Help with solution to partial differential equation

In summary, the conversation discusses solving a partial differential equation using the method of separation of variables. The equation is satisfied by a specific solution in the form of \psi(r,\phi)=R(r)\Phi(\phi). The conversation also mentions getting two second order differential equations with nonconstant coefficients and trying to solve them with a solution of the form Rm. The final solution is given in the form of f(R) = \frac{A}{R} + BR + CR^2 + DR^4, but it is unclear how to get to this solution. The conversation ends with a mention of an additional equation that may be necessary for solving the original equation.
  • #1
tiredryan
51
0
I am reading through a textbook and came across this part of the solution. I am wondering if anyone can give me a suggestion on how one goes from the top equation to the bottom equation. Is this something that is found in some book of derivatives or is it solved by hand? I am completely confused how this specific partial derivative is solved. Thanks

"[...]

[tex]
\left[\frac{\partial}{\partial R^2}+\frac{sin \phi}{R^2}\frac{\partial}{\partial\phi}\left(\frac{1}{sin\phi}\frac{\partial}{\partial\phi}\right)\right]^2 \psi = 0
[/tex]

This is satisfied by

[tex]
\psi = sin^2 \phi f(R)
[/tex]

if

[tex]
\left(\frac{d^2}{dR^2}-\frac{2}{R^2}\right)^2 f(R) = 0
[/tex]

The solution of the prior equation is

[tex]
f(R) = \frac{A}{R} + BR + CR^2 + DR^4
[/tex]

[...]"
 
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  • #2
You assume a solution of the form [itex]\psi(r,\phi)=R(r)\Phi(\phi)[/itex] and plug it into the equation. It'll separate into chunks that depend only on one variable, allowing you to write down two differential equations, one for each function.
 
  • #3
vela said:
You assume a solution of the form [itex]\psi(r,\phi)=R(r)\Phi(\phi)[/itex] and plug it into the equation. It'll separate into chunks that depend only on one variable, allowing you to write down two differential equations, one for each function.

Thanks vela. Here is what I have so far. I am reading through the method of separation of variables. I am getting two second order ordinary differential equations with nonconstant coefficients I don't know how to solve. I am doing the separation of variables, and do you have any suggestions? Thanks.

Here is the original equation

[tex]
\left[\frac{\partial}{\partial R^2}+\frac{sin \phi}{R^2}\frac{\partial}{\partial\phi}\left(\frac {1}{sin\phi}\frac{\partial}{\partial\phi}\right)\right]^2 \psi = 0
[/tex]

First I had to break up the following equation into a simpler form by solving the differential with respect to psi. This yielded the following equation.

[tex]
\left[\frac{\partial}{\partial R^2}-\frac{1}{R^2 tan\phi}\frac{\partial}{\partial\phi}+\frac{1}{R^2}\frac{\partial^2}{\partial\phi^2}\right]^2 \psi = 0
[/tex]

From there I separated psi into X which is only dependent on R and Y which is only depend on phi.

[tex]
\psi(r,\phi)=X(R)Y(\phi)
[/tex]

Replacing psi with X and Y in the the equation yeilds

[tex]

0 = X''Y - \frac{1}{R^2 tan\phi}XY' + \frac{1}{R^2}Y''X

[/tex]

Dividing the previous equation by XY yeilds

[tex]

0 = \frac{X''}{X} - \frac{1}{R^2 tan\phi}\frac{Y'}{Y} + \frac{1}{R^2}\frac{Y''}{Y}

[/tex]

Moving the X(R) to the left and Y(psi) to the right yields

[tex]

-\frac{X''}{X} = - \frac{1}{R^2 tan\phi}\frac{Y'}{Y} + \frac{1}{R^2}\frac{Y''}{Y}

[/tex]

Multiplying by R^2 on both sides yields

[tex]

-R^2\frac{X''}{X} = - \frac{1}{tan\phi}\frac{Y'}{Y} + \frac{Y''}{Y}

[/tex]

Now I set the left side and right side equation to "k" yeilding two ordinary differential equations

[tex]
k = -R^2\frac{X''}{X}
[/tex]

[tex]
k = - \frac{1}{tan\phi}\frac{Y'}{Y} + \frac{Y''}{Y}
[/tex]

Multiplying the first equation by X and the second equation by Y yeilds

[tex]
X k = -R^2 X''
[/tex]

[tex]
Y k = - \frac{1}{tan\phi}Y' + Y''
[/tex]

Now I get two second order differential equations with nonconstant coeffients I don't know how to solve.

