Let $\mathbb{B}(0,1) \subset \mathbb{R}^3$ denote the unit ball. Let $\Gamma = \{x_3=0\}$. Let us assume $f \in L^2(B)$ .Consider the problem $ \triangle u = f $ in $\mathbb{B}$ in the weak sense such that $ u|_{\Gamma}=0$. Does there exist a bounded linear operator $G: L^2 \to L^2$ such that $u = Gf$
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$\begingroup$ Is B a the ball or the sphere? Where does the equation $\Delta u=f$ hold? Where is $f$ defined? $\endgroup$– Alexandre EremenkoCommented Oct 15, 2017 at 17:14
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$\begingroup$ the equation holds weakly so in the interior of the ball. F is an L^2 function in the ball. $\endgroup$– AliCommented Oct 15, 2017 at 18:32
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$\begingroup$ Then why does your problem $\Delta u=f,\; u\vert_\Gamma=0$ have a solution? Or if it has, why is it unique? $\endgroup$– Alexandre EremenkoCommented Oct 15, 2017 at 20:20
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$\begingroup$ I am asking for the existense of a bounded right inverse. (The solution clearly is not unique if anything, this is more a question of existense ) $\endgroup$– AliCommented Oct 15, 2017 at 20:34
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$\begingroup$ I don't get the close and negative comments?! The problem is quite clear $\endgroup$– AliCommented Oct 15, 2017 at 21:10
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1 Answer
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The problem may have no solution at all!
If $\Delta u = f$ in $B$ and $G_B(x,y)$ is the Green's function of a ball, then $$h = u + G_B f$$ is harmonic in $B$ (here $G_B u(x) = \int_B G_B(x,y) u(y) dy$). In particular, $h$ is real analytic.
On the other hand, $G_B f$ need not be smooth on $\Gamma$. If this is the case, the condition $u = 0$ on $\Gamma$ would mean that $h = G_B f$ on $\Gamma$, a contradiction.
The answer may change if we know in advance that $f$ is real analytic near $\Gamma$. In this case I believe the solution exists at least locally near $\Gamma$. In order to prove this, I would first try to show that any homogeneous polynomial $P$ of degree $l$ in $2$ variables can be extended to a homogeneous (a.k.a. solid) harmonic polynomial of the same degree $l$ in $3$ variables, with some control on the size of the coefficients; and then apply this procedure to the Taylor expansion of the function $G_B f$. I did not attempt to do that, though.
answered Oct 16, 2017 at 10:58
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$\begingroup$ If $f$ is real analytic near $\Gamma$, then the local existence of a solution follows from Cauchy-Kovalevskaya. $\endgroup$ Commented Oct 16, 2017 at 14:50