Separation of Variables — Question 7

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Question 7

On 1<x<e1<x<e, t≥0t\ge 0, consider the variable-coefficient equation 1xut=(xux)x,u(1,t)=u(e,t)=0.\frac 1x u_t=(xu_x)_x,\qquad u(1,t)=u(e,t)=0. For a nonzero product u=X(x)T(t)u=X(x)T(t), use the separation convention (xX′)′+λX/x=0(xX')'+\lambda X/x=0, T′=−λTT'=-\lambda T. Introduce s=log⁡xs=\log x and Z(s)=X(es)Z(s)=X(e^s).

Tasks

  1. Derive the factor equations from the PDE and verify, by the chain rule, the transformed spatial equation and its endpoint conditions in ss.

  2. Determine every eigenvalue and spatial factor and verify the resulting separated fields in the original xx equation.

  3. Derive an integral identity that proves eigenvalue positivity. Identify the weight in the squared spatial norm; explain why an unweighted denominator would not follow from this equation.

  4. Prove weighted orthogonality of distinct spatial modes by subtracting their ODE identities. Evaluate their squared weighted norms and locate the interior zeros of the third mode in the original coordinate.

Original worksheet page 1: question and worked solution for 9-4-007
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Question 7 – Solution

Strategy. A change of spatial coordinate can simplify separation, but the original equation determines the correct integration weight.

Step 1: Transform the spatial equation carefully. Substitution gives XT′/x=(xX′)′TXT'/x=(xX')'T. Nonzero anchors yield (xX′)′=−λX/x(xX')'=-\lambda X/x and T′=−λTT'=-\lambda T. Since X′=Zs/xX'=Z_s/x, one has xX′=ZsxX'=Z_s and (xX′)′=Zss/x(xX')'=Z_{ss}/x. Multiplying the spatial equation by xx gives Zss+λZ=0,Z(0)=Z(1)=0.\boxed{Z_{ss}+\lambda Z=0,\qquad Z(0)=Z(1)=0.} The physical endpoints 1,e1,e become 0,10,1, not an interval of length e−1e-1.

Step 2: Recover the modes in the original variable. The transformed Dirichlet problem has λn=n2π2\lambda_n=n^2\pi^2 and Zn(s)=sin⁡(nπs)Z_n(s)=\sin(n\pi s) for n≥1n\ge 1. Thus un(x,t)=Csin⁡(nπlog⁡x)e−n2π2t.\boxed{u_n(x,t)=C\sin(n\pi\log x)e^{-n^2\pi^2t}.} Indeed xXn′=nπcos⁡(nπlog⁡x)xX_n'=n\pi\cos(n\pi\log x) and (xXn′)′=−n2π2Xn/x(xX_n')'=-n^2\pi^2X_n/x. Multiplication by the time factor gives exactly ut/xu_t/x. At x=1,ex=1,e the sine values vanish. Zero and negative eigenvalues produce only the trivial Dirichlet factor.

Step 3: Derive the weighted energy quotient. Multiplication of −(xX′)′=λX/x-(xX')'=\lambda X/x by XX and integration give λ∫1eX2xdx=∫1ex(X′)2dx.\boxed{\lambda\int_1^e\frac{X^2}{x}\,dx=\int_1^e x(X')^2\,dx.} The endpoint term vanishes. For a nonzero factor the denominator is positive; zero numerator would force a constant and then the zero field. Hence λ>0\lambda>0. The weight 1/x1/x comes from the coefficient of utu_t; dropping it would change the identity and the associated eigenvalue quotient.

Step 4: Verify orthogonality, norms and nodes. For distinct modes m,nm,n, subtracting the two integrated spatial identities cancels the derivative terms and yields (λn−λm)∫1eXnXmdx/x=0(\lambda_n-\lambda_m)\int_1^e X_nX_m\,dx/x=0. Thus their weighted inner product vanishes. With s=log⁡xs=\log x, ∫1eXn2xdx=∫01sin⁡2(nπs)ds=12.\int_1^e\frac{X_n^2}{x}\,dx=\int_0^1\sin^2(n\pi s)\,ds=\frac 12. For n=3n=3, the interior zeros occur at s=1/3,2/3s=1/3,2/3, hence at x=e1/3,e2/3x=e^{1/3},e^{2/3}. Equal spacing in ss is not equal spacing in xx.

Original worksheet page 2: question and worked solution for 9-4-007

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