[tex]
0 = -R^2 X'' - X k
[/tex]

[tex]
0 = - \frac{1}{tan\phi}Y' + Y'' - Y k
[/tex]
 
Last edited:
  • #4
tiredryan said:
Now I get two second order differential equations with nonconstant coeffients I don't know how to solve.

[tex]0 = -R^2 X'' - X k[/tex]

[tex]0 = - \frac{1}{tan\phi}Y' + Y'' - Y k[/tex]
The first one you can solve assuming a solution of the form Ra. I'm not sure about the second, but sin^2(phi) does satisfy it for k=-2.

The fact that the operator is squared seems to complicate things. If it were just applied once, the problem is straightforward to solve via separation.

EDIT: The second equation is close to the Legendre differential equation, so a change of variable like x=cos(phi) might help.
 
Last edited:
  • #5
Thanks. I have another question that is bolded towards the end.

I am trying to solve the first one with the solution of the form Rm. After reading your post I found that the first equation is in the form of the Cauchy-Euler equation. The general form of this equation is

[tex]
x^2 \frac{d^2y}{dx^2} + ax \frac{dy}{dx} + by = 0
[/tex]

From this I began with the following equation

[tex]
0 = R^2 X'' + X k
[/tex]

Assuming the solution is X(R)=Rm

[tex]
X(R) = R^m
[/tex]

Therefore the first and second differentials are as follows

[tex]
\frac{dX}{dR} = mR^{(m-1)}
[/tex]

[tex]
\frac{d^2X}{dR^2} = (m-1)(m)R^{(m-2)}
[/tex]

Plugging into the differential equation yields

[tex]
0 = R^2 (m-1)(m)R^{(m-2)} + kR^m
[/tex]

The "R"s are canceled out yielding

[tex]
0 = m^2-m + k
[/tex]

You mentioned that k = -2 so that the m can be solved with m = -1 and m = 2

This yields the solution of the differential equation in the general form

[tex]
X(R) = \frac{A}{R} + CR^2
[/tex]


I am wondering how does one get to the solution as stated into the book? Thanks.

[tex]

f(R) = \frac{A}{R} + BR + CR^2 + DR^4

[/tex]

PS: I am thinking about it again, and it might have something to do with the book's statement as follows. I do not know why the solution of psi requires this additional equation that I mentioned in the first post.

"[The partial differential equation] is satisfied by

[tex]
\psi = sin^2 \phi f(R)
[/tex]

if

[tex]
\left(\frac{d^2}{dR^2}-\frac{2}{R^2}\right)^2 f(R) = 0
[/tex]

"

vela said:
The first one you can solve assuming a solution of the form Ra. I'm not sure about the second, but sin^2(phi) does satisfy it for k=-2.

The fact that the operator is squared seems to complicate things. If it were just applied once, the problem is straightforward to solve via separation.

EDIT: The second equation is close to the Legendre differential equation, so a change of variable like x=cos(phi) might help.
 
Last edited:
  • #6
tiredryan said:
This yields the solution of the differential equation in the general form

[tex]X(R) = \frac{A}{R} + CR^2[/tex]

I am wondering how does one get to the solution as stated into the book? Thanks.

[tex]f(R) = \frac{A}{R} + BR + CR^2 + DR^4[/tex]
Your solution is to the differential equation

[tex]\left[\frac{\partial}{\partial R^2}+\frac{\sin\phi}{R^2}\frac{\partial}{\partial\phi}\left(\frac{1}{\sin\phi}\frac{\partial}{\partial\phi}\right)\right] X(R)Y(\phi)= 0[/tex]

But the original differential equation applies the operator in the square brackets twice. In other words, after you solved for X(R) and Y(phi), you'd still have to solve the equation

[tex]\left[\frac{\partial}{\partial R^2}+\frac{\sin\phi}{R^2}\frac{\partial}{\partial\phi}\left(\frac{1}{\sin\phi}\frac{\partial}{\partial\phi}\right)\right] \psi = X(R)Y(\phi)[/tex]

to find all the solutions to the original differential equation.

