Laplace equation in cylindrical coordinates

In summary, In this conversation, Arildno is trying to solve the Laplace equation in cylindrical coordinates, but is getting stuck. He has found that the two ordinary diff.eqs can be solved using a trial solution, but the more problematic solution, (*), is more difficult. He has also found that the solution to (*) is of the form: p(r,z) = A[cosh(\lambda z)-sinh(\lambda z)]J_{0}(\lambda r) with J'_{0}(0\lambda, r=0) being zero. He is still trying to figure out how to use this information to solve for p in terms of r.
  • #1
alexrao
4
0
Can anyone help with the solution of the Laplace equation in cylindrical coordinates
[tex]\frac{\partial^{2} p}{\partial r^{2}}[/tex] [tex]+[/tex] [tex]\frac{1}{r}[/tex] [tex]\frac{\partial p}{\partial r}[/tex] [tex]+[/tex] [tex]\frac{\partial^{2} p}{\partial z^{2}}[/tex] [tex]= 0 [/tex]

with Neumann no-flux boundaries:
[tex]\frac{\partial p}{\partial r}[/tex] [tex]\left(0,z\right)[/tex] [tex] = 0[/tex]
[tex]\frac{\partial p}{\partial r}[/tex] [tex]\left(Rmax,z\right)[/tex] [tex] = 0[/tex]
[tex]\frac{\partial p}{\partial z}[/tex] [tex]\left(r,Zmax\right)[/tex] [tex] = 0[/tex]

and a Dirichlet upper boundary:
[tex]p(r,0) = f(r)[/tex]
 
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  • #2
With a trial solution using separation of variables, p(r,z)=P(r)*Q(z), we get:
[tex]\frac{P''(r)+\frac{1}{r}P'(r)}{P(r)}=\frac{1}{Q(z)}Q''(z)[/tex]

Thus, we get two ordinary diff.eqs,
[tex]P''(r)+\frac{1}{r}P'(r)=CP(r) (*)[/tex]
[tex]Q''(z)=CQ(z)(**)[/tex]

You might try to work with these two, (*) clearly being the most problematic one.
 
  • #3
arildno said:
With a trial solution using separation of variables, p(r,z)=P(r)*Q(z), we get:
[tex]\frac{P''(r)+\frac{1}{r}P'(r)}{P(r)}=\frac{1}{Q(z)}Q''(z)[/tex]

Thus, we get two ordinary diff.eqs,
[tex]P''(r)+\frac{1}{r}P'(r)=CP(r) (*)[/tex]
[tex]Q''(z)=CQ(z)(**)[/tex]

You might try to work with these two, (*) clearly being the most problematic one.

But maybe not too problematic. Multiply it through by r2 and you have a form of Bessel's equation.
 
  • #4
Thanks for getting me started, and pardon the ignorance of this geochemist who hasn't taken a pde class. I'm trying to follow you, and can't understand how you got that 2nd ode. Shouldn't (**) be

[tex]Q''(z) = -C [/tex] [tex]Q(z)[/tex] [tex] ?[/tex]
 
  • #5
You are BOTH right. Mea culpa.
 
  • #6
Moving along here...

So the solution to (**):

[tex] Q(z) = Acosh(\lambda z)+Bsinh(\lambda z)[/tex]

As Arildno predicted, the solution to (*) is more problematic, since this is my first experience with Bessel functions. Here's where I am so far...

[tex]P(r) = CJ_{0}(\lambda r)+DY_{0}(\lambda r)[/tex]

And here I set the second term must go to zero for the solution to be bounded.

So the final solution is of the form:
[tex]p(r,z) = [Acosh(\lambda z)+Bsinh(\lambda z)]J_{0}(\lambda r)[/tex]

Using the 2nd Neumann BC:
[tex]\frac{\partial p}{\partial r}(r=Rmax=9.5) = 0[/tex]
This means that
[tex]J'_{0}(9.5\lambda) = \frac{\partial J_{0}}{\partial r}= 0[/tex]
But I'm not sure how this helps me right now, so I'll keep it in mind...

Using the 3rd Neumann BC:
[tex]\frac{\partial p}{\partial z}(z=Zmax=20) = 0[/tex]
This means that
[tex][Acosh(20\lambda)+Bsinh(20\lambda)]=0[/tex]
Ah, this is more helpful. I learned that A=-B, and I think at the bottom boundary, this whole term drops out.

