Find the most general u=u(r) satisfying the…

Find the most general $u=u(r)$ satisfying the three-dimensional Laplace equation $\frac{\partial^2u}{\partial x^2}+\frac{\partial^2u}{\partial y^2}+\frac{\partial^2u}{\partial z^2}=0$ under the substitution $r=x^2+y^2+z^2$.

Step 1: Chain Rule Derivatives

With $\partial r/\partial x=2x$ (and likewise for $y$, $z$), the same computation as in the two-variable case gives

\[ \frac{\partial u}{\partial x} = 2x\,u', \qquad \frac{\partial^2u}{\partial x^2} = 2u' + 4x^2u'', \]

and by symmetry

\[ \frac{\partial^2u}{\partial y^2} = 2u' + 4y^2u'', \qquad \frac{\partial^2u}{\partial z^2} = 2u' + 4z^2u''. \]

Step 2: Substitution into the PDE

Summing the three second partials: \[\frac{\partial^2u}{\partial x^2}+\frac{\partial^2u}{\partial y^2}+\frac{\partial^2u}{\partial z^2} = 6u' + 4(x^2+y^2+z^2)u'' = 6u' + 4r\,u'' = 0.\]

Dividing by $2$ gives the ODE in $r$: $2r\,u'' + 3u' = 0$.

Step 3: Solving the ODE

This is separable in $v=u'$: from $2r\,v'=-3v$, \[\frac{dv}{v} = -\frac{3}{2r}\,dr \;\Longrightarrow\; \ln|v| = -\frac32\ln r + \mathrm{const} \;\Longrightarrow\; v = u' = C_1\,r^{-3/2}.\]

Integrating once more, $u = C_1\cdot\frac{r^{-1/2}}{-1/2} + C_2 = -2C_1\,r^{-1/2} + C_2$.

Renaming $-2C_1\to C_1$: $u = \frac{C_1}{\sqrt{r}} + C_2$.

Step 4: Back-Substitution

Since $r=x^2+y^2+z^2$, $u = \frac{C_1}{\sqrt{x^2+y^2+z^2}} + C_2$.

Step 5: Verification

Write $s=x^2+y^2+z^2$, so $u_1=s^{-1/2}$. Then $u_{1,x}=-x\,s^{-3/2}$, and differentiating again, $u_{1,xx} = -s^{-3/2} + 3x^2s^{-5/2}$.

with the analogous expressions for $u_{1,yy}$ and $u_{1,zz}$. Summing all three, \[u_{1,xx}+u_{1,yy}+u_{1,zz} = -3s^{-3/2} + 3(x^2+y^2+z^2)s^{-5/2} = -3s^{-3/2} + 3s\cdot s^{-5/2} = -3s^{-3/2}+3s^{-3/2} = 0.\ \checkmark\]

The additive constant $C_2$ trivially contributes zero to every derivative, so the full solution satisfies the equation identically away from the origin.

\[u = \frac{C_1}{\sqrt{x^2+y^2+z^2}} + C_2\]