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

In summary, solving Poisson's equation in spherical coordinates requires knowledge of Greens functions and Sturm-Liouville theory.
  • #1
sukmeov
10
2
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?
 
Physics news on Phys.org
  • #2
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## ?
 
  • Like
  • Haha
Likes Astronuc, PhDeezNutz, Delta2 and 3 others
  • #3
My surname's funny? Many thanks.
 
  • #4
sukmeov said:
Also it's not a nickname it's my surname.
Ah ... Sorry Ukan !
 
  • Like
Likes sukmeov
  • #5
This is a linear non-homogeneous equation so that you can first find the general solution to the associated homogeneous equation, [tex]U_{xx}+ U_{yy}+ U_{zz}= 0[/tex], then add anyone solution, such as [tex]U= \frac{C}{2}x^2[/tex], to the entire equation to get the general solution to the entire equat0ion.
 
  • Like
Likes Delta2 and roam
  • #6
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.
 
  • #7
sinse ##C## is a constant it is just an ODE in the spherical coordinates
 
  • #8
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.
 
  • #9
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.
 
  • Like
  • Haha
Likes Delta2, Chestermiller and etotheipi
  • #10
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.
 
  • Like
Likes PhDeezNutz
  • #11
Solving [tex]
\frac1{r^2} \frac{d}{dr}\left(r^2\frac{du}{dr}\right) = C[/tex] subject to [itex]u(1) = 0[/itex] 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 [itex](\theta,\phi)[/itex] then the solution probably doesn't, that the boundary condition at the origin is that [itex]u[/itex] is finite, and that the uniqueness theorem guarantees that if we find a solution then it must be the solution.
 
  • Like
Likes PhDeezNutz
  • #12
pasmith said:
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.
 
  • #13
Did we all notice OP wasn't seen since June 28 (this year) ? :rolleyes:
 
  • Like
Likes PhDeezNutz and Delta2

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

1. What is the general approach to solving Uxx+Uyy+Uzz=C when C is non-zero?

The general approach to solving this type of equation is to use the method of separation of variables. This involves separating the equation into three ordinary differential equations, one for each variable, and then solving each equation separately.

2. How do I determine the boundary conditions for this type of equation?

The boundary conditions for this type of equation are typically given as part of the problem or can be determined by considering the physical or mathematical constraints of the problem. These conditions are necessary for finding a unique solution to the equation.

3. Can this type of equation be solved analytically?

Yes, this type of equation can be solved analytically using separation of variables. However, in some cases, the solution may involve complex or special functions that cannot be expressed in closed form.

4. Is there a specific method or technique for solving this type of equation?

As mentioned before, the method of separation of variables is commonly used to solve this type of equation. However, there are other techniques such as the method of eigenfunction expansion or the method of integral transforms that can also be used.

5. What are some real-life applications of this type of equation?

This type of equation is commonly used in physics and engineering to model various physical phenomena such as heat conduction, diffusion, and wave propagation. It can also be used in finance and economics to model diffusion processes or in biology to model population growth and diffusion of substances in cells.

Similar threads

Replies
1
Views
2K
Replies
4
Views
2K
  • Differential Equations
Replies
2
Views
2K
  • Topology and Analysis
Replies
3
Views
2K
  • Differential Equations
Replies
2
Views
630
  • Differential Equations
Replies
2
Views
1K
  • Differential Equations
Replies
5
Views
919
Replies
3
Views
1K
Replies
65
Views
2K
  • Differential Equations
Replies
16
Views
1K
Back
Top