# MATH COURSE NOTES - CLASS MEETING # Introduction to PDEs, Fall 2011 Professor: Jared Speck

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1 MATH 8.52 COURSE NOTES - CLASS MEETING # Introduction to PDEs, Fall 20 Professor: Jared Speck Class Meeting # 6: Laplace s and Poisson s Equations We will now study the Laplace and Poisson equations on a domain (i.e. open connected subset) Ω R n. Recall that (0.0.) The Laplace equation is def = n i= 2 i. (0.0.2) u(x) = 0, x Ω, while the Poisson equation is the inhomogeneous equation (0.0.3) u(x) = f(x). Functions u C 2 (Ω) verifying (0.0.2) are said to be harmonic. (0.0.2) and (0.0.3) are both second order, linear, constant coefficient PDEs. As in our study of the heat equation, we will need to supply some kind of boundary conditions to get a well-posed problem. But unlike the heat equation, there is no timelike variable, so there is no initial condition to specify!. Where does it come from?.. Basic examples. First example: set t u 0 in the heat equation, and (0.0.2) results. These solutions are known as steady state solutions. Second example: We start with Maxwell equations from electrodynamics. interest are E = (E (t, x, y, z), E 2 (t, x, y, z), E 3 (t, x, y, z)) is the electric field B = (B (t, x, y, z), B 2 (t, x, y, z), B 3 (t, x, y, z)) is the magnetic induction J = (J (t, x, y, z), J 2 (t, x, y, z), J 3 (t, x, y, z)) is the current density ρ is the charge density Maxwell s equations are The quantities of (..) (..2) t E B = J, E = ρ, t B + E = 0, B = 0. Recall that is the curl operator, so that e.g. B = ( y B 3 z B 2, z B x B 3, x B 2 y B ). Let s look for steady-state solutions with t E = t B 0. Then equation (..2) implies that

2 2 MATH 8.52 COURSE NOTES - CLASS MEETING # 6 (..3) E = 0, so that by the Poincaré lemma, there exists a scalar-valued function φ(x, y, z) such that (..4) E(x, y, z) = φ(x, y, z). The function φ is called an electric potential. Plugging (..4) into the second of (..), and using the identity φ = φ, we deduce that (..5) φ(x, y, z) = ρ(x, y, z). This is exactly the Poisson equation (0.0.3) with inhomogeneous term f = ρ. Thus, Poisson s equation is at the heart of electrostatics..2. Connections to complex analysis. Let z = x + iy (where x, y R) be a complex number, and let f(z) = u(z) + iv(z) be a complex-valued function (where u, v R). We recall that f is said to be differentiable at z 0 if f(z) f(z 0 ) (.2.) lim z z 0 z z 0 exists. If the limit exists, we denote it by f (z 0 ). A fundamental result of complex analysis is the following: f is differentiable at z 0 = x 0 + iy 0 (x 0, y 0 ) if and only if the real and imaginary parts of f verify the Cauchy-Riemann equations at z 0 (.2.2) (.2.3) u x (x 0, y 0 ) = v y (x 0, y 0 ), u y (x 0, y 0 ) = v x (x 0, y 0 ). Differentiating (.2.2) and using the symmetry of mixed partial derivatives (we are assuming here that u(x, y) and v(x, y) are C near (x 0, y 0 )), we have (.2.4) (.2.5) u def = u xx + u yy = v yx v xy = 0, v def = v xx + v yy = u yx + u xy = 0. Thus, the real and imaginary parts of a complex-differentiable function are harmonic! 2. Well-posed Problems Much like in the case of the heat equation, we are interested in well-posed problems for the Laplace and Poisson equations. Recall that well-posed problems are problems that i) have a solution; ii)the solutions are unique; and iii)the solution varies continuously with the data. Let Ω R n be a domain with a Lipschitz boundary, and let ˆN denote the unit outward normal vector to Ω. We consider the PDE (2.0.6) u(x) = f(x), x Ω, supplemented by some boundary conditions. The following boundary conditions are known to lead to well-posed problems:

