Tough pde

1. May 19, 2014

joshmccraney

hey pf!

i was wondering if you could help me out with a pde, namely $$\alpha ( \frac{z}{r} \frac{\partial f}{\partial r} + \frac{\partial f}{\partial z} ) = \frac{2}{r} \frac{\partial f}{\partial r} + \frac{\partial^2 f}{\partial r^2} + \frac{\partial^2 f}{\partial z^2} + 2 \frac{z}{r} \frac{\partial^2 f}{\partial r \partial z}$$
i won't list the boundary conditions, as i'm just trying to find a general solution for now. i tried the substitution $r^2 = z^2 + y^2$ which changed the equation to separable (it's not currently separable, i've tried). i then used a bessel function and an exponential but could not fit them to the boundary conditions. i know an analytical solution exists, but i'm not sure how to get there. the solution seems to be separable in $r$ and $z$ but the equation is not. this is ironic because if i turn the equation into a separable PDE the solution is not separable.

please help me out!

thanks!

2. May 19, 2014

Simon Bridge

Why would you try $r^2=z^2+y^2$?
If this were cylindrical-polar coordinates, then I'd expect $r^2=x^2+y^2$.
Where does the PDE come from, what is the context of the problem?

Note: IRL it is normal for the solution to refuse to come out untangled.
What did you get?

3. May 21, 2014

Sunfire

This is one unfortunate PDE
Not only there is a mixed derivative and second-order derivatives, but also first-order ones. And some of the coefficients have the form $\frac{z}{r}$ so one cannot group them into $F(r,\frac{\partial f}{\partial r}, \frac{\partial ^2 f}{\partial r^2}, \frac{\partial ^2 f}{\partial r \partial z})$ and $F(z,\frac{\partial f}{\partial z}, \frac{\partial ^2 f}{\partial z^2}, \frac{\partial ^2 f}{\partial r \partial z})$.

Let's say, that for any reason, you have introduced the substitution $r^2 - z^2 = y^2$.
Could you write the PDE you are getting as a result?

4. May 21, 2014

joshmccraney

the problem arises from a mass transfer problem. originally, mass transfer is governed by the following equation: $$\frac{D f}{D t} = \gamma \nabla^2 f$$ where $f$ is the mass, $\gamma$ is a diffusivity constant, and $\frac{D}{Dt}$ is the substantial, or material, derivative. now i have a flow in the $z$ direction, and mass is being injected into the flow from a point. if we are in cylindrical coordinates and make the substitution that $s^2 = r^2 + z^2$ we arrive at the equation i have posted after some physical simplifications (the letters i have used in my previous post are different, but you get the idea). because of the boundary conditions, before the substitution we cannot solve by separation of variables. after the substitution, the solution, in terms of $s$ and $z$ is separable, but we cannot separate the equation. my question is, how can we solve the posted pde (which is after we apply the substitution, since the solution is analytic).

to satisfy any curiosity, the posted solution is $$\frac{C_1}{s} e^{-C_2(s-z)}$$ where $C_1$ and $C_2$ are constants determined by boundary conditions, which i can post if you would like.

i'm not sure what you mean here?

thanks so much for your help. i appreciate your continued help!

5. May 21, 2014

joshmccraney

i know! and it's killing me that there is an analytic solution and i cant get there! but i'm not too sure what you are saying in your post...

6. May 21, 2014

Staff: Mentor

If the equation in #4 is the correct solution, one obvious thing to try is to change variables from r and z to s and z.

Chet

7. May 21, 2014

joshmccraney

Hey Chet! Yea, I've done this and the above differential equation is what results (the one in my first post). I'm sorry for confusing everyone with the variables. But, although an analytic solution exists, how to get it??

8. May 21, 2014

joshmccraney

In other words, the $s$ in the solution is the $r$ in the initial pde.

9. May 21, 2014

Staff: Mentor

Did you substitute the solution to confirm?

