Math 412-501Theory of Partial Differential EquationsLecture 2-11: Review for Exam 2.Heat conduction in a rectangleInitial-boundary value problem:∂u∂t= k∂2u∂x2+∂2u∂y2(0 < x < L, 0 < y < H),u(x, y, 0) = f (x, y ) (0 < x < L, 0 < y < H),∂u∂x(0, y, t) =∂u∂xu(L, y, t) = 0 (0 < y < H),∂u∂y(x, 0, t) =∂u∂yu(x, H, t) = 0 (0 < x < L).We search for the solution u(x, y, t) as asuperposition of solutions with separated variablesthat satisfy the boundary conditions.Separation of variables: u(x, y, t) = φ(x)h(y)G (t).Substitute this into the heat equation:φ(x)h(y)dGdt= kd2φdx2h(y)G (t) + φ(x)d2hdy2G (t).Divide both sides by k · φ(x)h(y)G (t) = k · u(x, y, t):1kG·dGdt=1φ·d2φdx2+1h·d2hdy2.It follows that1φ·d2φdx2= −λ,1h·d2hdy2= −µ,1kG·dGdt= −λ −µ, where λ and µ are separationconstants.The variables have been sep arated:dGdt= −(λ + µ)kG ,d2φdx2= −λφ,d2hdy2= −µh.Proposition Suppose G , φ, and h are solutions ofthe above ODEs for the same values of λ and µ.Then u(x, y, t) = φ(x)h(y)G (t) is a solution of theheat equation.Boundary conditions∂u∂x(0, y, t) =∂u∂x(L, y, t) = 0hold if φ′(0) = φ′(L) = 0.Boundary conditions∂u∂y(x, 0, t) =∂u∂y(x, H, t) = 0hold if h′(0) = h′(H) = 0.1st eigenvalue problem: φ′′= −λφ, φ′(0) = φ′(L) = 0.Eigenvalues: λn= (nπL)2, n = 0, 1, 2, . . .Eigenfunctions: φ0= 1, φn(x) = cosnπxL, n ≥ 1.2nd eigenvalue problem: h′′= −µh, h′(0) = h′(H) = 0.Eigenvalues: µm= (mπH)2, m = 0, 1, 2, . . .Eigenfunctions: h0= 1, hm(y) = cosmπyH, m ≥ 1.Dependence on t:G′(t) = −(λ + µ)kG (t) =⇒ G (t) = C0e−(λ+µ)ktSolution of the boundary value problem:u(x, y, t) = e−(λn+µm)ktφn(x)hm(y)= exp−((nπL)2+ (mπH)2)kt· cosnπxL· cosmπyH.We are looking for the solution of theinitial-boundary value problem as a superposition ofsolutions with separated variables.u(x, y, t) =X∞n=0X∞m=0Cn,me−(λn+µm)ktφn(x)hm(y)=∞Xn=0∞Xm=0Cn,mexp−((nπL)2+ (mπH)2)kt· cosnπxL· cosmπyHHow do we find coefficients Cn,m?From the initial condit ion u(x, y , 0) = f (x, y).f (x, y) =X∞n=0X∞m=0Cn,mcosnπxLcosmπyH(double Fourier cosine series)How do we solve the heat conduction problem?∂u∂t= k∂2u∂x2+∂2u∂y2(0 < x < L, 0 < y < H),u(x, y, 0) = f (x, y ) (0 < x < L, 0 < y < H),∂u∂x(0, y, t) =∂u∂x(L, y, t) = 0 (0 < y < H),∂u∂y(x, 0, t) =∂u∂y(x, H, t) = 0 (0 < x < L).• Expand f into the double Fourier cosine series:f (x, y) =X∞n=0X∞m=0Cn,mcosnπxLcosmπyH.