Seperation of Variables (double check please

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Homework Help Overview

The discussion revolves around the method of separation of variables applied to the second-order wave equation \( u_{tt} = u_{xx} \). The original poster explores the substitution \( u = A(x)B(t) \) and attempts to derive solutions based on the resulting ordinary differential equations for \( A \) and \( B \).

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to confirm their understanding of the separation of variables method and the completeness of the solutions derived. Some participants question the assumption of the constant value being \(-\omega^2\) and the implications of boundary conditions on the solutions. Others discuss the nature of linear combinations of solutions and their validity.

Discussion Status

The discussion is ongoing, with participants providing insights and raising questions about the assumptions made in the problem. There is an exploration of the implications of boundary conditions and the nature of the solutions, indicating a productive exchange of ideas.

Contextual Notes

The original poster notes a lack of extensive coverage on separation of variables in their course, which may contribute to their uncertainty regarding the topic. The absence of boundary conditions is also highlighted as a point of confusion in the discussion.

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separation of Variables (double check please :)

I have a final coming up, and I want to make sure I have this method down.

Q: For the second-order wave equation [itex]u_{tt}=u_{xx}[/itex], the substitution of [itex]u=A(x)B(t)[/itex] will give second-order equations for A nd B when the x and t variables are separation. From [itex]B''/B=A''/A=\omega^2[/itex], find all solutions of the form [itex]u=A(x)B(t)[/itex]

Assume: [tex]u(x,t)=A(x)B(t)[/tex]

[tex]\frac{A(x)B''(t)}{A(x)B(t)}=\frac{A''(x)B(t)}{A(x)B(t)}[/tex]

[tex]\frac{B''}{B}=\frac{A''}{A}=-\omega^2[/tex] is a second order ODE of the form:

[tex]B''+\omega^2 B = 0[/tex]

Solving yields:
[tex]B(t)=c_1 \cos \omega t + c_2 \sin \omega t[/tex] with the assumption that [tex]\omega > 0[/tex]

and:
[tex]A(x)=d_1 \cos \omega x + d_2 \sin \omega x[/tex]

therefore:
[tex]u(x,t) = A(x)B(t)= (c_1 \cos \omega t + c_2 \sin \omega t)(d_1 \cos \omega x + d_2 \sin \omega x)[/tex]

And this is simply all the solutions right? It seems really straightforward, but sometimes when I think it is... it totally isn't. Thanks
 
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First, why are you assuming that the constant value is [itex]-\omega^2[/itex]? Is there a boundary condition you didn't mention?

No, that is not "all the solutions". Any linear combination of solutions, for different [itex]\omega[/itex] is also a solution. The general sum would involve a sum over all possible values of [itex]\omega[/itex]. What the possible values are goes back to that boundary condition I mentioned.
 
The question is verbatim from the text. I used [itex]-\omega^2[/itex] as the assumpition, as I was told to in the question itself.

I'm not exactly sure what you are saying that any linear combination satisfies the equation. I see that any [itex]\omega[/itex] value satisfies u_tt = u_xx. Since there is no boundary condition can't [itex]\omega[/itex] be anything?
 
Ahh.. Well there really isn't anything else in my book about separation of variables, but I've read up a little more on them. I understand what you mean by the linear combination and the factor. Since the "terms" each will differentiate to 0 they can each have a constant in front and are indpendent of each other. Interesting.

My professor did not go into separation of variables for more than 25 minutes the whole semester. We only had three problems to do also...

so anyways. Thanks for the comments man, it allowed me to read up on some interesting stuff.
 

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