Solving the Heat Equation — Question 9

Question 9 – Solution

Strategy. Each forward-decaying mode becomes an exponentially amplified mode when time is reversed.

Step 1: Characterize exact inversion. Forward evolution gives $d_n=c_ne^{-n^2T}$, so the only possible coefficients are $\boxed{c_n=e^{n^2T}d_n}$. Completeness implies at most one initial field. It exists in $L^2$ exactly when $$\boxed{\sum_{n\ge 1}e^{2n^2T}|d_n|^2<\infty.}$$ Indeed this condition constructs an $L^2$ sine expansion whose forward solution has the given coefficients. Arbitrary $L^2$ final data need not satisfy this stronger condition.

Step 2: Disprove continuity despite uniqueness. Take $f_n=\phi_n$, whose initial norm is one. Its final field has norm $e^{-n^2T}\to 0$. Thus final data tend to zero while their uniquely determined initial fields do not. The backward map on its attainable-data domain is not continuous at zero in the stated norms. No contradiction arises with the stable forward problem, which damps these same modes.

Step 3: Bound noise after truncating the inverse. Let $\eta_n$ be the noise coefficients, so $\sum|\eta_n|^2\le\delta^2$. The retained reconstructed noise satisfies $$\left\|\sum_{n=1}^N e^{n^2T}\eta_n\phi_n\right\|_2^2 =\sum_{n=1}^N e^{2n^2T}|\eta_n|^2\le e^{2N^2T}\delta^2.$$ Taking square roots gives the bound, attained by noise $\delta\phi_N$.

Step 4: Choose a noise cutoff and state its limitation. The requirement is $e^{0.1N^2}10^{-6}<10^{-2}$, or $N^2<40\log 10$. Thus $\boxed{N=9}$ is the largest certified integer: the bounds for $9,10$ are approximately $0.003295$ and $0.022027$. The plot uses discrete stems, since $n$ is an integer; their heights are $0.1n^2/\log 10$. The omitted initial tail has norm $(\sum_{n>N}|c_n|^2)^{1/2}$, for which no numerical bound was supplied. A noise cutoff controls amplification but requires additional prior information to control that approximation error.