$L^\infty$ stability property of an ODE

Suppose we have the initial-value problem on $$(0,L)$$:

$$\frac{d u(x)}{d x} = f(x) u(x),\, \qquad x\in\Omega,\,~~ u(0) = u_0,$$

I am reading a claim that says if we multiply the ODE by $$u$$ and integrate over $$(0,L)$$, we have

$$\frac{1}{2}u^2(L) - \frac{1}{2} u^2_0 = \int_{0}^{L} f(x)u^2(x) \,dx$$

"from which the $$L^\infty$$-stability of the solution follows." I agree that the equation is correct, but:

1. Why does this guarantee stability?
2. What exactly is meant by $$L^\infty$$-stability? I interpret stability in the context that the numerical solution will remain bounded as the the step size is reduced, but here, we do not have a discretization yet...

This discussion is given in the context of discontinuous Galerkin methods.

• Can you provide the reference? – nicoguaro Jun 8 '19 at 16:33
• The $L^\infty$ norm of a function is its maximum absolute pointwise value, as in $\|f\|_\infty = \max_{x\in\Omega} |f(x)|$. Stability here means that if $\|f\|_\infty < \infty$, then $\|u\|_\infty < \infty$. The reasoning is a bit circular to me, because you can't just multiply pointwise by $u$ if $\|u\|_\infty = \infty$, but I can't immediately think of a counterexample. – Richard Zhang Jun 8 '19 at 16:49

Expanding on my comment. The $$L^\infty$$ norm of a function $$f:\Omega\to\mathbb{R}$$ is the maximum absolute pointwise value over its range, as in $$\|f\|_\infty = \max_{x\in\Omega} |f(x)|.$$ Using this norm, we can define the function space $$L^\infty$$ for all functions with bounded $$L^\infty$$ norm, as in $$f \in L^\infty \iff \|f\|_\infty < \infty.$$
Now, the ODE is said to be "stable in $$L^\infty$$" if every right-hand side $$f\in L^\infty$$ yields a solution $$u\in L^\infty$$. Loosely, the ODE is stable if a bounded $$f$$ yields a bounded $$u$$. Bounded input gives bounded output.
The claim in the prompt is definitely correct, but I'm not totally sold on the proof. With a minor change of notation, we get the one-dimensional linear time-varying (LTV) equation $$\frac{\mathrm{d}x(t)}{\mathrm{d}t} = f(t)x(t), \qquad t\in(0,L),$$ and this has a well-known closed-form solution $$x(t) = x_0\exp \left(\int_0^t f(t') \mathrm{d} t' \right), \qquad t \in (0,L).$$ (This is a special case of the Magnus series, but as an exercise, take the ansatz $$x(t)=C \exp(g(t))$$ and observe that $$\mathrm{d}g/\mathrm{d}t = f$$.) We see the claim is true because every $$f\in L^\infty$$ is integrable, and that the exponential map is analytic.
• Interesting. So the algebra given is not enough to actually make the claim? The author's using additional information about this particular ODE? I had thought the idea was something to do with the derivative of energy $\frac{d}{dt}||u^2_h|| \leq 0$, but it was unclear to me why the RHS had to be less than or equal to zero, since no conditions are given on the forcing function $f(x)$. – user3482876 Jun 8 '19 at 18:11
• So by asserting $\int_0^y u (du/dx) dx = (1/2)u(y)^2 - (1/2)u(0)^2$, you're already implicitly assuming that $u (du/dx)$ is integrable over $[0,y]$, i.e. it already excludes $u(x) = 1/x$ from consideration. It's easier to be stable when the possible choices of $u$ have already been significantly narrowed down. – Richard Zhang Jun 8 '19 at 18:26