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TAMU MATH 412 - Lecture 3-1web

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Math 412-501Theory of Partial Differential EquationsLecture 3-1:Heat equation in an arbitrary domain.Spectrum of Laplace’s operator.Heat conduction in an arbitrary domainInitial-boundary value problem:∂u∂t= k∂2u∂x2+∂2u∂y2, (x, y) ∈ D,u(x, y , 0) = f (x, y), (x, y ) ∈ D,Boundary condition: u|∂D= 0,i.e., u(x, y , t) = 0 for (x, y ) ∈ ∂D.(Dirichlet condition)Alternative boundary condition:∂u∂n∂D= 0,where∂u∂n= ∇u · n is the normal derivative.(Neumann condition)Mixed boundary condition:∂D = γ1⊔ γ2(disjoint union),u|γ1= 0,∂u∂nγ2= 0.Boundary condition of the third kind:∂u∂n+ αu∂D= 0, where α is a function on ∂D.We search for the solution u(x, y , t) as asuperposition of solutions with separated variablesthat satisfy the boundary conditions.For a general domain, we can only separate the timevariable from the others.Separation of variables: u(x, y , t) = φ(x, y)G (t).Substitute this into the heat equation:φ(x, y)dGdt= k∂2φ∂x2+∂2φ∂y2G (t).Divide both sides by k · φ(x, y )G (t) = k · u(x, y , t):1kG·dGdt=1φ·∂2φ∂x2+∂2φ∂y2.It follows that1kG·dGdt=1φ·∂2φ∂x2+∂2φ∂y2= −λ,where λ is a separation constant.The time variable has been separated:dGdt= −λkG , ∇2φ = −λφ.Proposition Suppose G and φ are solutions of theabove differential equations for the same value of λ.Then u(x, y , t) = φ(x, y)G (t) is a solution of theheat equation.Boundary condition u|∂D= 0 holds if φ|∂D= 0.Boundary condition∂u∂n∂D= 0 holds if∂φ∂n∂D= 0.Eigenvalue problem:∇2φ = −λφ, φ|∂D= 0.(Dirichlet Laplacian)Alternative eigenvalue problem:∇2φ = −λφ,∂φ∂n∂D= 0.(Neumann Laplacian)We assume that there are eigenvalues λ1, λ2, . . .and corresponding eigenfunctionsφ1(x, y), φ2(x, y), . . . .Dependence on t:G′(t) = −λkG (t) =⇒ G(t) = C0e−λktSolution of the boundary value problem:u(x, y , t) = e−λnktφn(x, y).We are looking for the solution of theinitial-boundary value problem as a superposition ofsolutions with separated variables.u(x, y , t) =X∞n=1cne−λnktφn(x, y)How do we find coefficients cn?Substitute the series into the initial conditionu(x, y , 0) = f (x, y).f (x, y) =X∞n=1cnφn(x, y)How do we solve the heat conduction problem?∂u∂t= k∂2u∂x2+∂2u∂y2, (x, y) ∈ D,u(x, y , 0) = f (x, y), (x, y ) ∈ D,u|∂D= 0.• Expand f into eigenfunctions of the DirichletLaplacian:f (x, y) =X∞n=1cnφn(x, y).• Write the solution:u(x, y , t) =X∞n=1cne−λnktφn(x, y).Spectrum of the LaplacianEigenvalue problem:∇2φ + λφ = 0 in D,αφ + β∂φ∂n∂D= 0,where α, β are piecewise continuous functions on∂D such that |α| + |β| 6= 0 everywhere on ∂D.We assume that ∂D is piecewise smooth.The PDE is called the Helmholtz equation.Boundary condition covers all cases considered.The eigenvalue problem is the many-dimensionalanalog of the Sturm-Liouville eigenvalue problem.The eigenvalues of the problem are eigenvalues ofthe negative Laplacian −∇2.The set of eigenvalues of an operator is called itsspectrum. Properties of eigenvalues andeigenfunctions are called spectral properties.The Laplacian has six important spectral properties.Property 1. All eigenvalues are real.Property 2. All eigenvalues can be arranged inthe ascending orderλ1< λ2< . . . < λn< λn+1< . . .so that λn→ ∞ as n → ∞.This means that:• there are infinitely many eigenvalues;• there is a smallest eigenvalue;• on any finite interval, there are only finitelymany eigenvalues.Remark. For the Dirichlet Laplacian, λ1> 0.For the Neumann Laplacian, λ1= 0.The set of eigenfunctions corresponding to aparticular eigenvalue λ together with zero functionform a linear space. The dimension of this space iscalled the multiplicity of λ.An eigenvalue is simple if it is of multiplicity 1.Then the eigenfunction is unique up tomultiplication by a scalar.Otherwise the eigenvalue is called multiple.Property 3. An eigenvalue λnmay be multiplebut its multiplicity is finite.Moreover, the smallest eigenvalue λ1is simple,and the corresponding eigenfunction φ1has no zerosinside the domain D.Property 4. Eigenfunctions corresponding todifferent eigenvalues are orthogonal relative to theinner producthf , gi =ZZDf (x, y)g(x, y) dx dy .That is, hφ, ψi = 0 whenever φ and ψ areeigenfunctions corresponding to differenteigenvalues.Property 5. Any eigenfunction φ can be relatedto its eigenvalue λ through the Rayleigh quotient:λ =−I∂Dφ∂φ∂nds +ZZD|∇φ|2dx dyZZD|φ|2dx dy.Property 6. There exists a sequence φ1, φ2, . . . ofpairwise orthogonal eigenfunctions that is completein the Hilbert space L2(D).Any square-integrable function f ∈ L2(D) isexpanded into a seriesf (x, y) =X∞n=1cnφn(x, y),that converges in the mean. The series is unique:cn=hf , φnihφn, φni.If f is piecewise smooth then the series convergespointwise to f at points of continuity.Example.∇2φ = −λφ in D = {(x, y ) | 0 < x < L, 0 < y < H},φ(0, y) = φ(L, y ) = 0, φ(x, 0) = φ(x, H) = 0.This problem can be solved by separation of variables.Eigenfunctions φnm(x, y) = sinnπxLsinmπyH, n, m ≥ 1.Corresponding eigenvalues: λnm= (nπL)2+ (mπH)2.Thus the double Fourier sine series is the expansionin eigenfunctions of the Dirichlet Laplacian in arectangle.Similarly, the double Fourier cosine series is theexpansion in eigenfunctions of the NeumannLaplacian in a


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