Solving Partial differential equation

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Homework Help Overview

The discussion revolves around solving a differential equation, initially presented as a partial differential equation (PDE), but later identified as an ordinary differential equation (ODE) involving the function ##\psi##, which is dependent on the variable ##y##. The equation includes boundary conditions and parameters such as ##E## and ##\omega##.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss various methods for addressing the equation, including integration techniques and the possibility of using series solutions. There are questions about the nature of the equation, with some suggesting substitutions to simplify it, while others express uncertainty about the implications of these substitutions.

Discussion Status

The conversation is ongoing, with participants exploring different interpretations of the equation and considering various approaches. Some guidance has been offered regarding the nature of the equation and potential methods, but there is no explicit consensus on a single approach or solution at this time.

Contextual Notes

Participants note the boundary conditions and the range of the variable involved, as well as the complexity introduced by the nonlinear aspects of the equation. There is a recognition of the challenges posed by the current form of the equation and the need for further exploration.

AHSAN MUJTABA
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Homework Statement
solve the equation for $$\frac{\partial^{2}\psi(\sqrt{g} y)}{\partial y^{2}}=(\frac{sin^{2}(\sqrt{g}y)}{g}-\frac{2E}{\omega})\psi(\sqrt{g}y)$$.
Relevant Equations
The boundary conditions are $$\psi(\frac{\pi}{2})=0$$ and $$\psi(\frac{-\pi}{2})=0$$. I know the values of ##\frac{E}{\omega}##
I have tried to do it in standard way by integrating in PDE's but it turned out that ##\psi## is a function of y, so now I have no clue to start this. I know the range of ##\sqrt {g}y## from ##\frac{-\pi}{2}## to ##\frac{\pi}{2}##
 
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AHSAN MUJTABA said:
Homework Statement:: solve the equation for $$\frac{\partial^{2}\psi(\sqrt{g} y)}{\partial y^{2}}=(\frac{sin^{2}(\sqrt{g}y)}{g}-\frac{2E}{\omega})\psi\sqrt{g}y$$.
Relevant Equations:: The boundary conditions are $$\psi(\frac{\pi}{2}=0$$ and $$\psi(\frac{-\pi}{2})=0$$. I know the values of ##\frac{E}{\omega}##

I have tried to do it in standard way by integrating in PDE's but it turned out that ##\psi## is a function of y, so now I have no clue to start this. I know the range of ##\sqrt {g}y## from ##\frac{-\pi}{2}## to ##\frac{\pi}{2}##
Since ##\phi## is apparently a function of y alone, it seems to me that you're not dealing with a partial differential equation -- just an ordinary differential equation. An integrating factor might work.
 
Mark44 said:
Since ##\phi## is apparently a function of y alone, it seems to me that you're not dealing with a partial differential equation -- just an ordinary differential equation. An integrating factor might work.
If I just double integrate this equation on y, you mean that?
 
I was thinking of some series solutions but I have to solve it properly
 
AHSAN MUJTABA said:
If I just double integrate this equation on y, you mean that?
No, that's not what I mean. I would first do a substitution -- Let ##u = \sqrt g y## -- and then get a DE that looks like this:
##\frac{d^2 \phi(u)}{du^2} - (\frac{\sin^2(u)} u - \frac {2E} \omega)\phi(u) = 0##
It won't look exactly like this, because there are some factors that I've omitted that come from the chain rule.

This is an ordinary differential equation, not a PDE, but it's nonlinear. I don't have any ideas for attacking it at the moment.
 
I have tried to attack it with the same strategy but since ##\phi(u)## at the end creates a high problem for solving this.
 
Can I substitute appropriately to make it a first order ode?
 
AHSAN MUJTABA said:
Can I substitute appropriately to make it a first order ode?
No, because you have ##\phi''## and ##\phi## in the equation.
 
Is there going to be a power series solution?
 

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