Can Laplace's Equation Be Solved in Cylindrical Coordinates on a 3-Sphere?

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In summary: Unfortunately, there is no simple way to solve the non-separable equation in cylindrical coordinates. You would need to use numerical methods to find the solutions.
  • #1
Old Smuggler
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I wish to find exact solutions of Laplace's equation in cylindrical coordinates on (a subset of) the 3-sphere.
This pde is linear but not separable. The potential [itex]{\Phi}(x,z)[/itex] must fulfil the following pde:

[tex]
(1-{\frac{x^2}{a^2}}){\frac{{\partial}^2}{{\partial}x^2}}{\Phi}(x,z)+
(1-{\frac{z^2}{a^2}}){\frac{{\partial}^2}{{\partial}z^2}}{\Phi}(x,z)+
{\frac{1}{x}}(1-{\frac{3x^2}{a^2}}){\frac{{\partial}}{{\partial}x}}{\Phi}(x,z)-
{\frac{2xz}{a^2}}{\frac{{\partial}^2}{{\partial}x{\partial}z}}{\Phi}(x,z)-
{\frac{3z}{a^2}}{\frac{{\partial}}{{\partial}z}}{\Phi}(x,z)=0
[/tex]

Here a is a constant (and [tex]x,z<a[/tex], [tex]z{\neq}0[/tex]). Does anyone know how to solve this equation?
(I'm aware that a transformation of this equation to spherical coordinates yields a separable pde, but
this gives a bunch of useless solutions blowing up near the origin.)
 
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  • #2
No, it doesn't. One of the "boundary conditions" for a problem like that is that it NOT go to infinity as r goes to 0.

The general solution to the Laplace's equation in spherical coordinates involves a double sum, over l and m, of things like (amlrl+ bml/rl+1)Ylm(theta, phi) where Ylm(theta, phi) are the spherical harmonics. If your region includes the origin, then one of the conditions must be that all bml are 0.
 
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  • #3
HallsofIvy said:
No, it doesn't. One of the "boundary conditions" for a problem like that is that it NOT go to infinity as r goes to 0.

The general solution to the Laplace's equation in spherical coordinates involves a double sum, over l and m, of things like (amlrl+ bml/rl+1)Ylm(theta, phi) where Ylm(theta, phi) are the spherical harmonics. If your region includes the origin, then one of the conditions must be that all bml are 0.

Let me state the problem in more detail. I wish to find the Newtonian potential outside a flat, very thin disk lying in
the surface z=0 in cylindrical coordinates. If one first solves Laplace's equation in cylindrical coordinates in ordinary
Euclidean space, the solution is given by (from MAPLE)
[tex]
{\exp}(-kz)[C_1J_0(kx)+C_2Y_0(kx)] , \qquad (z>0),
[/tex]
where [tex] k, C_1, C_2 [/tex] are constants and [tex] J_0, Y_0 [/tex] are Bessel functions of the first and second kind, respectively ([tex] C_2 [/tex] is set to zero for physical reasons). Since [tex] k [/tex] is a continuous parameter, solutions can be summed by integration over [tex] k [/tex], and using Gauss theorem over the disk, and moreover Hankel transforms, we can find the potential in the disk for any given density distribution. All this is standard stuff. On the other hand, if one changes to spherical coordinates, the solution is
[tex]
{\Big [}C_3r^{-1/2+{\sqrt{1+4C}}/2}+C_4r^{-1/2-{\sqrt{1+4C}}/2}]{\Big ]}
{\Big [}C_5P^0_{-1/2+{\sqrt{1+4C}}/2} ({\cos}{\theta})+C_6Q^0_{-1/2+{\sqrt{1+4C}/2}}({\cos}{\theta}){\Big ]}
[/tex],
where [tex] C, C_3,C_4, C_5, C_6 [/tex] are constants, and [tex] P^{\mu}_{\nu}, Q^{\mu}_{\nu} [/tex] are Legendre functions of the first and second kind, respectively. However, if one reinserts cylindrical coordinates into this solution, it is not clear to me how to get back the above solution involving Bessel functions in a simple manner, and how the constants [tex] k [/tex] and [tex] C [/tex] are related.

Now for the similar problem on a subset of the 3-sphere. Laplace's equation on a subset of the 3-sphere is separable in spherical coordinates, and again the solution involves Legendre functions (but this time with non-zero order).
However, in cylindrical coordinates the equation is non-separable. If there is a simple way of taking the solution for the equation in spherical coordinates, reinsert cylindrical coordinates into this solution and then get something that resembles the flat space solution involving Bessel functions, that would be great. However I don't know how to do that even in the Euclidean case. So it seems that I have to find solutions of the non-separable pde after all.
 
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  • #4
Guys I need help to solve this differential equation:

[tex] (1-x^2)~\frac{\partial ^2 F(x,y)}{\partial x^2}+\left(\frac{1-2x^2}{x}\right)~\frac{\partial F(x,y)}{\partial x}-\left[ax^2(1-2y^2)-c\right]~F(x,y)=0 [/tex] where a and c are constants and x and y are the variables

Does this have a solution in terms of well known functions? Mathematica could not find any so I proposed a non separable series solution of the form :

[tex] \sum_{n}\sum_{p}C_{n,p} x^n y^p [/tex]

But the problem is when I tried to find the eigenvalues of the formed matrix it got too complicated and c (which is a constant in terms of the energy of the system) turned out to be complex in some cases! I want real values as the system would not be physical..can anyone help me in this matter please.

Thanks
 

Related to Can Laplace's Equation Be Solved in Cylindrical Coordinates on a 3-Sphere?

1. What is a non-separable, linear PDE?

A non-separable, linear partial differential equation (PDE) is a type of mathematical equation that describes the relationship between a function of multiple variables and its partial derivatives. It is considered non-separable when the coefficients of the partial derivatives cannot be separated into individual functions of each variable.

2. What are the characteristics of a non-separable, linear PDE?

A non-separable, linear PDE has a linear relationship between the dependent variable and its partial derivatives, meaning that the coefficients of the derivatives are constants or functions of the independent variables. It also cannot be solved by separating variables, as the coefficients cannot be separated into individual functions of each variable.

3. How are non-separable, linear PDEs solved?

Non-separable, linear PDEs are typically solved using methods such as the method of characteristics, separation of variables, or the use of integral transforms. These methods involve transforming the PDE into a simpler form that can be solved using standard techniques.

4. What are some applications of non-separable, linear PDEs?

Non-separable, linear PDEs are commonly used in many areas of science and engineering to model physical phenomena, such as heat transfer, fluid dynamics, and quantum mechanics. They are also used in image and signal processing, as well as financial modeling and economics.

5. What is the difference between a separable and non-separable PDE?

A separable PDE has coefficients that can be separated into individual functions of each variable, allowing for a solution to be found by separating variables. On the other hand, a non-separable PDE has coefficients that cannot be separated, making it more difficult to solve and requiring alternative methods such as the method of characteristics or integral transforms.

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