# Trouble Solving Partial Differential Equation [closed]

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.

• Relevant meta post about some previous answers and possible close reason Commented Mar 8, 2020 at 4:02

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.

• Both methods lead to the same results. I just personally prefer not to complex numbers in such problems. Commented Mar 6, 2020 at 23:34
• Yes, its definitely a personal preference. I've always preferred exponentials to trig. Commented Mar 6, 2020 at 23:35
• That's fine. But, in this problem (and ones like it), the answer is in terms of trig functions. Commented Mar 7, 2020 at 1:26
• Plus, shouldn't the pressure term in your equation also have an $r^2$? And, rather than the dynamic viscosity $\mu$, shouldn't this be the kinematic viscosity $\nu$? Commented Mar 7, 2020 at 4:59
• Is there any meaning to the substitution $x=\sqrt{i\omega/\mu}$ in terms of natural units of the problem? Commented Mar 7, 2020 at 21:52

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 $$\nu\left[B''+\frac{B'}{r}\right]+\omega A=0\tag{4}$$

• So I took your suggestion and plugged in that representation of u and got the following: $\mu A''(r) + \frac{\mu}{r}A'(r)-\omega B(r) = -\frac{\Delta p}{\rho L}$ and $mu B'(r) + \frac{\mu}{r}B'(r) + \omega A'(r) = -\frac{\Delta p i}{\rho L}$ I'm not sure how to continue from here as my experience with these kind of ODEs is limited. Would you have any advice as to how to continue? Thanks! Commented Mar 6, 2020 at 16:10
• Only one of the equations should have a pressure term , and the 2nd equation should not have an A’. Solve the 1st Eqn for B, and substitute into the 1st equation. Commented Mar 6, 2020 at 16:27
• It should be just an A. Commented Mar 6, 2020 at 17:48
• See the extended version of my answer. Commented Mar 6, 2020 at 21:35