partial Helmholtz differential equation in two dimensions

Question

The partial Helmholtz differential equation in two dimensions has in polar coordinates $(r,\varphi)$ $x=r\cos(\varphi)$, $y=r\sin(\varphi)$ the form:

$u_{rr}(r,\varphi)+\frac1r u_r(r,\varphi)+\frac1{r^2}u_{\varphi\varphi}(r,\varphi)+u(r,\varphi)=0$

I am told to make the approach to set $u(r,\varphi)=v(r)w(\varphi)$ to seperate $r$ and $\varphi$ in the equation and solve the resulting differential equations.

I got:

$v''(r)w(\varphi)+\frac1rv'(r)w(\varphi)+\frac1{r^2}w''(\varphi)v(r)+w(\varphi)v(r)=0$

But how do I seperate now $\varphi$ and $r$?

The resulting differential equations, in $v$ and $w$, are supposed to have 'names'. Which commen differential equations do you get?

Thanks in advance.

Answer 1

Divide both sides by $vw$: $$\frac{v''}{v}+\frac{1}{r}\frac{v'}{v}+\frac{1}{r^2}\frac{w''}{w}+1=0$$ $$r^2\frac{v''}{v}+r\frac{v'}{v}+r^2=-\frac{w''}{w}$$ The RHS depends on $\varphi$ only, while the LHS depends on $r$ only. It can happen only if both of them are constant: $$-\frac{w''}{w}=k$$ $$r^2\frac{v''}{v}+r\frac{v'}{v}+r^2=k$$ Or if we know the sign of k, we can get 'nicer' equations: If $k>0$, then let $k=a^2$, so we will have that $$w''=-a^2w$$ $$r^2v''+rv'+v(r^2-a^2)=0$$ The second equation is called Bessel differential equation.
If $k<0$, let $-b^2=k$, so we have that $$w''=b^2w$$ $$r^2v''+rv'+v(r^2+b^2)=0$$ And I don't know the name of this one.

Answer 2

$$v''(r)w(\varphi)+\frac1rv'(r)w(\varphi)+\frac1{r^2}w''(\varphi)v(r)+w(\varphi)v(r)=0$$ $$\frac{v''(r)}{v(r)}+\frac1r \frac{v'(r)}{v(r)}+\frac1{r^2}\frac{w''(\varphi)}{w(\varphi)}+1=0$$ $$r^2\frac{v''(r)}{v(r)}+r \frac{v'(r)}{v(r)}+\frac{w''(\varphi)}{w(\varphi)}+r^2=0$$ $$r^2\frac{v''(r)}{v(r)}+r \frac{v'(r)}{v(r)}+r^2=-\frac{w''(\varphi)}{w(\varphi)}=\lambda=\text{constant}$$ $$\begin{cases} r^2\frac{v''(r)}{v(r)}+r \frac{v'(r)}{v(r)}+r^2=\lambda \\ \frac{w''(\varphi)}{w(\varphi)}=-\lambda \end{cases}$$ The first ODE involves Bessel functions.