Math 412-501Theory of Partial Differential EquationsLecture 4-5:Uniqueness of solutions of PDEs.The maximum principle.Uniqueness of solutions of PDEsPrincipal idea: under some natural, non-restrictiveconditions the initial/boundary value problems forthe heat, wave, and Laplace’s equations have uniquesolutions.Theorem The initial-boundary value problem forthe heat equation∂u∂t= k∂2u∂x2+ Q(x, t) (0 < x < L, 0 < t < T ),u(x, 0) = f (x) (0 < x < L),u(0, t) = A(t), u(L, t) = B(t) (0 < t < T )has at most one solution that is twice differentiableon [0, L] × [0, T ].Proof: Suppose u1and u2are two solutions.Let w = u1− u2. Then∂w∂t= k∂2w∂x2(0 < x < L, 0 < t < T ),w(x, 0) = 0, w(0, t) = w(L, t) = 0.Let E (t) =RL0|w(x, t)|2dx. ThenE′(t) = 2ZL0w∂w∂tdx = 2kZL0w∂2w∂x2dx= 2kw∂w∂xLx=0−2kZL0∂w∂x2dx ≤ 0.Since E (t) ≥ 0 and E (0) = 0, it follows thatE = 0 =⇒ w = 0 =⇒ u1= u2.The maximum principleTheorem Let D ⊂ R2be a bounded domain andu :D → R be a continuous function on the closureD = D ∪ ∂D.If u is twice differentiable in D and ∇2u ≥ 0 thenmaxx∈Du(x) = maxx∈∂Du(x).(the maximum is attained on the boundary)Corollary 1 If ∇2u = 0 then the maximum andthe minimum of u inD are both attained on theboundary ∂D.Proof: Since ∇2u = 0, the theorem applies to bothu and −u.Corollary 2 If ∇2u = 0 and u = 0 on theboundary ∂D, then u = 0 in D as well.Corollary 3 Given functions Q : D → R andf : ∂D → R, the boundary value problem∇2u = Q in the domain D,u = f on the boundary ∂Dhas at most one solution that is twice differentiablein D and continuous on the closure D.Proof: Suppose u1and u2are two solutions. Letw = u1− u2. Then ∇2w = 0 in D and w = 0 on∂D. By Corollary 2, w = 0 in D, i.e., u1= u2.Proof of the maximum principleLemma 1 Let f : (a, b) → R be a twicedifferentiable function. If f has a local maximum ata point c ∈ (a, b), then f′(c) = 0, f′′(c) ≤ 0.Lemma 2 Let u be a twice differentiable functionon the domain D. If ∇2u > 0 then u has no localmaximum in D.Proof: Suppose that u has a local maximum atsome point (x0, y0) ∈ D. Then the functionf (x) = u(x, y0) has a local maximum at x0whileg(y) = u(x0, y ) has a local maximum at y0.By Lemma 1, f′′(x0) ≤ 0, g′′(y0) ≤ 0. Then∇2u(x0, y0) = f′′(x0) + g′′(y0) ≤ 0, a contradiction.Proof of Theorem: Suppose u is continuous on Dand ∇2u ≥ 0 in D. Let w(x, y) = x2+ y2.Then ∇2w = 4. For any ε > 0 let uε= u + εw.Then ∇2uε= ∇2u + 4ε > 0 in D.By Lemma 2, uεhas no local maximum in D. HencesupDuε≤ max∂Duε.But max∂Duε≤ max∂Du + ε max∂Dw andsupDu ≤ supDuε. ThereforesupDu ≤ max∂Du + ε max∂Dw.Since ε can be chosen arbitrarily small, we havesupDu ≤ max∂Du.Mean value theoremThe value of a harmonic function at any point P isthe average of its values along any circle centered at P.∇2u = 0 in D =⇒ u(P) =12πr0IC (P ,r0)={x:|x−P|=r0}u(x) dsProof: Introduce the polar coordinates r, θ withthe origin at P. Let f (θ) = u(r0, θ), −π < θ ≤ π.Then u is the solution of the boundary value problem∇2u = 0 (0 ≤ r < r0),u(r0, θ) = f (θ).Solution:u(r, θ) = a0+X∞n=1rr0n(ancos nθ + bnsin nθ),where a0+P∞n=1(ancos nθ + bnsin nθ) is theFourier series of f (θ).Now u(P) = u(0, θ) = a0=12πZπ−πf (θ)
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