Trouble Solving Partial Differential Equation I'm solving the velocity profile of a fluid flow for a circular channel with an oscillating pressure gradient $\frac{dp}{dx}=\frac{\Delta p}{\rho L}e^{-i\omega t}$. I plugged in to the Navier Stokes equations and am having trouble figuring out how to approach the solution to the partial differential equation below for u(r,t).
$$ \frac{\partial u}{\partial t}  - \frac{\Delta p}{\rho L} e^{-i\omega t} = \mu\left(\frac{\partial ^2}{\partial r^2} + \frac{1}{r}\frac{\partial}{\partial r}\right)u$$
I know if the $\frac{\Delta p}{\rho L} e^{-i\omega t}$ term weren't in the equation, I could use separation of variables, but the addition of that term seems to throw a wrench in things.
Any advice would be greatly appreciated.
 A: I disagree with Chet, I think complex numbers are the way to go here.
As ever in these kind of problems, given that we have an oscillating pressure gradient, it makes sense to look for an oscillating  velocity profile,
$$
u(r,t) = \hat{u}(r)e^{-i\omega t}.
$$
Then your PDE reduces to the ODE
$$
r^2\frac{d^2\hat{u}}{dr^2} + r\frac{d\hat{u}}{dr} + \frac{i\omega r^2}{\mu}\hat{u} + \frac{\Delta p r^2}{\rho L\mu} = 0.
$$
Now make the change of variables $\hat{u} = U(r) - \frac{\Delta p}{\rho L i \omega}$ and $x = r\sqrt{\frac{i\omega}{\mu}}$. Denoting derivatives with respect to $x$ with primes, this becomes
$$
x^2U'' + xU' + x^2U=0.
$$
This is Bessel's Equation, which has solutions given by Bessel functions.
A: I never use complex numbers on a problem like this.  I would represent u as $$u=A(r)\cos{\omega t}+B(r)\sin{\omega t}\tag{1}$$  This leads to two coupled ODEs in r for A and B.
So I would start with $$ \frac{\partial u}{\partial t}  - \frac{\Delta p}{\rho L} \cos{\omega t} = \nu\left(\frac{\partial ^2}{\partial r^2} + \frac{1}{r}\frac{\partial}{\partial r}\right)u\tag{2}$$where $\nu$ is the kinematic viscosity.  And substituting Eqn. 1 into this equation, I would obtain:  $$\nu\left[A''+\frac{A'}{r}\right]-\omega B=-\frac{\Delta P}{\rho L}\tag{3}$$and $$\nu\left[B''+\frac{B'}{r}\right]+\omega A=0\tag{4}$$