3 MATH 8.52 COURSE NOTES - CLASS MEETING # 6 3 () Dirichlet data: specify a function g(x) defined on Ω such that u Ω (x) = g(x). (2) Neumann data: specify a function h(x) defined on Ω such that ˆNu(x) Ω (x) = h(x). (3) Robin-type data: specify a function h(x) defined on Ω such that ˆNu(x) Ω (x)+αu Ω (x) = h(x), where α > 0 is a constant. (4) Mixed conditions: for example, we can divide Ω into two disjoint pieces Ω = S D S N, where S N is relatively open in Ω, and specify a function g(x) defined on S D and a function h(x) defined on S N such that u SD (x) = g(x), ˆNu SN (x) = h(x). (5) Conditions at infinity: When Ω = R n, we can specify asymptotic conditions on u(x) as x. We will return to this kind of condition later in the course. 3. Uniqueness via the Energy Method In this section, we address the question of uniqueness for solutions to the equation (0.0.3), supplemented by suitable boundary conditions. As in the case of the heat equation, we are able to provide a simple proof based on the energy method. Theorem 3.. Let Ω R n be a smooth, bounded domain. Then under Dirichlet, Robin, or mixed boundary conditions, there is at most one solution of regularity u C 2 (Ω) C (Ω) to the Poisson equation (0.0.3). In the case of Neumann conditions, any two solutions can differ by at most a constant. Proof. If u and v are two solutions to (0.0.3) with the same boundary data, then we can subtract them (aren t linear PDEs nice?!...) to get a solution w def = u v to the Poisson equation with 0 data: (3.0.7) w = 0. Let s perform the usual trick of multiplying (3.0.7) by w, integrating over Ω, and integrating by parts via the divergence theorem: (3.0.8) 0 = Ω w w d n x = Ω w w d n x = Ω w 2 d n x + Ω w ˆNw dσ. In the case of Dirichlet data, w Ω = 0, so the last term in (3.0.8) vanishes. Thus, in the Dirichlet case, we have that (3.0.9) Ω w 2 = 0. Thus, w = 0 in Ω, and so w is constant in Ω. Since w is 0 on Ω, we have that w 0 in Ω, which shows that u v in Ω. Similarly, in the Robin case (3.0.0) which implies that Ω w ˆNw dσ = α Ω w 2 dσ 0, (3.0.) Ω w 2 = 0, and we can argue as before conclude that w 0 in Ω.

4 4 MATH 8.52 COURSE NOTES - CLASS MEETING # 6 Now in the Neumann case, we have that ˆNw Ω = 0, and we can argue as above to conclude that w is constant in Ω. But now we can t say anything about the constant, so the best we can conclude is that u = v + constant in Ω. 4. Mean value properties Harmonic functions u have some amazing properties. Some of the most important ones are captured in the following theorem, which shows that the pointwise values of u can be determined by its average over solid balls or their boundaries. Theorem 4. (Mean value properties). Let u(x) be harmonic in the domain Ω R n, and let B R (x) Ω be a ball of radius R centered at the point x. Then the following mean value formulas hold: (4.0.2a) n u(x) = ω n R n u(y) d n y, BR (x) (4.0.2b) u(x) = ω n R n u(σ) dσ, BR (x) where ω n is the area of B (0) R n, that is, the area of the boundary of the unit ball in R n. Proof. Let s address the n = 2 case only; the proof is similar for other values of n. Let s also assume that x is the origin; as we will see, we will be able to treat the case of general x by reducing it to the origin. We will work with polar coordinates (r, θ) on R 2. For a ball of radius r, we have that the measure dσ corresponding to B r (0) is dσ = r dθ. Note also that along B r (0), we have that r u = u ˆN = ˆNu, where ˆN(σ) is the unit normal to B r (0). For any 0 r < R, we define g(r) def = 2πr u(σ) dσ = 2π B r(0) 2πr ru(r, θ) dθ = 2π (4.0.3) θ=0 2π θ=0 We now note that since u is continuous at 0, we have that u(r, θ) dθ. (4.0.4) u(0) = lim g(r). r 0 + Thus, we would obtain (4.0.2b) in the case x = 0 if we could show that g (r) = 0. Let s now show this. To this end, we calculate that g (r) = 2π 2π r u(r, θ) dθ = 2π (4.0.5) θ=0 2π ˆNu(r, θ) dθ = θ=0 2π ˆN(σ) u(σ) dσ. B (0) By the divergence theorem, this last term is equal to (4.0.6) 2π u(y) d 2 y. B (0) But u = 0 since u is harmonic, so we have shown that (4.0.7) g (r) = 0, and we have shown (4.0.2b) for x = 0. To prove (4.0.2a), we use polar coordinate integration and (4.0.2b) (in the case x = 0) to obtain

