Simple Partial Differential Equation

In summary: One way to approach this would be to differentiate the equation with respect to r and then use the Chain Rule to integrate.Another approach would be to use the Laplace Transform.Is that correct?Not really. If u(r)=dV/dr then the equation becomes du/dr=(-2/r)u. Doesn't it? Solve that.Ahh, I'm an idiot. u=r^{-2}
  • #1
jaydnul
558
15

Homework Statement


This is actually an electromagnetism problem but all the physics is done, I just don't remember how to solve the PDE:
[itex]\frac{d^2V}{dr^2}=-\frac{2}{r}\frac{dV}{dr}[/itex]
The d's should be del's, just don't know how to do that...

Homework Equations



Not sure.

The Attempt at a Solution


[/B]
Don't know where to start.
 
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  • #2
There are no other independent variables in the problem besides r. Assume V can be written as a product of functions of all the independent variables. I.e. it's separable. V(r)=u(r)*v(theta)*w(phi). The treat it as an ODE. Do you remember those? I would try putting dV/dr=u(r) and separating variables.
 
Last edited:
  • #3
Ahh I think that's just what I needed. So

[itex]V_{rr}=-k^2u(r)[/itex] where k=sqrt{2/r}
[itex]u(r)=Ae^{ikr}+Be^{-ikr}[/itex]
[itex]V(r)=\frac{-iA}{k}e^{ikr}+\frac{iB}{k}e^{-ikr}[/itex]
then sub in for k

Is that correct?
 
  • #4
Jd0g33 said:
Ahh I think that's just what I needed. So

[itex]V_{rr}=-k^2u(r)[/itex] where k=sqrt{2/r}
[itex]u(r)=Ae^{ikr}+Be^{-ikr}[/itex]
[itex]V(r)=\frac{-iA}{k}e^{ikr}+\frac{iB}{k}e^{-ikr}[/itex]
then sub in for k

Is that correct?

Not really. If u(r)=dV/dr then the equation becomes du/dr=(-2/r)u. Doesn't it? Solve that.
 
  • #5
Ahh, I'm an idiot. [itex]u=r^{-2}[/itex]

Thanks Dick!
 
  • #6
Jd0g33 said:
Ahh, I'm an idiot. [itex]u=r^{-2}[/itex]

Thanks Dick!

You're welcome but don't forget the arbitrary constant when you integrate.
 
  • #7
Jd0g33 said:

Homework Statement


This is actually an electromagnetism problem but all the physics is done, I just don't remember how to solve the PDE:
[itex]\frac{d^2V}{dr^2}=-\frac{2}{r}\frac{dV}{dr}[/itex]
The d's should be del's, just don't know how to do that...

Homework Equations



Not sure.

The Attempt at a Solution


[/B]
Don't know where to start.

Note that
[tex] \frac{d}{dr}\left( r^k \frac{dF(r)}{dr} \right) = r^k \frac{d^2F(r)}{dr^2}+ k r^{k-1} \frac{dF(r)}{dr}[/tex]
Can you see how that fact helps?
 

1. What is a Simple Partial Differential Equation?

A Simple Partial Differential Equation (PDE) is a mathematical equation that involves multiple independent variables and their partial derivatives. It describes the relationship between a function and its partial derivatives with respect to these variables.

2. What are the types of Simple Partial Differential Equations?

The types of Simple Partial Differential Equations include linear and nonlinear PDEs. Linear PDEs have a linear relationship between the function and its partial derivatives, while nonlinear PDEs have a nonlinear relationship.

3. What are some real-life applications of Simple Partial Differential Equations?

Simple Partial Differential Equations have various applications in physics, engineering, and other scientific fields. They are used to model heat transfer, diffusion, and wave propagation, among other phenomena.

4. How are Simple Partial Differential Equations solved?

Solving Simple Partial Differential Equations involves finding a function or set of functions that satisfy the given equation. This can be done analytically using techniques such as separation of variables or numerically using computer algorithms.

5. What are the challenges in solving Simple Partial Differential Equations?

Solving Simple Partial Differential Equations can be challenging due to their complex nature and the presence of multiple independent variables. Additionally, nonlinear PDEs often do not have exact solutions, requiring the use of numerical methods.

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