Using the 1st Neumann BC: (I guess I could have done these in order...)
[tex]\frac{\partial p}{\partial r}(r=0) = 0[/tex]
This means that
[tex]J'_{0}(0\lambda) = \frac{\partial J_{0}}{\partial r}= 0[/tex]
Again I'm not really sure how this helps me right now, so I'll keep it in mind...

So now we have:
[tex]p(r,z)=A[cosh(\lambda z)-sinh(\lambda z)]J_{0}(\lambda r)[/tex]

with
[tex]J'_{0}(0\lambda, r=0) = J'_{0}(9.5\lambda, r=9.5) = 0[/tex]

Now here's where it gets tricky and I get stuck...

Using the fourth Dirichlet upper (z=0) boundary condition, I get:

[tex]p(r,0)=f(r)=\sum_{\infty}^{n=1}A_{n}[cosh(0)-sinh(0)]J_{0}(\lambda _{n}r)=\sum_{\infty}^{n=1}A_{n}J_{0}(\lambda _{n}r)[/tex]

Unfortunately, f(r) is not a constant, but rather an ugly empirical function of r.

So I have written:

[tex]f(r)=\sum_{\infty}^{n=1}A_{n}J_{0}(\lambda _{n}r)[/tex]

And I have no idea where to go from here. I gather this will eventually end up in Matlab or R, but first I need a better understanding of the Bessel function J0 and A. I've read that the Bessel function of the first kind of order n (in my case n=0, right?) can be expressed as:

[tex]J_{n}(x) = \sum_{\infty}^{k=1}\frac {(-1)^{k}(x/2)^{n+2k}}{k!\Gamma (n+k+1)}}[/tex]

Again, any help would be greatly appreciated.
 
  • #7
Now, assuming that the class of J_0's, with scaled arguments, represents a COMPLETE BASIS for functions on R (i.e, that any function is representable as a linear combination of these J_0's), then the A_n's are simply the required coefficients.
 
  • #8
I haven't checked all your details, but your P equation and boundary conditions appear to be a Sturm-Liouville system which answers questions about orthogonality of the eigenfunctions, gives formulas for the eigenfunction expansion coefficients and settles convergence. For example, see:

http://www.efunda.com/math/ode/Sturm_liouville.cfm
 
  • #9
Thanks again. Arildno, you mentioned that the [tex]A_{n}[/tex]'s are the required coefficients. But if I want to define a pressure distribution in this cylinder, don't I also need to figure out [tex]\lambda[/tex]? And is it true that [tex]\lambda[/tex] in P(r) and [tex]\lambda[/tex] in Q(z) are not the same?
 
  • #10
No, the [itex]\lambda_{n}[/itex] MUST be the same numbers; otherwise, your diff.eq won't be satisfied. (Remember, it is directly related to your C's!)
 

1. What is the Laplace equation in cylindrical coordinates?

The Laplace equation in cylindrical coordinates is a partial differential equation that describes the variation of a scalar field in three-dimensional space. It is expressed as: ∇²u = 0, where u is the scalar field and ∇² is the Laplace operator.

2. What are the advantages of using cylindrical coordinates in solving the Laplace equation?

Cylindrical coordinates are particularly useful for solving problems with cylindrical symmetry, such as those involving cylinders, pipes, or rotating objects. They also simplify the Laplace equation, making it easier to solve compared to other coordinate systems.

3. How is the Laplace equation in cylindrical coordinates derived?

The Laplace equation in cylindrical coordinates can be derived from the Laplace equation in Cartesian coordinates using the chain rule and the transformation equations for converting between the two coordinate systems. The resulting equation is then simplified by taking advantage of the cylindrical symmetry.

4. What are some applications of the Laplace equation in cylindrical coordinates?

The Laplace equation in cylindrical coordinates has many applications in physics and engineering, including electrostatics, fluid dynamics, heat transfer, and quantum mechanics. It is used to model and solve problems involving charged particles, fluid flow, temperature distribution, and quantum systems with cylindrical symmetry.

5. Are there any limitations to using the Laplace equation in cylindrical coordinates?

While the Laplace equation in cylindrical coordinates is useful for solving problems with cylindrical symmetry, it is limited to problems that can be simplified to two dimensions. It also assumes that the scalar field being studied is continuous and has a well-defined value at every point in space.

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