PARTIAL DIFFERENTIAL EQUATIONS ( S11) LECTURE 12

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1 PATIAL DIFFEENTIAL EQUATIONS (0.47S) LECTUE 2 CHENGBO WANG Last time: (Derive) wave equations: 2 u t 2 k u Q(x, t) Boundary conditions Today: Maximum principle, Uniqueness. Introduction: Uniqueness ecall that we have discussed two methods of solving PDEs (IBVP for wave/heat equations, BVP for Laplace equation) () method of separation of variables (MSV) find the solutions in terms of Fourier series existence of PDE (2) method of eigenfunction expansion assuming existence of the solutions with certain smooth property, then expand it as series of eigenfunctions, determine the coefficients (uniqueness of PDE) after that, check existence and the assumption (existence of PDE) So, we have discussed one method to prove uniqueness. However, for the method of eigenfunction expansion, Pros () prove existence/uniqueness at the same time (2) express the solution explicitly Cons () rely heavily on the knowledge of the eigenfunctions (2) apply only to the problems with regular domains (like rectangle) In this lecture, we want to present a general method to prove uniqueness for the Laplace/heat equations. For simplicity, here, we will discuss only the 2-dimensional Laplace equation. 2. Maximum Principle Theorem (Maximum Principle). Let D be a connected (regular) bounded open set in 2. Assuming u u(x, y) is a solution of the Laplace equation u 0 in D and continuous on D D D ( D is the boundary of D). Then the maximum and minimum values of u are attained on D, that is, x M, x m D, s.t. u( x m ) u( x) u( x M ) x D

2 2 CHENGBO WANG Proof. ) If we have maximum value in D at, say, (x 0, y 0 ), then we have u xx 0, and u yy 0, at (x 0, y 0 ) Since u u xx + u yy 0, we see that u xx u yy 0 at (x 0, y 0 ). This does not give us enough information. 2) Instead, for any ɛ > 0, consider v(x, y) u(x, y) + ɛ(x 2 + y 2 ) v(x, y) u(x, y) + ɛ (x 2 + y 2 ) 0 + 4ɛ > 0 ecall that v 0 at interior maximum points, this tells us that there is no interior maximum points in D. 3) Since u is continuous on D D D, v is also continuous on D and so has maximum value at x 0 D. Thus x 0 D, max v(x, y) v( x 0) max v(x, y) (x,y) D (x,y) D 4) Now we are ready to prove the Maximum Principle. By the construction of v, we see that for any (x, y) D, u(x, y) v(x, y) v( x 0 ) u( x 0 ) + ɛ x 0 2 max u( x) + ɛl x D where l max x D x 2 <. Letting ɛ 0, we have and so 5) Similar proof yield u(x, y) max x D u( x) max u( x) max u( x) x D x D min u( x) min u( x) x D x D 3. Application of Maximum Principle: Uniqueness and Stability With Maximum Principle, it will be easy to prove the uniqueness and stability of the solutions for the Poisson equations. Theorem 2 (Uniqueness). Considering the boundary value problem (BVP) of the Poisson equation posed in a connected (regular) bounded open set D 2 { u F ind (3.) u f on D there exists at most one solution (uniqueness). Proof. Let u, v be any two solutions to (3.). Let w u v, then by linearity of the problem, { w u v F F 0 ind w u v f f 0 Applying Maximum Principle to w, we see that on D 0 min w(x) min w(x) max w(x) max w(x) 0, D D D D that is w 0 and so u v.

3 PDE (0.47S) 3 emark. A similar proof will yield the stability of the problem with respect of the boundary conditions. That is, if v is a solution for { v F ind then v g on D u v max f g. D 4. Poisson s formula and strong maximum principle In fact, as we will see, we can have even more strong form of the maximum principle. ecall that when we have solved the Laplace equation on disk, by using polar coordinates, u 0 r 2 x 2 + y 2 < 2 u(, θ) f(θ) (4.) u(0, θ) < singularbc u(, ) u(, π), θ u(, ) θ u(, π) periodicbc we find the solutions are of the form (by MSV or eigenfunction expansion) (4.2) u(r, θ) a 0 + a n r n cos(nθ) + b n r n sin(nθ) with a 0 π n f(θ)dθ a n π n π π n f(θ) cos(nθ)dθ, n b n π n f(θ) sin(nθ)dθ n. It turns out that we can simplify the formula (4.2). Theorem 3 (Poisson s formula). The solution of (4.) is given by (4.3) u(r, θ) 2 r 2 π f(α) 2 2r cos(θ α) + r 2 dα. Proof. By the formulas of the coefficients and (4.2), we see that u(r, θ) π π f(θ)dθ + ( π n f(α) cos(nα)dα)r n cos(nθ) n π + ( π n f(α) sin(nα)dα)r n sin(nθ) n π r n π f(θ)dθ + π n f(α)[cos(nα) cos(nθ) + sin(nα) sin(nθ)]dα n [ ] π 2r n f(α) + cos(n(α θ)) dα. πn n Question: what is + n 2rn π n cos(n(α θ))? ecall again the Euler s identity e ix cos x + i sin x, 2 cos nα e inα + e inα,

4 4 CHENGBO WANG thus + n 2r n π n cos(nα) + ( r ) n (e inα + e inα ) n ( r eiα) n ( r + e iα) n n0 n0 r + eiα r e iα 2 2 r cos(α) r eiα r cos(α) 2 r cos(α) + ( ) r 2 ) 2 ( r 2 r cos(α) + ( ) r 2 2 r 2 2 2r cos(α) + r 2 Substituting this identity to the previous identity gives us (4.3). With Poisson s formula, it will be easy to obtain a remarkable property of the solution for the Laplace equation. Corollary 4 (mean value property). Let u 0 in an open disk with radius, B, and u be continuous function in B B, then the value of u at the center is the same as the average value of u on the circle B. Proof. Without loss of generality, assume B center at the origin. Then an application of (4.3) with r 0 gives us u(0) π u(, θ)dθ. As an application of the mean value property, it will be easy to prove a stronger form of the maximal principle. Theorem 5 (Strong maximum principle). Let D be a connected (regular) bounded open set in 2. Assuming u u(x, y) is a solution of the Laplace equation u 0 in D and continuous on D D D. Then there is no maximum or minimum points of u in D, unless u is a constant. Proof. The proof is straightforward application of the mean value property. By contradiction, if there is maximum points of u in D at, say (x 0, y 0 ) u(x 0, y 0 ) max u(x, y) M. D Then there must exist a 0 > 0 such that B 2 (x 0, y 0 ) D for any < 0. Notice that u(x 0, y 0 ) average of u on B (x 0, y 0 ) M

5 PDE (0.47S) 5 with the equality if and only if u M for any (x, y) B (x 0, y 0 ). This forces u M for any (x, y) B 0 (x 0, y 0 ). By connectivity of D, we see that u(x, y) M for any (x, y) D, which completes the proof. In fact, for the mean value property, we can have another simple proof, without using the Poisson s formula. Another simple proof of the mean value property. Let Then φ(r) : φ (r) π u( y)ds( y) π u((x, y) + r(cos α, sin α))dα. (cos α, sin α) ( u)((x, y) + r(cos α, sin α))dα B( x,r) Hence φ is a constant, and so y x r u n ds( y) u( y)d y 0. φ(r) lim φ(t) lim t 0 t 0 t which is the mean value property. ( u)( y)ds( y) B( x,t) u( y)ds( y) u( x), Department of Mathematics, Johns Hopkins University, Baltimore, Maryland 228 address: cwang@math.jhu.edu