# A Global Estimates for Linear Elliptic PDE

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?

• 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 $f$. – Fan Zheng Dec 7 '15 at 18:55

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.

• 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 – User4966 Dec 8 '15 at 7:20
• @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. – username Dec 8 '15 at 8:22
• Now done, as announced by @FanZheng earlier, $H^1$ is sufficient. – username Dec 8 '15 at 20:54

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 $a$.