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.

  • $\begingroup$ Can you provide the reference? $\endgroup$ – nicoguaro Jun 8 '19 at 16:33
  • 1
    $\begingroup$ 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. $\endgroup$ – Richard Zhang Jun 8 '19 at 16:49
  • $\begingroup$ A reference is: ima.umn.edu/sites/default/files/1921.pdf and the specific statement is on page 2. I also don't understand the argument, but it is important to how the author formulates stability of the DG methods within the paper. $\endgroup$ – cm2 Apr 7 at 21:11

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