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Ordinary Differential Equations

1. The problem statement, all variables and given/known data
Find all eigenvalues and eigenfunctions:
[itex] -y'' (x) = \lambda y(x) , x \in (a,b) [/itex]
[itex] y(a)= y(b) =0 [/itex].


2. Relevant equations
[itex] sin x = \frac{e^{ix} + e^{-ix} }{2i} [/itex].

3. The attempt at a solution
So actually the only problem I have is to find the eigenfunctions (which should be something like [itex] sin \{ \frac{n \pi (x-a) }{(b-a)} \} [/itex] ) .
I received the eigenvalues are: [itex] \lambda_n = \frac{n^2 \pi^2 }{(b-a)^2} [/itex].
But how can I solve the two equations system I receive when applying these eigenvalues to the solution: [itex] y= c_1 e^{i \sqrt{\lambda_n} x} + c_2 e^{-i \sqrt{\lambda_n} x} [/itex].

I hope someone will be able to help me solve this two equations-system.

Thanks in advance!
1. The problem statement, all variables and given/known data



2. Relevant equations



3. The attempt at a solution
 

LCKurtz

Science Advisor
Homework Helper
Insights Author
Gold Member
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1. The problem statement, all variables and given/known data
Find all eigenvalues and eigenfunctions:
[itex] -y'' (x) = \lambda y(x) , x \in (a,b) [/itex]
[itex] y(a)= y(b) =0 [/itex].


2. Relevant equations
[itex] sin x = \frac{e^{ix} + e^{-ix} }{2i} [/itex].

3. The attempt at a solution
So actually the only problem I have is to find the eigenfunctions (which should be something like [itex] sin \{ \frac{n \pi (x-a) }{(b-a)} \} [/itex] ) .
I received the eigenvalues are: [itex] \lambda_n = \frac{n^2 \pi^2 }{(b-a)^2} [/itex].
But how can I solve the two equations system I receive when applying these eigenvalues to the solution: [itex] y= c_1 e^{i \sqrt{\lambda_n} x} + c_2 e^{-i \sqrt{\lambda_n} x} [/itex].

I hope someone will be able to help me solve this two equations-system.

Thanks in advance!
Write the problem as ##y'' +\lambda y = 0## and consider the case where ##\lambda = \mu^2 > 0## so ##y''+\mu^2y=0##. This gives you the solution pair ##\{e^{i\mu x},e^{-i\mu x}\}##. But by using appropriate linear combinations of those you could use the linearly independent pair ##\{\sin(\mu x),\cos(\mu x)\}## instead. Even better, you can use any appropriate linear combination of those to get something easier to work with. Try using this pair: ##\{\cos(\mu(x-a)),\sin(\mu(x-a))\}##. Take a general linear combination of those two and apply your boundary conditions. It will all drop out. At the end, remember ##\lambda = \mu^2.##
 
Is there any way of solving the two equations without having to do the process you mentioned?

Thanks !
 

HallsofIvy

Science Advisor
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One can, by direct substitution, show that [itex]y(x)= Acos(\lambda (x- a)+ Bsin(\lambda (x- a))[/itex] is the general solution to the differential equation [itex]y''+ \lambda^2 y= 0[/itex], for any number a. Of course, then [itex]y(0)= A= 0[/itex].
Now [itex]y(b)= B sin(\lambda(b- a))= 0[/itex]. You know, I presume that sin(x) is 0 if and only if x is a multiple of [itex]\pi[/itex] so you use that to determine possible values for the eigenvalue, [itex]\lambda[/itex].
 
Great ! Thanks ! But how can I prove that this is the general solution ? (i.e.- it obviously solves our problem...but why is it the general solution? why is it the only form of our eigenfunctions? )

Thanks!
 

LCKurtz

Science Advisor
Homework Helper
Insights Author
Gold Member
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Is there any way of solving the two equations without having to do the process you mentioned?

Thanks !
I suppose so, but I don't know why you would want to. Certainly you would use ##y = A\cos(\mu x) + B\sin(\mu x)## instead of the exponential version. I will show you how to get the solution from there, assuming you didn't think of the ##(x-a)## version. Applying the boundary conditions you get:$$A\cos(\mu a) + B\sin(\mu a)=0$$ $$
A\cos(\mu b) + B\sin(\mu b)=0$$For this to have a non-trivial solution for ##A## and ##B## the determinant of coefficients must be 0:$$
\left |\begin{array}{cc}
\cos(\mu a) & \sin(\mu a)\\
\cos(\mu b) & \sin(\mu b)
\end{array}\right | =\cos(\mu a)\sin(\mu b)-\cos(\mu b)\sin(\mu a)= 0$$ $$
\sin(\mu(b-a))=0$$
This gives ##\mu(b-a) = n\pi## or ##\mu_n = \frac{n\pi}{b-a}##

Under this condition the system of equations is dependent, so just using the first one we can write ##A\cos(\mu a) = -B\sin(\mu a)## so ##A=-\frac{B\sin(\mu a)}{\cos(\mu a)}##. Substituting that for the ##A## in the general solution for ##y## gives $$
y_n = -B\frac{\sin(\mu_n a)}{\cos(\mu_n a)}\cos(\mu_n x) + B\sin(\mu_n x)
=(\frac{B}{\cos(\mu_n a)})(-\sin(\mu_n a)\cos(\mu_n x)+\sin(\mu_n x)\cos(\mu_n a))
=(\frac{B}{\cos(\mu_n a)})\sin(\mu_n(x-a))$$Since any constant times a solution is a solution, we can ignore the constant ##\frac{B}{\cos(\mu_n a)}## and take$$
y_n = \sin(\mu_n(x-a))$$So what I showed you in my first post is something worth learning. It saves a lot of work.
 
Last edited:
Thanks a lot !!!
 

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