3
$\begingroup$

Let $\Omega$ be a bounded smooth region in $R^n$ and $u$ satisfy

$-\Delta u+a(x)u=f, \ \ u|_{\partial \Omega}=0$,

where $a(x)\geq 0$ and $f(x)$ are smooth functions. I wonder if the following estimate holds

$||a(x)u||_{H^1(\Omega)} \leq C ||a||\ \||f||_{H^1(\Omega)}$,

where $C$ is independent of $u$ and $a(x)$. What norm of $a(x)$ should appear in the right hand side of the above inequality?

$\endgroup$
1
  • 1
    $\begingroup$ My intuition is that you probably need $\|a\|_{H^1}$. Clearly this suffices as you can then pull $a$ out and use standard elliptic regularity. OTOH, it is probable that $u$ wiggles at the $H^3$ level, so $\Delta u$ wiggles at the $H^1$ level, which precisely cancels out the wiggle of $a$ at the $H^1$ level, leaving a rather smooth $f$. $\endgroup$ – Fan Zheng Dec 7 '15 at 18:55
2
$\begingroup$

I disagree with the accepted answer. You do not need a Lipschitz bound on $a$. I assume for simplicity that $a\in L^\infty(\Omega)$.

Step 1. The solution $u$ exists and is unique in $H^1_0(\Omega)$, by Lax-Milgram for example.

Step 2. Write $g:=f-au$. Then $u$ satisfies $-\Delta u = g$ in $\Omega$, so provided $\Omega$ is $C^1$ for example, $u\in H^2(\Omega)\cap H^1_0(\Omega)$. So

2.a If $n$ the dimension of the ambient space verifies $n<4$ then $u\in C^{0,1/4}(\Omega)$ and therefore $$ au \in H^{1}(\Omega) \mbox{ when } a\in H^1(\Omega). $$ 2.b If $n>4$ then $u\in L^{\frac{2n}{n-4}}(\Omega)$ and therefore $$ au \in H^{1}(\Omega) \mbox{ when } a\in W^{1,n/2}(\Omega). $$ 2.c If $n=4$, $a\in W^{1,2+\epsilon}(\Omega)$ with $\epsilon>0$ is enough.

3.a. Suppose $n \geq 6$, and $a\in H^{1}\cap L^\infty$. If we allow $f\in H^1$, as suggested in the question, then $f\in L^{\frac{2n}{n-2}}$, $au \in L^{\frac{2n}{n-4}}$ so $u \in W^{2,\frac{2n}{n-4}}$ and as $\frac{2}{n-4}< 2$, $u$ is Holder, therefore bounded and we are back to 2.a

3.b and 3.c, n=5 and n=4, argue as in 3.a but it bootstrap twice instead of once.

Therefore in all dimension, $a\in H^{1}\cap L^\infty$ is enough.

$\endgroup$
3
  • $\begingroup$ This leads to the estimate ||a(x)u||H1(Ω)≤(1+C||a||W1,pn) ∥|f||L2(Ω), and not ||a(x)u||H1(Ω)≤C||a|| ∥|f||L2(Ω). The second inequality gives a better estimate for small a $\endgroup$ – User4966 Dec 8 '15 at 7:20
  • $\begingroup$ @MathStudent I don't see what you mean. But 2.b is not optimal: if we allow $f\in H^1$ (as suggested) then it can be improved (by bootstrap) to get to $a$ in a slightly bigger space (I think). I'll edit it. $\endgroup$ – username Dec 8 '15 at 8:22
  • 1
    $\begingroup$ Now done, as announced by @FanZheng earlier, $H^1$ is sufficient. $\endgroup$ – username Dec 8 '15 at 20:54
1
$\begingroup$

Writing $$ \langle-\Delta u + au, u\rangle_{L^2(\Omega)}=\langle f, u\rangle_{L^2(\Omega)}, $$ using the Dirichlet boundary condition, you get $$ \Vert \nabla u\Vert_{L^2(\Omega)}^2+\langle au, u\rangle_{L^2(\Omega)} \le \Vert f\Vert_{L^2(\Omega)}\Vert u\Vert_{L^2(\Omega)}, $$ so that you control the $H^1$ norm of $u$ by the $L^2$ norm of $f$. To get a control of the $H^1$ norm of $au$, you will need some Lipschitz norm of $a$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.