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On the closed upper half space of ${\mathbb R}^{3}$ i.e.

$\quad\forall\ x, y$, and $z\geq 0\quad$ find functions $\quad u, v\quad $ satisfying:

$$\Delta u = 1\text{ and }u(x, 0)=0$$

and

$$\Delta v = \delta(0, 0, 1).$$

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user115573
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  • Your condition $z>0$ actually gives an open half space, not a closed half space. Could you also give some idea of where you are stuck with this problem. – Old John Dec 16 '13 at 16:32
  • Hi new user! $\color{red}{\Large\text{Welcome to Math.SE!}}$ Please show your working :) – Shaun Dec 16 '13 at 16:33
  • Up to now we have only been working with Laplace equation (0 right hand side) which has a closed form solution. Adding the non-homogenous term on the right is throwing me off since my technique goes out the window. Thus I believe there should be an extension or I think we should be able to reformulate the problem with the right hand side being an initial condition of some sort. – user115573 Dec 16 '13 at 16:41
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    It is absurd to vote to close questions like this as "off-topic". It's just newbie-biting. – Michael Hardy Dec 16 '13 at 17:52
  • any hints? not sure what to try next – user115573 Dec 16 '13 at 22:31
  • I agree with @MichaelHardy . – Felix Marin Sep 11 '14 at 21:07
  • What is $\delta\left(,0,0,1,\right)$ ?. – Felix Marin Sep 11 '14 at 21:12

1 Answers1

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Constant Laplacian corresponds to the function being a quadratic polynomial. (Indeed, the Laplacian of quadratic form $x\mapsto x^TAx$ is twice the trace of $A$). To satisfy the zero boundary condition, take $u$ to be a multiple of $z^2$.

As for $v$, just shift the fundamental solution of $\Delta$ by one unit up. You'll get some multiple of $(x^2+y^2+(z-1)^2)^{-1/2}$.

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