Partial differential equation

  • Thread starter JustinLevy
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  • #1
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I'm having trouble solving an equation. Can anyone help?

There is a function V(x,y,z) such that:
[tex]\nabla^2 V = \delta(x)\delta(y)\delta(z) \frac{q}{\epsilon_0} [/tex]
with boundary conditions at infinity V=0.

Since there is spherical symmetry, I was able to rewrite it just in terms of the radial distance, and got:
[tex]\nabla^2 V = \frac{q}{\epsilon_0 r} = \frac{q}{\epsilon_0 \sqrt{x^2 + y^2+z^2}}[/tex]

But now I'm dealing with a system with an extra term:
[tex]\nabla^2 V = \delta(x)\delta(y)\delta(z) \frac{q}{\epsilon_0} - (\partial_x V) \frac{a}{1+ax}[/tex]
a is a constant, and the boundary condition is that V->0 when y or z go to infinity.


I am very rusty with differential equations. Can someone tell me how to attack this?

Also, is this still considered linear since there is no term with just V in it? If I knew what to call this class of equations, maybe I could find some solutions in a table or something.
 

Answers and Replies

  • #2
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Hey,

I am not really versed in differential equations myself, so I should apologize beforehand in case my comments are utterly irrelevant to the problem at hand. As you said, the function has spherical symmetry - and therefore, the most sensible thing to do, in my opinion, would be to reformulate the equation in spherical coordinates . Since V = V(r), and you have an explicit expression for the Laplacian operator in polar coordinates, won't this problem simply be reduced into an ODE?
 
  • #3
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Thank you for your reply, but I don't think you read the entire post. The first differential equation has spherical symmetry and I already solved that. The second differential equation reduces to the first in the a->0 limit, but I have no idea how to solve it since when 'a' is non-zero there is no longer spherical symmetry.

I looked it up and this is indeed considered a linear partial differential equation, but I've yet to find a book or table that helps explain how to solve it because most things I find only refer to functions of one variable.

I'm really stumped.
 
  • #4
HallsofIvy
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I'm having trouble solving an equation. Can anyone help?

There is a function V(x,y,z) such that:
[tex]\nabla^2 V = \delta(x)\delta(y)\delta(z) \frac{q}{\epsilon_0} [/tex]
with boundary conditions at infinity V=0.

Since there is spherical symmetry, I was able to rewrite it just in terms of the radial distance, and got:
[tex]\nabla^2 V = \frac{q}{\epsilon_0 r} = \frac{q}{\epsilon_0 \sqrt{x^2 + y^2+z^2}}[/tex]
No, that's not at all the same thing! You can't use "spherical symmetry" because the right hand side, [itex]\delta(x)\delta(y)\delta(z)[/itex] is not spherically symmetric.

But now I'm dealing with a system with an extra term:
[tex]\nabla^2 V = \delta(x)\delta(y)\delta(z) \frac{q}{\epsilon_0} - (\partial_x V) \frac{a}{1+ax}[/tex]
a is a constant, and the boundary condition is that V->0 when y or z go to infinity.


I am very rusty with differential equations. Can someone tell me how to attack this?

Also, is this still considered linear since there is no term with just V in it? If I knew what to call this class of equations, maybe I could find some solutions in a table or something.
 
  • #5
894
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No, that's not at all the same thing! You can't use "spherical symmetry" because the right hand side, [itex]\delta(x)\delta(y)\delta(z)[/itex] is not spherically symmetric.
The right hand side is zero everywhere except at the origin. Why do you feel that is not spherically symmetric?

Regardless, do you have any suggestions on how to tackle the other differential equation:
[tex]\nabla^2 V = \delta(x)\delta(y)\delta(z) \frac{q}{\epsilon_0} - (\partial_x V) \frac{a}{1+ax}[/tex]
'a' is a constant, and the boundary condition is that V->0 when y or z go to infinity.
 

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