5 MATH 8.52 COURSE NOTES - CLASS MEETING # 6 5 u(0)r 2 R /2 = ru(0) dr = R 2π (4.0.8) 0 2π ru(r, θ) dθ dr = 0 θ=0 2π u(y) d 2 y. B R (0) We have now shown (4.0.2a) and (4.0.2b) when x = 0. To obtain the corresponding formulas for non-zero x, define v(y) def = u(x + y), and note that y v(y) = ( y u)(x + y) = 0. Therefore, using what we have already shown, (4.0.9) u(x) = v(0) = 2 ω n R 2 v(y) d 2 y = BR (0) 2 ω 2 R 2 u(x + y) d 2 y = BR (0) 2 ω 2 R 2 u(y) d 2 y, BR (x) which implies (4.0.2a) for general x. We can similarly obtain (4.0.2b) for general x. 5. Maximum Principle Let s now discuss another amazing property verified by harmonic functions. The property, known as the strong maximum principle, says that most harmonic functions achieve their maximums and minimums only on the interior of Ω. The only exceptions are the constant functions. Theorem 5. (Strong Maximum Principle). Let Ω R n be a domain, and assume that u C(Ω) verifies the mean value property (4.0.2a). Then if u achieves its max or min at a point p Ω, then u is constant on Ω. Therefore, if Ω is bounded and u C(Ω) is not constant, then for every x Ω, we have (5.0.20) u(x) < max u(y), u(x) > min u(y). Proof. We give the argument for the min in the case n = 2. Suppose that u achieves its min at a point p Ω, and that u(p) = m. Let B(p) Ω be any ball centered at p, and let z be any point in B(p). Choose a small ball B r (z) of radius r centered z with B r (z) B(p). Note that by the definition of a min, we have that (5.0.2) u(z) m. Using the assumption that the mean value property (4.0.2a) holds, we conclude that (5.0.22) m = B(p) u(y) d 2 y = B(p) B(p) { u(y) d 2 y + Br(z) u(y) d 2 y} B/Br(z) = B(p) { B r(z) u(z) + u(y) d 2 y} B/Br(z) B(p) { B r(z) u(z) + m( B(p) B r (z) )}. Rearranging inequality (5.0.22), we conclude that (5.0.23) u(z) m. Combining (5.0.2) and (5.0.23), we conclude that (5.0.24) u(x) = m

6 6 MATH 8.52 COURSE NOTES - CLASS MEETING # 6 holds for all points x B(p). Therefore, u is locally constant at any point where it achieves its min. Since Ω is open and connected, we conclude that u(x) = m for all x Ω. The next corollary will allow us to compare the size of two solutions to Poisson s equation if we have information about the size of the source terms and about the values of the solutions on Ω. The proof is based on Theorem 5.. Corollary Let Ω R n be a bounded domain and let f C(Ω). Then the PDE (5.0.25) { u = 0, x Ω, u(x) = f(x), x Ω, has at most one solution u f C 2 (Ω) C(Ω). Furthermore, if u f and u g are the solutions corresponding to the data f, g C(Ω), then () (Comparison Principle) If f g on Ω and f g, then u f > u g in Ω. (2) (Stability Estimate) For any x Ω, we have that u f (x) u g (x) max f(y) g(y). Proof. We first prove the Comparison Principle. Let w = u f u g. Then by subtracting the PDEs, we see that w solves (5.0.26) { w = 0, x Ω, u(x) = f(x) g(x) 0, x Ω, Since w is harmonic, since f(x) g(x) 0 on Ω, and since f g, Theorem 5. implies that w is not constant and that for every x Ω, we have (5.0.27) w(x) > max f(y) g(y) 0. This proves the Comparison Principle. For the Stability Estimate, we perform a similar argument for both ±w, which leads to the estimates (5.0.28) w(x) > max f(y) g(y), (5.0.29) w(x) > max f(y) g(y). Combining (5.0.28) and (5.0.29), we deduce the Stability Estimate. The at most one statement of the corollary now follows directly from applying the Stability Estimate to w in the case f = g. address:

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