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Let's say I have an equation of the form $\Delta A = J$ where $J=u\nabla u + A|u|^2$ (Clarification: We are on $\mathbb{R}^3$ and $u$ is assumed to be in $H^1(\mathbb{R}^3)$). Then I could simply infer from Hardy-Littlewood-Sobolev and Hölder that $$\|A\|_6 \leq \|J\|_{6/5} \leq \|u\|_3\|\nabla u\|_2 +\|A\|_6 \|u\|_3^2$$ and then from Sobolev $$\cdots \leq \|u\|_{H^1}^2 + \|A\|_6 \|u\|_{H^1}^2$$ Can I somehow infer from that that $\|A\|_6$ is controlled by some norm of $u$?

EDIT: A thought that came to mind: Using $2ab \leq \varepsilon a^2 +\frac{1}{\varepsilon} b^2$ we could write $$\cdots \leq \varepsilon \|A\|^2_6 + (1+ C_{\epsilon}\|u\|_{H^1}^2) \|u\|_{H^1}^2$$ for $\epsilon >0$. Is this sensible?

EDIT 2: The problem that arises is that from the inequality above we obtain $$\|A\|_6(1-\|u\|_{H^1}^2) \leq \|u\|_{H^1}^2$$ so if $||u||^2_{H^1} = 1 $ we get no information on $\|A\|_6$. Also if $||u||^2_{H^1} > 1 $ Then the LHS becomes negative.

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  • $\begingroup$ Which domain are you considering? If it is non compact, what are the decay assumptions for u? $\endgroup$
    – Romain Gicquaud
    Commented Nov 30, 2020 at 11:40
  • $\begingroup$ I'm sorry I forgot to add that information. We're on whole space in $\mathbb{R}^3$. $u$ is assumed to be in $H^1$. $\endgroup$
    – Jakob Möller
    Commented Nov 30, 2020 at 12:43
  • $\begingroup$ To understand: is $u$ given and you look for $A$? Or you already know that $A$ exists and verifies some estimates in other norms? $\endgroup$
    – Giorgio Metafune
    Commented Nov 30, 2020 at 14:44
  • $\begingroup$ @GiorgioMetafune Yes, $u$ is given in $H^1(\mathbb{R}^3)$ (whence the use of Sobolev embedding). I'm looking for $A$. In fact, $\Delta A=J$ is coupled to a magnetic heat-Schrödinger equation $\partial_t u = (i+\epsilon) (\Delta +2iA\nabla +|A|^2)u$ and I want to set up a contraction whence I need control over the nonlinearity (i.e. Lipschitz) in terms of $u$. $\endgroup$
    – Jakob Möller
    Commented Nov 30, 2020 at 14:50

1 Answer 1

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This is a way to get an a-priori estimate, if I understood correctly the question. Multiply by $A$ and integrate by parts the left-hand-side. Then $$\int_{R^3}(A^2u^2+|\nabla A|^2)=-\int_{R^3}Au\nabla u\le \|Au\|_2\|\nabla u\|_2 $$ and then both $\|Au\|_2, \|\nabla A\|_2 \le \|\nabla u\|_2$. Since $2^*=6$, $\|A\|_6 \le C\|\nabla A\|_2 \le C\|\nabla u\|_2$.

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  • $\begingroup$ I'm being a bit obtuse, but how does the integral give a bound $\lVert \nabla A \rVert_2$? The same trick as for $\lVert Au \rVert_2$ (dividing both sides) isn't available. $\endgroup$
    – Leo Moos
    Commented Nov 30, 2020 at 15:40
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    $\begingroup$ First estimate $\|Au\|_2$ neglecting $\nabla A$ which has the same sign. Then insert this estimate in the RHS. $\endgroup$
    – Giorgio Metafune
    Commented Nov 30, 2020 at 15:46
  • $\begingroup$ Oops, thank you Giorgio! $\endgroup$
    – Leo Moos
    Commented Nov 30, 2020 at 16:36
  • $\begingroup$ A little question (maybe trivial): Why can we perform integration by parts? $\endgroup$
    – Jakob Möller
    Commented Nov 30, 2020 at 19:43
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    $\begingroup$ Well, the apriori bound must be turned into an existence proof. This should work (details are omitted): approximate $u$ by $u_\epsilon$, smooth and with compact support in the $H^1$ norm and solve $\Delta A_\epsilon-A_\epsilon (u^2_\epsilon+\epsilon)=u_\epsilon\nabla u_\epsilon$ using Lax-Milgram to get existence in $H^1$. Then the above computations give the boundedness of $\|\nabla A_\epsilon\|_2 $ and $\|A_\epsilon\|_6$. Lettting $\epsilon \to 0$ you get existence in $L^6$ (with gradient in $L^2$) $\endgroup$
    – Giorgio Metafune
    Commented Nov 30, 2020 at 22:48

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