The backwards heat equation is not well posed

Question

I have some sort of issue on this problem. Here it is: Show that the backwards heat equation, $\frac {\partial u}{\partial t} = -k \frac {\partial^2 u}{\partial x^2}$, subject to $u(0,t) = u(L,t) = 0, u(x,0) = f(x)$ is not well posed.

Here is my attempt at the solution. I used the Method Of Separation Of Variables to work on this problem, $[u(x,t) = F(x)G(t)]$. By separation of variables, the P.D.E. turns out to be $\frac {1}{G} \frac{\partial G}{\partial T} = \frac {-k}{F} \frac{\partial^2 F}{\partial x^2} = \lambda$. I only focused on the conditions for $\lambda$ only for the function $F(x)$. For $\lambda =0$ and $\lambda <0$, the solution would be trivial. For $\lambda > 0$, I found the general solution as $F(x) = B_n \sin (\frac{n \pi x }{L}) $ which $B_n$ are arbitrary constants for $n = 1,2,3,......$
Now for this condition on $\lambda$ on $G(t)$, I had that $G(t) = C_n e^{k\frac{n \pi x}{L}t}$ which $u(x,t)= \sum_{i=1}^{n}b_n e^{k\frac{n \pi x}{L}t}\sin (\frac{n \pi x }{L}) $ which $b_n = B_n*C_n$. When I initialize the initial condition, the general solution would be $u(x,0)= \sum_{i=1}^{n}b_n\sin (\frac{n \pi x }{L}) = f(x) $. I know this a Fourier Series representation but I do not understand how to find out if this is well-posed or not. A well posed solution is if the solution is unique and if the solution depends continuously on the data.

Am I on the right track?

Answer 1

Writing down the formula $$u(x,t)= \sum_{i=1}^{n}b_n e^{k\frac{n \pi }{L}t}\sin (\frac{n \pi x }{L}) \tag{1}$$ [NB: I dropped extraneous "x" in the exponent] is a good idea, but you should connect it to the goal: proving that the $u(x,t)$ does not depend continuously on the initial data $f$. The goal is a bit vague because to make sense of "continuous" we need some topology on the set of functions that are used as initial values. It may happen that the problem is well-posed in one topology but not in another. Topology is usually defined by a norm in this context.

One possibility is the uniform norm $\|f\| = \sup_{[0,L]} |f|$. To show the problem is ill-posed with respect to this norm, we need a sequence $f_n$ such that $\|f_n\|\to 0$ but the corresponding solutions do not converge to zero (again, in what sense? I'll go with uniform norm again).

Looking at (1), we see than higher frequencies (greater $n$) get magnified more, by virtue of $\exp(k n\pi t/L)$. So... why not take $$f_n(x) =\exp( - k n\pi /L) \sin (n\pi x/L)$$ this converges to zero in many different senses but $$u_n(x,1) = \sin (n\pi x/L) $$ does not become small in the uniform norm. To make it even worse, $$u_n(x,2) = \exp( k n\pi /L) \sin (n\pi x/L) $$ becomes rather large.

Answer 2

It seems the problem is recently solved. Please see the following papers where (easy to check) necessary and sufficient conditions for the solvability of the Cauchy problem of the backward heat equation are found. Similar condition also works in case of bounded domains with Dirichlet boundary condition.

1. arXiv:1905.05845

A note on time analyticity for ancient solutions of the heat equation

2. arXiv:1907.01687

Time analyticity for the heat equation and Navier-Stokes equations

Answer 3

Indeed, the topology is crucial. If $u$ is an element of $L^{2}( \mathbb{R},dx)$ we can show that $-\partial _{x}^{2}$ extends to a non-negative self-adjoint operator $p^{2}$ and we can write

\begin{equation*} u(t)=\exp [kp^{2}t]u(0). \end{equation*} In this case $u(t)$ is well defined for all $t$.