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 $r=\sqrt{x^2+y^2+z^2}$.

Step 1: Chain Rule Derivatives

With $\partial r/\partial x=x/r$ (and likewise for $y$, $z$), the same computation as in Problem 208 gives

\[ \frac{\partial^2u}{\partial x^2} = \frac{u'}{r} + \frac{x^2u''}{r^2} - \frac{x^2u'}{r^3}, \]

with the analogous expressions, by symmetry, for $\partial^2u/\partial y^2$ and $\partial^2u/\partial z^2$.

Step 2: Substitution into the PDE

Summing all three second partials and using $x^2+y^2+z^2=r^2$: \[\sum\frac{\partial^2u}{\partial x^2} = \frac{3u'}{r} + \frac{r^2u''}{r^2} - \frac{r^2u'}{r^3} = \frac{3u'}{r} + u'' - \frac{u'}{r} = u'' + \frac{2u'}{r}.\]

The PDE therefore reduces to

\[ u'' + \frac{2u'}{r} = 0. \]

Multiplying by $r^2$: $r^2u'' + 2r\,u' = 0, \qquad\text{i.e.}\qquad \frac{d}{dr}\bigl(r^2u'\bigr) = 0$.

Step 3: Solving the ODE

The equation is exact as written: integrating once, $r^2u' = C_1 \;\Longrightarrow\; u' = \frac{C_1}{r^2}$.

Integrating a second time, $u = -\frac{C_1}{r} + C_2$.

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

Step 4: Back-Substitution

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

Step 5: Verification

This is the identical function verified in Problem 209: with $s=x^2+y^2+z^2$ and $u_1=s^{-1/2}$, \[u_{1,xx}+u_{1,yy}+u_{1,zz} = -3s^{-3/2} + 3s\cdot s^{-5/2} = -3s^{-3/2}+3s^{-3/2} = 0,\ \checkmark\]

and the constant $C_2$ again contributes nothing to any derivative, so $u=C_1/\sqrt{x^2+y^2+z^2}+C_2$ satisfies the 3D Laplace equation identically for $(x,y,z)\neq(0,0,0)$.

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