The Question:

Let $V=\{(x,y,z) \in \Bbb R^3:x^2+y^2+z^2<1\} $. Find an extremal $u(x,y,z)$ for minimizing the integral

$$\iiint_V \biggl[ \biggl( \frac{\partial u}{\partial x} \biggr)^2+\biggl( \frac{\partial u}{\partial y} \biggr)^2 +\biggl( \frac{\partial u}{\partial z} \biggr)^2\biggr]dxdydz$$

subject to the constraints

$$\iiint_Vu \, dxdydz=4\pi \qquad , \qquad u=1 \quad \text{on} \; \partial V$$

My Attempt:

The Euler-Lagrange equation in this case is

$$\biggl(\frac{\partial}{\partial x}\frac{\partial}{\partial u_x}+\frac{\partial}{\partial y}\frac{\partial}{\partial u_y}+\frac{\partial}{\partial z}\frac{\partial}{\partial u_z} \biggr)\bigl(F-\lambda G \bigr)=\frac{\partial}{\partial u} \bigl(F-\lambda G \bigr)$$

where $F=\biggl( \dfrac{\partial u}{\partial x} \biggr)^2+\biggl( \dfrac{\partial u}{\partial y} \biggr)^2 +\biggl( \dfrac{\partial u}{\partial z} \biggr)^2$ and $G=u$. Plugging it in, we get

\begin{align} & 2\frac{\partial^2u}{\partial x^2}+2\frac{\partial^2u}{\partial y^2}+2\frac{\partial^2u}{\partial z^2}=-\lambda \\ \implies & \vec \nabla^2u=-\frac{\lambda}{2} \end{align}

How on earth do I solve this inhomogeneous Laplace equation?

Or have I made some sort of mistake before that?

Any hints will be much appreciated.

  • $\begingroup$ Hint: Use Green's function on the ball. $\endgroup$ – Jacky Chong May 6 '18 at 20:25

Since nothing in the problem depends on the angles, there's no harm in seeing if a radial solution will work. In polar coordinates in 3D, the radial part of the Laplacian is $$ \frac{1}{r^2} \frac{\partial}{\partial r} r^2 \frac{\partial}{\partial r}, $$ so if $u=u(r)$, the equation becomes $$ (r^2u')' = -\frac{\lambda}{2}r^2. $$ Integrating once, $$ r^2u' = -\frac{\lambda}{6}r^3+A, $$ but $u$ is harmonic, so $r^2u' \to 0$ as $u \to 0$, so $A=0$. Dividing by $r^2$ and integrating again, $$ u = -\frac{\lambda}{12}r^2 + B, $$ and setting $r=1$, we conclude that $B=1+\lambda/12$, so $$ u = -\frac{\lambda}{12}r^2 + 1+\frac{\lambda}{12}. $$ It remains to calculate the integral over the ball.

EDIT: Apparently I can't remember the volume of a ball correctly. Integrating over the ball gives $$ 4\pi \int_0^r r^2 u \, dr = 4\pi \left( \frac{1}{3}+\frac{\lambda}{90} \right), $$ so $\lambda = 60$, and the final answer is the slightly anticlimactic $$ u(r) = 6-5r^2. $$

  • $\begingroup$ Shouldn't the integral be $$\iiint_V1\, dxdydz=\frac{4\pi}{3}$$ since it is the volume of the unit sphere? $\endgroup$ – glowstonetrees May 7 '18 at 14:33
  • $\begingroup$ But anyway, thanks for your answer $\endgroup$ – glowstonetrees May 7 '18 at 14:34
  • $\begingroup$ Yes, a shortcut too far. Fixed. $\endgroup$ – Chappers May 7 '18 at 19:25

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