑡𝑡 − 𝑢 𝑥𝑥 = 𝑓 in ℝ × (0, ∞) 𝑢 = 0, 𝑢 𝑡 = 0 on ℝ × {𝑡 = 0}. Describe the shape of 𝑢 for times 𝑡 > 3 . (J. G. Kingston, SIAM Review 30 (1988), 645–649) 24. (Equipartition of energy) Let 𝑢 solve the initial-value problem for the wave equation in one dimension: { 𝑢 𝑡𝑡 − 𝑢 𝑥𝑥 = 0 in ℝ × (0, ∞) 𝑢 = 𝑔, 𝑢 𝑡 = ℎ on ℝ × {𝑡 = 0}. Suppose 𝑔 , ℎ have compact support. The kinetic energy is 𝑘(𝑡) ≔ 1 2 ∫ ∞ −∞ 𝑢 2 𝑡 (𝑥, 𝑡) 𝑑𝑥 and the potential energy is 𝑝(𝑡) ≔ 1 2 ∫ ∞ −∞ 𝑢 2 𝑥 (𝑥, 𝑡) 𝑑𝑥 . Prove (a) 𝑘(𝑡) + 𝑝(𝑡) is constant in 𝑡 , (b) 𝑘(𝑡) = 𝑝(𝑡) for all large enough times 𝑡 . 2.6. REFERENCES Section 2.2: A good source for more on Laplace’s and Poisson’s equations is Gilbarg–Trudinger [ G-T , Chapters 2-4]. The proof of analyticity is from Mikhailov [ M ]. J. Cooper helped me with Green’s functions. Section 2.3: See John [ J2 , Chapter 7] or Friedman [ Fr1 ] for further informa- tion concerning the heat equation. Theorem 3 is due to N. Watson (Proc. London Math. Society 26 (1973), 385–417), as is the proof of link to page 69 link to page 72 link to page 711 link to page 74 link to page 714 link to page 78 link to page 714 link to page 80 link to page 709 link to page 716 link to page 98 link to page 103 2.6. References 87 Theorem 4. Theorem 6 is taken from John [ J2 ], and Theorem 8 fol- lows Mikhailov [ M ]. Theorem 11 is from Payne [ Pa , §2.3]. Section 2.4: See Antman (Amer. Math. Monthly 87 (1980), 359–370) for a careful derivation of the one-dimensional wave equation as a model for a vibrating string. The solution of the wave equation presented here follows Folland [ F1 ], Strauss [ St2 ]. Section 2.5: J. Goldstein contributed Problem 24. Front Cover Contents Preface to the second edition Preface to the first edition Chapter 1. Introduction 1.1. Partial Differential Equations 1.2. Examples 1.3. Strategies for Studying PDE 1.4. Overview 1.5. Problems 1.6. References Part I. Representation Formulas for Solutions Chapter 2. Four Important Linear Partial Differential Equations 2.1. Transport Equation 2.2. Laplace’s Equation 2.3. Heat Equation 2.4. Wave Equation 2.5. Problems 2.6. References Chapter 3. Nonlinear First-Order PDE 3.1. Complete Integrals, Envelopes 3.2. Characteristics 3.3. Introduction to Hamilton–Jacobi Equations 3.4. Introduction to Conservation Laws 3.5. Problems 3.6. References Chapter 4. Other Ways to Represent Solutions 4.1. Separation of Variables 4.2. Similarity Solutions 4.3. Transform Methods 4.4. Converting Nonlinear into Linear PDE 4.5. Asymptotics 4.6. Power Series 4.7. Problems 4.8. References Part II. Theory for Linear Partial Differential Equations Chapter 5. Sobolev Spaces 5.1. Hölder Spaces 5.2. Sobolev Spaces 5.3. Approximation 5.4. Extensions 5.5. Traces 5.6. Sobolev Inequalities 5.7. Compactness 5.8. Additional Topics 5.9. Other Spaces of Functions 5.10. Problems 5.11. References Chapter 6. Second-Order Elliptic Equations 6.1. Definitions 6.2. Existence of Weak Solutions 6.3. Regularity 6.4. Maximum Principles 6.5. Eigenvalues and Eigenfunctions 6.6. Problems 6.7. References Chapter 7. Linear Evolution Equations 7.1. Second-Order Parabolic Equations 7.2. Second-Order Hyperbolic Equations 7.3. Systems of Hyperbolic First-Order Equations 7.4. Semigroup Theory 7.5. Problems 7.6. References Part III. Theory for Nonlinear Partial Differential Equations Chapter 8. The Calculus of Variations 8.1. Introduction 8.2. Existence of Minimizers 8.3. Regularity 8.4. Constraints 8.5. Critical Points 8.6. Invariance, Noether’s Theorem 8.7. Problems 8.8. References Chapter 9. Nonvariational Techniques 9.1. Monotonicity Methods 9.2. Fixed Point Methods 9.3. Method of Subsolutions and Supersolutions 9.4. Nonexistence of Solutions 9.5. Geometric Properties of Solutions 9.6. Gradient Flows 9.7. Problems 9.8. References Chapter 10. Hamilton–Jacobi Equations 10.1. Introduction, Viscosity Solutions 10.2. Uniqueness 10.3. Control Theory, Dynamic Programming 10.4. Problems 10.5. References Chapter 11. Systems of Conservation Laws 11.1. Introduction 11.2. Riemann’s Problem 11.3. Systems of Two Conservation Laws 11.4. Entropy Criteria 11.5. Problems 11.6. References Chapter 12. Nonlinear Wave Equations 12.1. Introduction 12.2. Existence of Solutions 12.3. Semilinear Wave Equations 12.4. Critical Power Nonlinearity 12.5. Nonexistence of Solutions 12.6. Problems 12.7. References Appendices Appendix A. Notation A.1. Notation for matrices A.2. Geometric notation A.3. Notation for functions A.4. Vector-valued functions A.5. Notation for estimates A.6. Some comments about notation Appendix B. Inequalities B.1. Convex functions B.2. Useful inequalities Appendix C. Calculus C.1. Boundaries C.2. Gauss–Green Theorem C.3. Polar coordinates, coarea formula C.4. Moving regions C.5. Convolution and smoothing C.6. Inverse Function Theorem C.7. Implicit Function Theorem C.8. Uniform convergence Appendix D. Functional Analysis D.1. Banach spaces D.2. Hilbert spaces D.3. Bounded linear operators D.4. Weak convergence D.5. Compact operators, Fredholm theory D.6. Symmetric operators Appendix E. Measure Theory E.1. Lebesgue measure E.2. Measurable functions and integration E.3. Convergence theorems for integrals E.4. Differentiation E.5. Banach space-valued functions Bibliography Index Back Cover