Inconsistency in Determinability of Solution for Linear First Order PDE

differential equationsnumerical-analysis-of-pde

Consider the following differential equation:

$$\frac{\partial u(x,t)}{\partial t} = – \frac{\partial u(x,t)}{\partial x} + u(x,t) \label{1}\tag{1}$$

with $u(x,0)=f(x)$. The solution of \eqref{1}, using MOC, is $e^{t}f(x-t)$.

I, however, would like to solve \eqref{1} numerically using the backward Euler method. By discretizing variable $t$:

$$\frac{u_1-f(x)}{h}=- \frac{\partial u_1}{\partial x} + u_1, \label{2}\tag{2}$$

where $u_1(x)=u(x,0+h)$, which is a simple first order linear ODE. The solution of \eqref{2} is also known analytically up to a constant.

The difficulty that I am facing is how to determine the value of this constant? It seems that I have to assume something extra (boundary condition maybe) about $u_1(x)$, which does not seem to be necessary from \eqref{1}.

Best Answer

Pedestrian answer.

In practice $h>0$ is a small number. In what follows, it will be enough that $h\in(0,1)$. Because of $$\left(e^{(\frac1h-1)x}u_1\right)'=\frac1h e^{(\frac1h-1)x}f,$$ we must have $$u_1=ce^{(\frac1h-1)x}+\int^x\frac1h e^{(\frac1h-1)(y-x)}f(y)dy.$$ There remains to determine the constant $c$, as well as the over limit point $\pm\infty$ in the integration. The key observation is that the function $$\phi_h(x):= \frac1h e^{(\frac1h-1)x}{\bf1}(x<0)$$ is integrable. A natural assumption being that the initial data $u_0(x)$ be in some $L^p$-space ($1\le p\le\infty$), we must assume as well $f\in L^p({\mathbb R})$. Then $\phi_h\star f\in L^p({\mathbb R})$, and this is the only solution belonging to $L^p({\mathbb R})$. Therefore the right choice is $$u_1=\int^x_{-\infty}\frac1h e^{(\frac1h-1)(y-x)}f(y)dy.$$ Notice that $$\|u_1\|_p\le\|\phi_h\|_1\|f\|_p=\frac1{1-h}\|f\|_p.$$ In other words $\|u^h(nh)\|_p\le(1-h)^{-n}\|u_0\|_p$. This is reminiscent to the expected $e^{-t}$ decay in the inviscid limit $h\to0+$.

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