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<t<T 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.

See also: Ill-posed elliptic PDE, Ill-posed hyperbolic PDE.

Source: http://www.phy.ornl.gov/csep/pde/node6.html

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