How does one solve Uxx+Uyy+Uzz=C when C is non-zero?

[sukmeov]

How does one solve the partial differential equation Uxx+Uyy+Uzz=C when C is non-zero. Here U is a function of x,y and z where (x,y,z) lies in the ball centered at 0 of radius 1 and U=0 on the boundary. Uxx, Uyy and Uzz denote second partial derivatives with respect to x, y and z.
Any hints on how to approach this?

 

[BvU]

Funny user name .

Google Poisson equation. Your equation describes the electric potential of a non conducting uniformly charged sphere.

Are you aware of the equation for electric potential $\Delta\phi = C$ ?

 

[sukmeov]

My surname's funny? Many thanks.

 

[BvU]

Also it's not a nickname it's my surname.

Ah ... Sorry Ukan !

 

[HallsofIvy]

This is a linear non-homogeneous equation so that you can first find the general solution to the associated homogeneous equation, $$U_{xx}+ U_{yy}+ U_{zz}= 0$$, then add anyone solution, such as $$U= \frac{C}{2}x^2$$, to the entire equation to get the general solution to the entire equat0ion.

 

[phyzguy]

When you say C is non-zero, do you mean that C is a non-zero constant or that C is also a function of x,y,z? In either case, the answer, as @BvU said, is that this is Poisson's equation. There are several methods known for solving this, depending on what you know about C.

 

[wrobel]

sinse $C$ is a constant it is just an ODE in the spherical coordinates

 

[peterwang]

given that you have a perfect boundary condition: U=0 at r=1. The Sturm-Lioiville eigenproblem is designed to deal with such proper boundary condition. The boundary condition combined with the spherical coordinate gives orthorgonal basis set, each function of the basis set is product of spherical function and the Bessel function. The solution can be expanded using this basis set, and coefficients of the expansion can be fixed using C.

 

[PhDeezNutz]

As stated in post #7 since you have a spherical boundary conditions you want to solve this in spherical coordinates. I think the best way to go about this is to use greens functions...and I just realized this thread is several months old.

 

[Delta2]

Another way to solve Poisson's Equation if there is symmetry in the source term (like in this example).

First solve for electric field $\mathbf{E}=\nabla U$ and $\nabla\cdot\mathbf{E}=C$ by using Gauss' law in integral form and taking advantage of the symmetry.

Then solve for U by using $$U=\int_{\mathcal{C}} \mathbf{E}\cdot d\mathbf{l}$$ where $\mathcal{C}$ is a suitable path of integration, depending of what the form of E is. In this example I expect E to be in the radial direction, hence a suitable path of integration is along the radial line from position r to infinity.

 

[pasmith]

Solving $$\frac1{r^2} \frac{d}{dr}\left(r^2\frac{du}{dr}\right) = C$$ subject to $u(1) = 0$ doesn't really require resort to Green's funcions or Sturm-Liouville theory.

Most of the work here is in reducing the problem to that: knowing to use spherical polars and understanding that if the boundary condition does not depend on $(\theta,\phi)$ then the solution probably doesn't, that the boundary condition at the origin is that $u$ is finite, and that the uniqueness theorem guarantees that if we find a solution then it must be the solution.

 

[Delta2]

understanding that if the boundary condition does not depend on (θ,ϕ) then the solution probably doesn't

Not only the boundary condition but the source term also doesn't depend on $\theta,\phi$ and hence the Laplacian operator $\nabla^2$ reduces to $\frac{1}{r^2}\frac{d}{dr}\left (r^2\frac{d}{dr}\right )$, something that wasn't obvious from the start , because the problem was expressed with the Laplacian operator in cartesian coordinates.

 

[BvU]

Did we all notice OP wasn't seen since June 28 (this year) ?