Laplace's Equation — Question 9

PDF ↗

Question 9

Let uu be regular and harmonic on the unit disk. Its boundary temperature is known to have the form g(θ)=A+Bcos⁡(Nθ),N≥1,g(\theta)=A+B\cos(N\theta),\qquad N\ge 1, where NN is known and A,BA,B are unknown. The temperature is measured on the circle r=ρr=\rho, where 0<ρ<10<\rho<1.

Tasks

  1. Derive exact recovery formulas for A,BA,B from the measured angular mean and cosine coefficient. Specify the consistency conditions on the other measured Fourier coefficients.

  2. For ρ=1/2\rho=1/2, N=6N=6, and measured trace 2+0.03cos⁡(6θ)2+0.03\cos(6\theta), recover the boundary data and the full field.

  3. If the measured NNth cosine coefficient has error at most η\eta, find the sharp error bound for the recovered BB. Derive the largest allowable mode number when the desired per-coefficient error is at most ε>η\varepsilon>\eta. Evaluate it for ρ=1/2,η=10−4,ε=10−2\rho=1/2,\eta=10^{-4},\varepsilon=10^{-2}.

  4. Contrast forward boundary stability with inverse instability when arbitrarily high modes are allowed. Give an explicit sequence showing why small interior-circle errors need not imply small boundary errors.

Original worksheet page 1: question and worked solution for 9-7-009
Show solutionHide solution

Question 9 – Solution

Strategy. Harmonic continuation damps a mode by ρN\rho^N; inversion must divide by exactly that small factor.

Step 1: Recover the two parameters. The regular harmonic extension is u=A+BrNcos⁡(Nθ)u=A+Br^N\cos(N\theta). For the measured trace h(θ)=u(ρ,θ)h(\theta)=u(\rho,\theta), define m=12π∫02πh(θ)dθ,cN=1π∫02πh(θ)cos⁡(Nθ)dθ.m=\frac 1{2\pi}\int_0^{2\pi}h(\theta)\,d\theta,\qquad c_N=\frac 1\pi\int_0^{2\pi}h(\theta)\cos(N\theta)\,d\theta. Orthogonality gives A=m,B=ρ−NcN.\boxed{A=m,\qquad B=\rho^{-N}c_N.} Consistency with the stated model requires every sine coefficient and every nonconstant cosine coefficient except cNc_N to vanish. Since ρ>0\rho>0, the two parameters are unique.

Step 2: Reconstruct the given field. Here ρN=1/64\rho^N=1/64, m=2m=2, and c6=0.03=3/100c_6=0.03=3/100, so B=48/25=1.92B=48/25=1.92. Thus g(θ)=2+4825cos⁡(6θ),u(r,θ)=2+4825r6cos⁡(6θ).\boxed{g(\theta)=2+\frac{48}{25}\cos(6\theta),\qquad u(r,\theta)=2+\frac{48}{25}r^6\cos(6\theta).} Substitution at r=1/2r=1/2 reproduces the measured coefficient 3/1003/100. The polar mode is a Cartesian harmonic polynomial and is regular at the center.

Step 3: Quantify inverse amplification. A coefficient error ee, |e|≤η|e|\le\eta, produces error ρ−Ne\rho^{-N}e in BB. Consequently |δB|≤ηρ−N|\delta B|\le\eta\rho^{-N}, with equality when |e|=η|e|=\eta. The desired bound is met exactly when N≤log⁡(ε/η)log⁡(1/ρ),Nmax=⌊log⁡(ε/η)log⁡(1/ρ)⌋.N\le\frac{\log(\varepsilon/\eta)}{\log(1/\rho)},\qquad \boxed{N_{\max}=\left\lfloor\frac{\log(\varepsilon/\eta)} {\log(1/\rho)}\right\rfloor.} If this integer is zero, no positive mode qualifies. The given numbers give Nmax=6N_{\max}=6: mode six has error bound 0.00640.0064, whereas mode seven has 0.01280.0128. This is a per-coefficient criterion, not a bound on a sum of many mode errors.

Step 4: Separate forward and inverse stability. For continuous boundary perturbations, the maximum principle gives ∥u1−u2∥∞,disk≤∥g1−g2∥∞,∂disk\|u_1-u_2\|_{\infty,{\mathrm{disk}}}\le\|g_1-g_2\|_{\infty,\partial{\mathrm{disk}}}. In contrast, take gn(θ)=cos⁡(nθ)g_n(\theta)=\cos(n\theta) and un=rncos⁡(nθ)u_n=r^n\cos(n\theta). Then ∥gn∥∞=1\|g_n\|_\infty=1 but ∥un(ρ,⋅)∥∞=ρn→0\|u_n(\rho,\cdot)\|_\infty=\rho^n\to 0. Thus no fixed bound can control boundary sup-norm errors by measured-circle sup-norm errors across all modes, although each exact finite-mode inversion is unique.

Original worksheet page 2: question and worked solution for 9-7-009

Original worksheet layout. Use Enlarge or open the PDF for a closer view.