Solving the Heat Equation: Question 9-5.7 & Solution

Question 7

For the zero-endpoint heat problem $u_t=u_{xx}$ on $(0,\pi)$ , let $f\in L^2(0,\pi)$ with $\|f\|_2\le M$ , where $M>0$ . Its solution for $t>0$ is $u=\sum_{n\ge 1}b_ne^{-n^2t}\sin(nx),\qquad u_N=\sum_{n=1}^N b_ne^{-n^2t}\sin(nx),\quad N\ge 1.$ You may use Parseval: $\sum b_n^2=2\|f\|_2^2/\pi$ , and Cauchy–Schwarz.

Tasks

Prove $\|u-u_N\|_2\le M e^{-(N+1)^2t}$ . Show that the bound is sharp over the stated data class for each $N,t$ .
Derive a uniform-in- $x$ error bound using Cauchy–Schwarz and the tail sum $\sum_{n>N}e^{-2n^2t}$ .
Bound that sum by a geometric series using $n=N+1+j$ , $j\ge 0$ , and obtain an explicit bound requiring no infinite summation.
For $M=1,t\ge 0.1$ , certify that six modes give a uniform error below $0.01$ . Explain why the available bound does not certify five modes, and why this does not prove that five modes fail for every particular initial field.

Original worksheet page 1: question and worked solution for 9-5-007

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Question 7 – Solution

Strategy. A truncation certificate needs a bound on the omitted coefficients, not just a graph of the retained modes.

Step 1: Prove the sharp norm estimate. Orthogonality gives $\|u-u_N\|_2^2=\frac\pi 2\sum_{n>N}b_n^2e^{-2n^2t} \le e^{-2(N+1)^2t}\|f\|_2^2.$ Taking square roots proves the bound. Equality is attained by $f=M\sqrt{2/\pi}\sin((N+1)x)$ : all its energy lies in the first omitted mode. Thus the rate and constant are sharp for the whole $L^2$ ball.

Step 2: Control pointwise errors uniformly. Using $|\sin nx|\le 1$ and Cauchy–Schwarz, $|u-u_N|\le\left(\sum_{n>N}b_n^2\right)^{1/2} \left(\sum_{n>N}e^{-2n^2t}\right)^{1/2} \le\sqrt{\frac 2\pi}M\left(\sum_{n>N}e^{-2n^2t}\right)^{1/2}.$ This bound is independent of $x$ and finite for $t>0$ . It also justifies uniform convergence of the positive-time solution for arbitrary $L^2$ data.

Step 3: Replace the tail by a geometric bound. For the integer $j\ge 0$ , $(N+1+j)^2\ge(N+1)^2+(2N+3)j$ , since $j^2\ge j$ . Therefore $\sum_{n>N}e^{-2n^2t}\le \frac{e^{-2(N+1)^2t}}{1-e^{-2(2N+3)t}},\qquad \boxed{\|u-u_N\|_\infty\le \sqrt{\frac 2\pi}\,\frac{M e^{-(N+1)^2t}}{\sqrt{1-e^{-2(2N+3)t}}}.}$ The denominator is positive because $t>0$ . Each factor makes this certificate decrease as time increases.

Step 4: Make a numerical guarantee with the correct scope. At $t=0.1$ , the bound is less than $0.00611$ for $N=6$ and greater than $0.0226$ for $N=5$ . Hence six modes certify the requested $0.01$ uniform error for every $t\ge 0.1$ and every datum with $\|f\|_2\le 1$ . The five-mode certificate is too large; it does not prove failure for each field. For example, a datum supported entirely in the first five modes has zero truncation error with five modes. The bound is a sufficient test, not an exact error formula for unknown data.

Original worksheet page 2: question and worked solution for 9-5-007

Solving the Heat Equation — Question 7

Question 7

Question 7 – Solution