Inconsistency in determinability of the solution of a linear first order PDE

Question

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}.

Answer 1

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 $h\to0+$.

Answer 2

It may help to review quickly why the backwards Euler method works for finite dimensional ODEs, before moving to the infinite dimensional case.

Finite Dimensional Backwards Euler

Consider the autonomous (only for convenience) ODE in $\mathbb{R}^n$

$$ u' = f(u) $$

The backwards Euler approximation asks you to solve the equation, for step size $h$,

$$ \frac{u_{n+1} - u_n}{h} = f(u_{n+1})$$

which can be written as

$$ u_{n+1} - h f(u_{n+1}) = u_n $$

and this requires you to invert the function $ u \mapsto u - h f(u)$. In typical applications, this mapping is invertible as a consequence of the implicit function theorem. More precisely, suppose that $f$ is continuously differentiable, then if we set $G(h,u) = u - h f(u)$, we see that

• $G(0,u_n) = u_n$
• $\partial_u G(0,u_n) = \mathrm{Id}$

and hence there is a continuously differentiable function $g: (-\epsilon,\epsilon) \to \mathbb{R}^n$ satisfying $G(h,g(h)) = u_n$, and this provides the desired inverse.

Incidentally, this also proves the approximation property: taking the $h$ derivative we find $$ \partial_h G(h, g(h)) + \partial_u G(h,g(h)) \cdot g'(h) = 0 $$ and so $$ - f(g(h)) + g'(h) - h f'(u) \cdot g'(h) = 0 $$ and evaluating at $h = 0$ you see that $g'(0) = f(g(0))$ and so for small $h$ $g(h) \approx g(0) + h g'(0)$ is approximates the true solution.

The infinite dimensional case

The same argument above would also work for ODEs on a Banach space $X$, provided the function $f:X\to X$ is a continuously differentiable function.

In your case, however, for typical function spaces, your mapping $f$ which involves the differentiation is either only densely defined, or is discontinuous, or both. So we can no longer rely on the standard argument.

On the other hand, formally we can carry through the argument as long as we have a well-defined inverse operation! Let's outline this a bit more precisely in the case where $f$ is given by a linear operator $L$. Suppose you are trying to solve

$$ u' = Lu $$

then formally the backwards Euler formula leads you to solve

$$ u_{n+1} - h L u_{n+1} = u_n $$

which requires only finding a right inverse $K:X\to X$ to the operator $\mathrm{Id} - hL$ (meaning $(\mathrm{Id} - hL)Ku = u$). Now, finding such an operator (essentially defining the resolvent of an operator) is very common in the study of PDEs. Of course, this only gives you a candidate for what the backwards Euler approximation could look like: unlike the finite dimensional case this doesn't actually prove that the candidate is a good approximation to the actual solution, because $L$ may be unbounded.

Your actual problem

In your case, your operator is $L = \frac{\partial}{\partial x} + \mathrm{Id}$. Supposing that you are working with solutions in the Banach space $X = C^1_b(\mathbb{R})$ of continuously differentiable functions with uniform bounds on the function and its derivatives. Then your goal is to find a right inverse $K_h:X\to X$ to the operator $$ \mathrm{Id} - hL = (1-h)\mathrm{Id} - h \frac{\partial}{\partial x} $$ As you already alluded to, we can solve this explicitly $$ (1-h) u - h u' = v \implies [e^{-(1-h)x/h} u]' = -e^{-(1-h)x/h} v $$ which implies $$ u= C e^{(1-h)x/h} + e^{(1-h)x/h} \int_{x}^{x_0} e^{-(1-h)y/h} v(y) ~dy $$ A valid choice is to take $x_0 = +\infty$, as then the integral $$ \int_x^{\infty} e^{\frac{1-h}{h} (x-y)} v(y) ~dy $$ is guaranteed to converge (due to the exponential decay as $y > x$), and that the integral is uniformly bounded in $x$ for any $v$ that is uniformly bounded.

In this case, we must also choose $C = 0$, because any other value of $C$ will leave the operator no longer a mapping from $C^1_b$ to itself (it generates an unbounded function). Note that this means the choice

$$ K_hv = \int_x^\infty e^{\frac{1-h}{h}(x-y)}v(y) ~dy $$

is also the ONLY reasonable choice you can take if you chose to work in any Banach space $\tilde{X}$ that is a subspace of $C^1_b$ (otherwise $K$ would land outside of $\tilde{X}$). Note also that you can extend the argument to show that the same holds if you work with the space of functions with moderate growth.

Of course, the argument doesn't prove

$$ \lim_{h\to 0} \frac{1}{h} [ K_h v - v] = Lv $$

This is something you will have to check by hand using the formula for $K_h$ and an appropriate sense (topology) of what the limits and equality means.