# Ngô Quốc Anh

## February 6, 2010

### An illustration of an ill-posed parabolic PDE

Filed under: PDEs — Ngô Quốc Anh @ 13:28

The classic example of an ill-posed parabolic PDE problem is the “backward-in-time heat equation”.

$\displaystyle\begin{gathered} {u_t} - \kappa {u_{xx}} = 0 \quad 0 < x < L,0 < t < T, \hfill \\ u(x,T) = f(x)\quad 0 < x < L, \hfill \\ u(0,t) = 0\quad 0 < t < T, \hfill \\ u(L,t) = 0\quad 0 < t < T. \hfill \\ \end{gathered}$

Here, if we think of $u(x,t)$ as the temperature in a one dimensional heat conduction rod, the condition $u(x,T)=f(x)$ can be thought of as giving the temperature distribution at some specific time $t=T$. The PDE problem calls for using this information, together with the heat balance equation and the boundary conditions to predict the temperature distribution at some earlier time, sat $t=0$. It can be shown (see Schaum’s Outline of PDE, solved problem 4.9) that

• if $f(x)$ is not infinitely continuously differentiable, then no solution to the problem exists.
• If $f(x)$ is infinitely continuously differentiable, then it is shown that the solution on $0 does not depend continuously on the data, namely $f(x)$.

For simplicity, choose $\kappa=1$.

If $f(x)$ is not infinitely continuously differentiable: Write $u_0(x) = u(x, 0)$, the initial state (temperature). Then the function

$\displaystyle u(x,t) = \sum\limits_{n = 1}^\infty {u_0^{(n)}{e^{ - {n^2}{\pi ^2}t}}\sin (n\pi x)} , \quad 0 < x < 1,0 < t < T$

where

$\displaystyle u_0^{(n)}(x) = 2\int_0^1 {{u_0}(x)\sin (n\pi x)dx} , \quad n \geqslant 1$

will solve the problem, provided the $u_0^{(n)}$ are such that

$\displaystyle f(x) = \sum\limits_{n = 1}^\infty {u_0^{(n)}{e^{ - {n^2}{\pi ^2}t}}\sin (n\pi x)} , \quad 0 < x < 1$.

But the above series converges uniformly to an infinitely differentiable function of $x$, whatever the $u_0^{(n)}$. It follows that no solution exists when $f(x)$ is not infinitely differentiable.

If $f(x)$ is infinitely continuously differentiable: We will show that the solution does not depend continuously on the data. If, for instance,

$\displaystyle f(x) = \frac{{\sin (N\pi x)}}{N}, \quad N \in \mathbb{Z}$

the unique solution to the problem is

$\displaystyle u(x,t) = \frac{1}{N}{e^{{N^2}{\pi ^2}(T - t)}}\sin (N\pi x), \quad 0 < x < 1,0 < t < T$.

For large $N$, on the one hand, $|f(x)|$ becomes uniformly small; that is, the data function differs by as little as we wish from the data function $f =0$, to which corresponds the solution $u= 0$. On the other hand, $|u(x, t)|$ grows with $N$; i.e., the solution does not remain close to $u = 0$. Thus, there is no continuity of dependence on the data.