trying to solve
$$\frac{d}{dr}\bigg(\frac{1}{r} \frac{d}{dr}(r\bar u_r)\bigg) - k^2 \bar u_r = \frac{k^2}{\omega^2}\phi(r) \bar u_r $$
subject to
$$R_1 \leq r \leq R_2,$$ $$\bar u_r(R_1) = \bar u_r(R_2) = 0,$$
where
$$\phi (r) = \frac{2V}{r^2}\frac{d}{dr}(rV).$$
I am trying to solve for $\bar u_r $ in the case where $V(r) = 0.$
My attempt has got me to
$$\frac{d^2 \bar u_r}{dr^2} + \frac{1}{r} \frac{d \bar u_r}{dr} - \bar u_r \bigg(\frac{1}{r^2}+k^2 \bigg).$$
When I try to solve this I end up with a zero solution? WolframAlpha is giving me solutions to this as Bessel functions. Could anyone point me in the right direction?
I should note, this is a fluid dynamics problem.
$$\frac{d^2 \bar u_r}{dr^2} + \frac{1}{r} \frac{d \bar u_r}{dr} - \bar u_r \bigg(\frac{1}{r^2}+k^2 \bigg)=0$$
Knowing the properties of the Bessel functions and related differential equations, it is obvious that the solution is : $$\bar u_r(r)=c_1 I_k(r)+c_2 K_k(r)$$ The functions $I_k(r)$ and $K_k(r)$ are the Modified Bessel functions of first and second kind respectively.
http://mathworld.wolfram.com/ModifiedBesselDifferentialEquation.html
http://mathworld.wolfram.com/ModifiedBesselFunctionoftheSecondKind.html
http://mathworld.wolfram.com/ModifiedBesselFunctionoftheFirstKind.html
There is nothing to prove since this is the direct consequence of the definition of the Bessel functions and of the basic properties.
Of course, $\bar u_r(r)=0$ is a particular solution which satisfies the ODE (case $c_1=c_2=0$ ).
In the above general solution $c_1$ and $c_2$ are arbitrary constants. Those constants can be determined if some well posed boundary conditions are specified.
If in trying to solve the PDE you end up with the zero solution only, several causes can be suspected. A possible one could be that you introduces some "hided" or "implicit" conditions which are not in the wording of the problem. All depends on what method of solving you use and what exactly is your calculus. Without checking all steps of your calculus and without seeing all your intermediate equations it is impossible to give a definitive diagnosis.