Find the most general u=u(r) satisfying (∂^2u)/(∂…

Find the most general $u=u(r)$ satisfying $\frac{\partial^2u}{\partial x^2}+\frac{\partial^2u}{\partial y^2} - \frac{4u}{x^2+y^2} = 0$ under the substitution $r=\sqrt{x^2+y^2}$.

Step 1: Chain Rule Derivatives

Here $\partial r/\partial x=x/r$ and $\partial r/\partial y=y/r$, so $\partial u/\partial x = xu'/r$. Differentiating again (product rule on $x$ and $u'/r$, then chain and quotient rule):

\[ \frac{\partial^2u}{\partial x^2} = \frac{u'}{r} + x\left(\frac{x\,u''}{r^2}-\frac{x\,u'}{r^3}\right) = \frac{u'}{r} + \frac{x^2u''}{r^2} - \frac{x^2u'}{r^3}. \]

By symmetry ($x\leftrightarrow y$), \[\frac{\partial^2u}{\partial y^2} = \frac{u'}{r} + \frac{y^2u''}{r^2} - \frac{y^2u'}{r^3}.\]

Step 2: Substitution into the PDE

Adding the two second partials and using $x^2+y^2=r^2$: \[\frac{\partial^2u}{\partial x^2}+\frac{\partial^2u}{\partial y^2} = \frac{2u'}{r} + \frac{r^2u''}{r^2} - \frac{r^2u'}{r^3} = \frac{2u'}{r} + u'' - \frac{u'}{r} = u'' + \frac{u'}{r}.\]

The equation becomes, with $x^2+y^2=r^2$ in the last term, $u'' + \frac{u'}{r} - \frac{4u}{r^2} = 0$.

Multiplying by $r^2$ puts it in equidimensional (Euler–Cauchy) form: $r^2u'' + r\,u' - 4u = 0$.

Step 3: Solving the ODE

Try the power ansatz $u=r^k$: substituting, $u'=kr^{k-1}$, $u''=k(k-1)r^{k-2}$, and

\[ k(k-1)r^k + k\,r^k - 4r^k = \bigl[k^2-4\bigr]r^k = 0 \;\Longrightarrow\; k^2=4 \;\Longrightarrow\; k=\pm2. \]

The two roots are real and distinct, giving the general solution

\[ u = C_1r^2 + \frac{C_2}{r^2}. \]

Step 4: Back-Substitution

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

Step 5: Verification

For $u_1=x^2+y^2$: $\partial^2u_1/\partial x^2=2$, $\partial^2u_1/\partial y^2=2$, so the left side is $2+2-\dfrac{4(x^2+y^2)}{x^2+y^2}=4-4=0.\ \checkmark$

For $u_2=1/(x^2+y^2)$, write $s=x^2+y^2$. Then $u_{2,x}=-2x\,s^{-2}$ and

\[ u_{2,xx} = -2s^{-2} + 8x^2s^{-3}, \qquad u_{2,yy} = -2s^{-2} + 8y^2s^{-3}\ \text{(by symmetry)}. \]

Summing, \[u_{2,xx}+u_{2,yy} = -4s^{-2} + 8(x^2+y^2)s^{-3} = -4s^{-2} + 8s^{-2} = 4s^{-2}.\]

Meanwhile $4u_2/s = 4s^{-1}/s = 4s^{-2}$, which exactly matches, so the residual is $4s^{-2}-4s^{-2}=0.\ \checkmark$ Both terms of the general solution satisfy the PDE independently, as expected of a linear equation.

\[u = C_1(x^2+y^2) + \frac{C_2}{x^2+y^2}\]