6
$\begingroup$

I am trying to numerically solve the initial value problem (see later discussion for ICs)

$$ x \frac{\partial f}{\partial t} = \frac{\partial}{\partial x} (1-x^2) \frac{\partial f}{\partial x} - f$$

for $$ f= f(x,t), \quad x \in X = [-1, 1] \quad \text{and} \quad t \in T = [0,1] $$ where $$ f(1, \ t), \ f(-1, \ t) \text{ are finite} \ \forall \ t \in T $$

and f is continuous for all x, t. Please note that my question only concerns solution using iterative methods, not a solution by eigenfunction expansion.

I attempted to solve this problem given an initial condition $$ \breve{f}(x, t = 0)$$ with a simple BTCS method. This leads to reverse diffusion for x < 0, hence the solution gets lost in oscillations. Question 1 - Is there any way to solve this problem this problem directly? How can I check the solution even exists?

I was recommended to solve reformulated problem, so that there are two piecewise initial conditions:

$$ \breve{f}_0(x_1, t = 0), \ x_1 \in [0, 1] \quad \breve{f}_1(x_0, t = 1), \ x_0 \in [-1, 0] $$

and march backwards in time for $x \in [-1, 0]$. I am however unsure how to deal with boundary conditions at $x = 0$. The solution is continuous for all $x, t$ which leads to the following requirement at the boundary

$$ \lim_{|x| \rightarrow 0} \frac{\frac{\partial ^ 2 f}{\partial x^2} - f}{x} \text { is finite } \forall \ t \in T $$

I cannot solve the both parts of X interval simultaneously however, which seems to prevent me from imposing this BC. Question 2 - How can I impose the boundary condition at $x = 0$ to ensure the final solution is continuous? Are there any restrictions on $\breve{f}_1(x_0, t = 1) \ \text{given} \ \breve{f}_0(x_1, t = 0)$ ?

Any other suggestions on solution are appreciated.

Thank you

$\endgroup$
2
  • $\begingroup$ You can treat this as the generator for a markov process in $(x,t)$ (introducing a separate random variable for the time parameter), from which you can see that there will be a unique solution with the piecewise initial conditions. $\endgroup$ Aug 29, 2013 at 16:54
  • $\begingroup$ ...as long as you also have boundary conditions specified at $x=\pm1$ $\endgroup$ Aug 29, 2013 at 17:03

2 Answers 2

2
$\begingroup$

This kind of problem has been discussed in J. Math. Phys. vol 22, page 954 and vol 24, page 1932.

$\endgroup$
2
  • $\begingroup$ Would you mind making your answer more complete? What was the result of this discussion? It would be great if you could summarize what is solved there and what is still unanswered. $\endgroup$ Aug 16, 2013 at 19:30
  • $\begingroup$ @AndrásBátkai, unfortunately, although I remember looking at this paper years ago, I never studied it enough to summarize it. $\endgroup$ Aug 17, 2013 at 21:41
0
$\begingroup$

It is not a well-posed problem, see http://en.wikipedia.org/wiki/Parabolic_partial_differential_equation#Backward_parabolic_equation.

Numerically you should see high-frequency instability.

$\endgroup$
2
  • $\begingroup$ But deciding which boundary/initial/final conditions to be specified seems to be a part of the problem itself. For instance, the OP was suggesting to solve the forward heat part with an initial condition, and the backward heat part with a final condition. $\endgroup$
    – timur
    Aug 29, 2013 at 12:32
  • $\begingroup$ user39289 - true. The original problem is indeed ill posed. That is the reason for the reformulation as then there is always correct sense of diffusion thanks to specifying final conditions on negative x half-domain, instead of initial ones. $\endgroup$ Sep 5, 2013 at 3:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.