How Can I Solve This Complicated Partial Differential Equation?

In summary: Again, I'm very rusty with differential equations, so can someone please tell me how to attack this equation?In summary, the function has spherical symmetry, but I can't use that because the right hand side is not spherically symmetric. I reformulated the equation in terms of radial distance and got: \nabla^2 V = \frac{q}{\epsilon_0 r} = \frac{q}{\epsilon_0 \sqrt{x^2 + y^2+z^2}}. However, I still have an equation with an extra term that I don't understand. Is this still considered a linear partial differential equation? No, because there is a
  • #1
JustinLevy
895
1
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.
 
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  • #2
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
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
JustinLevy said:
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
HallsofIvy said:
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.
 

1. What is a partial differential equation (PDE)?

A partial differential equation is a mathematical equation that involves multiple independent variables and their partial derivatives. It is used to describe physical phenomena that vary in space and time, such as heat flow, fluid dynamics, and quantum mechanics.

2. What is the difference between a partial differential equation and an ordinary differential equation?

The main difference between a partial differential equation and an ordinary differential equation is the number of independent variables. PDEs involve multiple independent variables, while ODEs only have one independent variable. Additionally, the derivatives in a PDE can be partial derivatives, while the derivatives in an ODE are ordinary derivatives.

3. How are partial differential equations solved?

There is no general method for solving all types of PDEs. The approach to solving a PDE depends on its type, boundary conditions, and initial conditions. Some common methods include separation of variables, Fourier transforms, and numerical methods.

4. What are some real-world applications of partial differential equations?

PDEs are used to model a wide range of physical phenomena, including heat transfer, fluid flow, electromagnetism, and quantum mechanics. They are also used in fields such as engineering, physics, chemistry, and finance to analyze and predict behaviors of complex systems.

5. What are the challenges of working with partial differential equations?

One of the main challenges of working with PDEs is that they can be difficult to solve analytically. This means that numerical methods, which require computer simulations, are often needed to find solutions. Additionally, PDEs can have multiple solutions or no solution at all, making it important to carefully consider boundary conditions and initial conditions when formulating a problem.

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