$L^\infty$ stability property of an ODE

Question

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.

Answer 1

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.