Chet

10. May 21, 2014

joshmccraney

yes, in latex. if you would like i can send you the pdf of my work?

i posted the pde as is (after the substitution) because i feel if the solution is in terms of $s$ and $r$ then the solution arose given the substitution. but how was the solution obtained?

the work i did regarding the bessel equations does not agree with the boundary conditions.

what is your take on how to proceed? (spherical coordinates are a real mess, and are not separable).

11. May 21, 2014

Staff: Mentor

Is this the same problem we have been corresponding about via email? It doesn't seem like it. If you fully convert to spherical coordinates, the exponent in the "solution" would be s (1-cosθ).
I don't know whether this would be of any help or not.

Chet

12. May 21, 2014

joshmccraney

Yes, this is the same problem. I have emailed you my work on the substitution. I'm still unsure how to obtain the answer, though.

13. May 21, 2014

Staff: Mentor

The solution in #4 is certainly a function of s times a function of z. So, how come when you applied the separation of variables technique, it didn't deliver this solution? Let's see the initial part of your setup. This solution obviously doesn't involve Bessel Functions.

Chet

14. May 21, 2014

joshmccraney

Notice the governed equation derives from $$\alpha \vec{v} \cdot \nabla f = \nabla ^2 f$$ such $\alpha := v_0 / \gamma$, the mainstream velocity parallel to the $z$ axis divided by the diffusive constant.

To proceed in spherical coordinates, recognize that $\vec{v} = - v_0 \cos \theta \hat{r} + v_0 \sin \theta \hat{\theta}$, where $\hat{x}$ is the unit vector in the $x$ direction. There is no flow in the polar angle. Taking the definition of del and the laplacian in spherical coordinates (where we have symmetry through $\phi$) yields:
$$\alpha \left( -v_0 \cos \theta \frac{\partial f}{\partial r} +v_0 \sin \theta \frac{1}{r} \frac{\partial f}{\partial \theta} \right) = \frac{2}{r} \frac{\partial f}{\partial r} + \frac{\partial^2 f}{\partial r^2} + \frac{1}{r^2 \tan \theta}\frac{\partial f}{\partial \theta} + \frac{1}{r^2}\frac{\partial^2 f}{\partial \theta^2}$$
but this is not separable. For example, if we try to free the last term of the above equation by multiplying by $r^2$, we then have $r$ appearing in the second term on the left.

this makes me think spherical is just as troublesome of an equation as is using the $s^2 = r^2 + z^2$ substitution, except I do have the boundary conditions for the substitution.

a good idea though!

15. May 21, 2014

joshmccraney

i just realized you asked to see my work for the transform involving $s^2 = r^2 + z^2$. i emailed this to you. what i have posted above is the spherical transform.

if simon and sunfire want my work for this substitution, i can post it.

16. May 21, 2014

Staff: Mentor

I got the opposite sign for the left hand side. I took θ to be measured with respect to the + z direction.

Chet

17. May 21, 2014

Simon Bridge

You are probably best to stay with Chet - I'll keep watching.

18. May 21, 2014

joshmccraney

Cool, we're doing something right then! But now what? It's not separable, so how to proceed? Any idea?

19. May 21, 2014

joshmccraney

Finally, this one is taken care of! Chet, I've emailed you a pdf. Thanks for your interest help and patience. And thanks for everyone else's interest!

20. May 22, 2014

Staff: Mentor

I looked over the solution you sent me, and it looks right. Now, the only question is, "how could anyone have come to this solution." It certainly doesn't look like anything one would think of immediately (i.e., the part with the betas).

The boundary conditions suggest that, at small s (r in this thread), the solution is going to have to approach f->W/(4πCs), so this suggests the s in the denominator of #4. But I wouldn't have thought of expressing the overall solution as this times a function of s-z.

I don't have time to discuss this further now, but I'll be back later after I have time to play with the equations a little.

(Another peculiar feature of the solution is that, along the z axis, at s = z, the solution is the same as the pure point source solution without convection).

Chet

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