Terminology — Question 9

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Question 9

A problem is well posed in specified function spaces if solutions exist, are unique, and depend continuously on the prescribed data. To examine the third requirement, consider the Cauchy data problem uxx+uyy=0,0<x<π,0<y≤1;u(x,0)=f(x),uy(x,0)=g(x),u_{xx}+u_{yy}=0,\quad 0<x<\pi,\ 0<y\le 1;\qquad u(x,0)=f(x),\quad u_y(x,0)=g(x), with u(0,y)=u(π,y)=0u(0,y)=u(\pi,y)=0. Measure the bottom data by ∥f∥∞+∥g∥∞\|f\|_\infty+\|g\|_\infty on [0,π][0,\pi] and the observed solution by ∥u(⋅,1)∥∞\|u(\cdot,1)\|_\infty. For integers n≥1n\ge 1, set un(x,y)=e−nnsin⁡(nx)sinh⁡(ny).u_n(x,y)=\frac{e^{-n}}{n}\sin(nx)\sinh(ny).

Tasks

  1. Verify the PDE and side conditions and find the exact bottom data fn,gnf_n,g_n.

  2. Compute the bottom data norm and top solution norm. Determine their limits as n→∞n\to\infty.

  3. Prove that the amplification ratio is unbounded and explain why the calculation rules out a continuous solution map at zero in the stated norms if the problem is uniquely solvable on a linear class containing these fields.

  4. Explain why convergence of un(⋅,1)u_n(\cdot,1) to zero does not rescue continuity. Rescale the fields to obtain data converging to zero while the top norm stays exactly one, and sketch the vertical amplitude profiles for three nn.

Original worksheet page 1: question and worked solution for 9-3-009
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Question 9 – Solution

Strategy. Small absolute outputs are not enough for stability; compare their size with the much smaller input, then normalize.

Step 1: Verify the supplied solutions. Two xx derivatives multiply unu_n by −n2-n^2; two yy derivatives multiply it by n2n^2. Their sum is zero, and sin⁡(nx)\sin(nx) vanishes on both sides. At the bottom, fn=0,gn=e−nsin⁡(nx)\boxed{f_n=0,\quad g_n=e^{-n}\sin(nx)}. These are smooth fields with compatible side and bottom values.

Step 2: Measure the input and output. For each integer nn, sup⁡[0,π]|sin⁡(nx)|=1\sup_{[0,\pi]}|\sin(nx)|=1. Hence the bottom norm is e−ne^{-n} and the top norm is Mn=e−nnsinh⁡n=1−e−2n2n.M_n=\frac{e^{-n}}{n}\sinh n=\frac{1-e^{-2n}}{2n}. Both tend to zero. Thus this sequence by itself does not contradict continuity; the relative amplification is the information needed for the next step.

Step 3: Expose unbounded amplification. The ratio is Mn/e−n=sinh⁡n/n→∞M_n/e^{-n}=\sinh n/n\to\infty. If uniqueness holds in a linear solution class containing these fields, solving the homogeneous linear PDE with data is a linear map. A continuous linear map between the stated normed spaces must obey a uniform bound ∥u(⋅,1)∥∞≤C(∥f∥∞+∥g∥∞)\|u(\cdot,1)\|_\infty\le C(\|f\|_\infty+ \|g\|_\infty) on its domain. The unbounded ratios rule out that bound. No existence or uniqueness theorem for arbitrary Cauchy data has been assumed.

Step 4: Give a direct discontinuity sequence. Set wn=un/Mn=sin⁡(nx)sinh⁡(ny)/sinh⁡nw_n=u_n/M_n=\sin(nx)\sinh(ny)/\sinh n. Its top norm is one, whereas its bottom norm is n/sinh⁡n→0n/\sinh n\to 0. The zero data have the zero solution; uniqueness would force the solution map to select these wnw_n, contradicting continuity at zero. The vertical envelopes are sinh⁡(ny)/sinh⁡n\sinh(ny)/\sinh n, not solution traces at one fixed xx for every nn.

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Original worksheet page 2: question and worked solution for 9-3-009

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