I have a PDE defined over $\mathbb{R}$, for which I don't have the exact solution, and I am to approximate it with finite differences so I need to input some BC.

Can anyone suggest any good references for dealing with infinite intervals as BCs?

Possible attempts:

  1. Input $[-M,M]$ as domain for large M and try to guess the exact solution at $x=\pm M$.

  2. Because of the symmetry of my IC ($\log\cosh x$), identify left and right boundaries, i.e. $u(-M,t)=u(M,t)$.

  3. Try some Neumann conditions.

  4. Try periodic boundaries.


The PDE of interest is $$u_{t}=a_{1}u_{xx}+a_{2}\Phi(t)u_{x}^{2}$$ with $u(0,x)=\log\cosh x$ , defined over $(x,t)\in \mathbb{R}\times [0,1]$, $\Phi(t)$ is the CDF of standard Gaussian and $a_{i}$ are constants.

This doesn't have an explicit analytic solution.

  • $\begingroup$ One other approach is to impose the asymptotically linear boundary condition, $u_{xx}(M,t) = u_{xx}(-M, t) = 0$. (This might raise a well-posedness issue, though.) $\endgroup$ – Kirill Jan 31 '16 at 22:13

You must have boundary conditions in addition to the PDE itself. For example, a common boundary condition for cases such as yours is $u(t,x)\rightarrow 0$ for $x\rightarrow \pm\infty$ for any $t$. I've seen other cases where the condition was of the form $x^{2n} u(t,x)\rightarrow 0$ for $x\rightarrow \pm\infty$.

I don't know of finite difference methods that can solve such equations directly (though there are "infinite element" formulations of the finite element method for it). But you can introduce a coordinate transform, for example $(-1,1) \ni y \mapsto x \in (-\infty,\infty)$ that transforms your equation into one that is posed on a finite subset of the real line.

To avoid some complications, let me assume that your original problem had been posed on the half-line $x\in (0,\infty)$ and that we want to transform it to one on the interval $y \in (0,1)$. Then we could (among many other options) choose $y = 1-e^{-x}$, which yields $\frac{\partial}{\partial x}=\frac{\partial y}{\partial x} \frac{\partial}{\partial y} = e^{-x}\frac{\partial}{\partial y} = (1-y)\frac{\partial}{\partial y}$.

You would then replace all derivatives with regard to $x$ by derivatives with regard to $y$ using this expression. This yields a PDE with spatially variable coefficients (depending on $y$) to which you can apply the finite difference method.


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