• Write the solution: u(x, y, t) ==∞Xn=0∞Xm=0Cn,mexp−((nπL)2+ (mπH)2)ktcosnπxLcosmπyH.How do we expand f into the double Fourier cosineseries?f (x, y) =X∞n=0X∞m=0Cn,mcosnπxLcosmπyH.If f (x, y) is smooth then the expansion exists.How do we determine coefficients?Multiply both sides by φN(x)hM(y) and integrateover t he rectangle.ZL0ZH0f (x, y)φN(x)hM(y) dx dy=∞Xn=0∞Xm=0Cn,mZL0ZH0φN(x)hM(y)φn(x)hm(y) dx dy=∞Xn=0∞Xm=0Cn,mZL0φN(x)φn(x) dxZH0hM(y)hm(y) dy.We have the orthogonality relations:ZL0φN(x)φn(x) dx = 0, n 6= N,ZH0hM(y)hm(y) dy = 0, m 6= M.HenceZL0ZH0f (x, y)φN(x)hM(y) dx dy= CN,MZL0φ2N(x) dxZH0h2M(y) dy.It remains to recall thatZL0φ20(x) dx = L,ZL0φ2N(x) dx =L2, N ≥ 1,ZH0h20(x) dx = H,ZH0h2M(x) dx =H2, M ≥ 1.How do we expand f (x, y ) into the double series?f (x, y) =X∞n=0X∞m=0Cn,mcosnπxLcosmπyH,whereC0,0=1LHZL0ZH0f (x, y) dx dy,Cn,0=2LHZL0ZH0f (x, y) cosnπxLdx dy, n ≥ 1,C0,m=2LHZL0ZH0f (x, y) cosmπyHdx dy, m ≥ 1,Cn,m=4LHZL0ZH0f (x, y) cosnπxLcosmπyHdx dy.Laplace’s equation in a semicircleSolve Laplace ’s equation inside a semicircle ofradius a (0 < r < a, 0 < θ < π) subject to theboundary c onditions: u = 0 on the diameter andu(a, θ) = g(θ).Laplace’s e quation in polar coordinates (r , θ):∂2u∂r2+1r∂u∂r+1r2∂2u∂θ2= 0.u = 0 on the diameter =⇒u(r, 0) = u(r, π) = 0 (0 < r < a),u(0, θ) = 0 (0 < θ < π).Separation of variables: u(r , θ) = h(r )φ(θ).Substitute this into Laplace ’s equation:d2hdr2φ(θ) +1rdhdrφ(θ) +1r2h(r)d2φdθ2= 0.Divide both sides by r−2h(r)φ(θ) = r−2u(r, θ):1h·r2d2hdr2+ rdhdr= −1φ·d2φdθ2.It follows that1h·r2d2hdr2+ rdhdr= −1φ·d2φdθ2= λ = const.The variables have been sep arated:r2d2hdr2+ rdhdr= λh,d2φdθ2= −λφ.Proposition Suppose h and φ are solutions of theabove ODEs for th e same value of λ. Thenu(r, θ) = h(r )φ(θ) is a solution of Laplace’sequation.Boundary conditions u(r, 0) = u(r, π) = 0 hold ifφ(0) = φ(π) = 0.Boundary condition u(0, θ) = 0 holds ifh(0) = 0.Euler’s (or equidimensional) equationr2d2hdr2+ rdhdr− λh = 0 (r > 0)λ > 0 =⇒ h(r) = C1rp+ C2r−p(λ = p2, p > 0)λ = 0 =⇒ h(r) = C1+ C2log rEigenvalue problem: φ′′= −λφ, φ(0) = φ(π) = 0.Eigenvalues: λn= n2, n = 1, 2, . . .Eigenfunctions: φn(θ) = sin nθ.Dependence on r:r2h′′+ rh′= λh, h(0) = 0.=⇒ h(r) = C0rp(p =√λ)Solution of Laplace’s eq uation:u(r, θ) = rnsin nθ, n = 1, 2, . . .A superposition of the solutions with separatedvariables is a seriesu(r, θ) =X∞n=1cnrnsin nθ,where c1, c2, . . . are constants.Substituting the series into the boundary condit ionu(a, θ) = g(θ), we getg(θ) =X∞n=1cnansin nθ.Hence cn= bna−n, n = 1, 2, . . . , whereg(θ) =X∞n=1bnsin nθ.Solution.u(r, θ) =X∞n=1bnransin nθ,whereX∞n=1bnsin nθis t he Fourier sine series of the function g (θ) on[0, π], that is,bn=2πZπ0g(θ) sin nθ dθ, n = 1, 2, . .