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Help with solution to partial differential equation

by tiredryan
Tags: differential, equation, partial, solution
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tiredryan
#1
Mar2-10, 10:43 AM
P: 51
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{\pa rtial}{\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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vela
#2
Mar2-10, 01:51 PM
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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.
tiredryan
#3
Mar2-10, 02:42 PM
P: 51
Quote Quote by vela View Post
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}{\p artial\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]

vela
#4
Mar2-10, 09:11 PM
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Help with solution to partial differential equation

Quote Quote by tiredryan View Post
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.
tiredryan
#5
Mar3-10, 08:52 AM
P: 51
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]

"

Quote Quote by vela View Post
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.
vela
#6
Mar3-10, 04:05 PM
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Quote Quote by tiredryan View Post
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}{\s in\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}{\s in\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).
tiredryan
#7
Mar4-10, 12:08 PM
P: 51
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}{\s in\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.


Quote Quote by vela View Post
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}{\s in\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}{\s in\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).


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