Find the most general u=u(r) satisfying Laplace's equation…

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

Step 1: Chain Rule Derivatives

Since $u$ depends on $x$ and $y$ only through $r=x^2+y^2$, with $\partial r/\partial x=2x$ and $\partial r/\partial y=2y$, the chain rule gives

\[ \frac{\partial u}{\partial x} = u'(r)\cdot\frac{\partial r}{\partial x} = 2x\,u'. \]

Differentiating again with the product rule (and the chain rule once more on $u'$): \[\frac{\partial^2u}{\partial x^2} = 2u' + 2x\cdot\frac{\partial u'}{\partial x} = 2u' + 2x\,(2x\,u'') = 2u' + 4x^2u''.\]

By the identical computation with $x$ and $y$ interchanged ($\partial r/\partial y=2y$), $\frac{\partial^2u}{\partial y^2} = 2u' + 4y^2u''$.

Step 2: Substitution into the PDE

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

Dividing by $4$ leaves an ordinary differential equation in $r$ alone: $r\,u'' + u' = 0, \qquad\text{i.e.}\qquad \frac{d}{dr}\bigl(r\,u'\bigr) = 0$.

Step 3: Solving the ODE

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

Integrating a second time, $u = C_1\ln r + C_2$.

Step 4: Back-Substitution

Replacing $r$ by $x^2+y^2$ recovers $u$ as a function of $x$ and $y$: $u = C_1\ln(x^2+y^2) + C_2$.

Step 5: Verification

Differentiating $u=C_1\ln(x^2+y^2)+C_2$ directly: \[\frac{\partial u}{\partial x} = \frac{2C_1x}{x^2+y^2}, \qquad \frac{\partial^2u}{\partial x^2} = \frac{2C_1(y^2-x^2)}{(x^2+y^2)^2},\]

\[ \frac{\partial u}{\partial y} = \frac{2C_1y}{x^2+y^2}, \qquad \frac{\partial^2u}{\partial y^2} = \frac{2C_1(x^2-y^2)}{(x^2+y^2)^2}. \]

Adding the two second partials, the numerators $(y^2-x^2)$ and $(x^2-y^2)$ cancel exactly, leaving $\partial^2u/\partial x^2+\partial^2u/\partial y^2\equiv0$ for every $(x,y)\neq(0,0)$, i.e. $\checkmark$.

\[u = C_1\ln(x^2+y^2) + C_2\]