PS: I am thinking about it again, and it might have something to do with the book's statement as follows. I do not know why the solution of psi requires this additional equation that I mentioned in the first post.

"[The partial differential equation] is satisfied by

[tex]\psi = \sin^2 \phi f(R)[/tex]

if

[tex]\left(\frac{d^2}{dR^2}-\frac{2}{R^2}\right)^2 f(R) = 0[/tex]"
Yes, you're right. The book is saying if you choose the solution that is of the form [itex]\psi = \sin^2 \phi f(R)[/itex], then f(R) satistfies that differential equation. You can plug that solution into the original differential equation and let the operator act on the angular part of psi, and you'll find that the second term will turn into -2/R2, which leaves you with the differential equation for f(R).
 
  • #7
Thanks vela. Do you have any suggestions on how to solve

[tex]\left[\frac{\partial}{\partial R^2}+\frac{\sin\phi}{R^2}\frac{\partial}{\partial\phi}\left(\frac{1}{\sin\phi}\frac{\partial}{\partial\phi}\right)\right] \psi = X(R)Y(\phi)[/tex]

Since the right side equals X(R)Y([itex]\phi[/itex]) and does not equation zero, I am having a hard time separating the variables. If I let W(R)Z([itex]\phi[/itex]) equation to [itex]\psi[/itex] then equation becomes as follows

[tex]
XY = W''Z - \frac{1}{R^2 tan\phi}WZ' + \frac{1}{R^2}W''Z
[/tex]

Dividing by WZ

[tex]
\frac{XY}{WZ} = \frac{W''}{W} - \frac{1}{R^2 tan\phi}\frac{Z'}{Z} + \frac{1}{R^2}\frac{Z''}{Z}
[/tex]

How do I now separate the equations into two sets of ordinary differential equations?
Thanks.
vela said:
Your solution is to the differential equation

[tex]\left[\frac{\partial}{\partial R^2}+\frac{\sin\phi}{R^2}\frac{\partial}{\partial\phi}\left(\frac{1}{\sin\phi}\frac{\partial}{\partial\phi}\right)\right] X(R)Y(\phi)= 0[/tex]

But the original differential equation applies the operator in the square brackets twice. In other words, after you solved for X(R) and Y(phi), you'd still have to solve the equation

[tex]\left[\frac{\partial}{\partial R^2}+\frac{\sin\phi}{R^2}\frac{\partial}{\partial\phi}\left(\frac{1}{\sin\phi}\frac{\partial}{\partial\phi}\right)\right] \psi = X(R)Y(\phi)[/tex]

to find all the solutions to the original differential equation.Yes, you're right. The book is saying if you choose the solution that is of the form [itex]\psi = \sin^2 \phi f(R)[/itex], then f(R) satistfies that differential equation. You can plug that solution into the original differential equation and let the operator act on the angular part of psi, and you'll find that the second term will turn into -2/R2, which leaves you with the differential equation for f(R).
 
Last edited:

1. What is a partial differential equation?

A partial differential equation is a mathematical equation that involves partial derivatives of an unknown function. It is used to describe physical phenomena that vary in space and time.

2. Why are partial differential equations important in science?

Partial differential equations are used to model and understand many natural phenomena, such as fluid dynamics, heat transfer, and quantum mechanics. They allow us to make predictions and solve complex problems in various fields of science.

3. How are partial differential equations solved?

Partial differential equations can be solved analytically or numerically. Analytical solutions involve finding a closed-form solution using mathematical techniques, while numerical solutions involve using computer algorithms to approximate the solution.

4. What types of boundary conditions are involved in solving partial differential equations?

Boundary conditions specify the behavior of the unknown function at the boundaries of the problem domain. They can be of different types, such as Dirichlet boundary conditions, which specify the value of the function at the boundary, or Neumann boundary conditions, which specify the derivative of the function at the boundary.

5. Can all partial differential equations be solved?

No, not all partial differential equations have analytical solutions. Some equations are too complex to solve using mathematical techniques and require numerical methods. Additionally, even with numerical methods, some equations may be too computationally intensive to solve accurately.

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