Seperation of Variables (double check please

In summary, this equation states that if the x and t variables are seperated, then the solutions are found by solving for B(t) and A(x) respectively. Any linear combination of these solutions will also satisfy the equation.
  • #1
FrogPad
810
0
Seperation 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 seperated. 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
 
Last edited:
Physics news on Phys.org
  • #2
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.
 
  • #3
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?
 
  • #4
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.
 

Related to Seperation of Variables (double check please

What is Separation of Variables?

Separation of Variables is a mathematical method used to solve certain types of differential equations. It involves separating the variables in the equation and solving each part separately before combining them to find the overall solution.

When is Separation of Variables used?

Separation of Variables is commonly used in physics and engineering to solve differential equations that involve multiple independent variables. It is also used in other areas of mathematics, such as in partial differential equations and Fourier series.

What are the steps involved in Separation of Variables?

The first step is to identify the variables in the equation and separate them into different parts. Next, each part is solved separately using standard techniques and methods. Finally, the solutions are combined to find the overall solution to the equation.

What are the benefits of using Separation of Variables?

Separation of Variables can simplify complex equations and make them easier to solve. It also allows for the use of known solutions to simpler equations, making it a useful tool in solving more difficult problems.

Are there any limitations to using Separation of Variables?

Yes, Separation of Variables can only be used on certain types of differential equations, specifically those that are linear and homogeneous. It also may not always provide the most general solution and may require additional techniques to find a complete solution.

Similar threads

  • Calculus and Beyond Homework Help
Replies
8
Views
453
  • Calculus and Beyond Homework Help
Replies
1
Views
902
  • Calculus and Beyond Homework Help
Replies
15
Views
908
  • Calculus and Beyond Homework Help
Replies
1
Views
830
  • Calculus and Beyond Homework Help
Replies
2
Views
423
  • Calculus and Beyond Homework Help
Replies
23
Views
1K
  • Calculus and Beyond Homework Help
Replies
1
Views
789
  • Introductory Physics Homework Help
Replies
10
Views
425
  • Calculus and Beyond Homework Help
Replies
2
Views
1K
  • Calculus and Beyond Homework Help
Replies
1
Views
991
Back
Top