Find the Eigen value and Eigen function of the BVP $ \ y''+\lambda^2 y=0 \ , \ y(\pi)=0, \ \ y'( 3\pi )=0 \ $

Question

Find the Eigen value and Eigen function of the BVP $ \ y''+\lambda^2 y=0 \ , \ y(\pi)=0, \ \ y'( 3\pi )=0 \ $.

Answer:

For $ \ \lambda =0 \ $ , we get trivial solution.

For $ \lambda \neq 0 \ $ , the general solution is

$ y(x)=A \cos (\lambda x)+B \sin (\lambda x) \ $ ,

where $ \ A,B \ $ are arbitrary constants

Now satisfying the condition $ \ y(\pi)=0 \ \ $, we get

$ A \cos (\lambda \pi)+B \sin (\lambda \pi)=0 \ .......(1) $

Similarly using $ y'(3 \pi)=0 \ $ , we get

$ -A \cos (\lambda \pi)+B \sin (\lambda \pi)=0 \ ,.......(2) $

But I can not eliminate $ \ A,B \ $ from (1) and (2) in order to find the eigen value and eigen function.

help me doing this.

Answer 1

For convenience, we'll shift the solution so that the first boundary condition is at the origin. Set $t = x - \pi$ to get

$$ y''(t) + \lambda^2 y(t) = 0, \ y(0) = 0, y'(2\pi) = 0 $$

This obviously gives

$$ y(t) = A\cos(\lambda t) + B\sin(\lambda t) $$

The first B.C. gives $$ y(0) = A = 0 $$

The remaining constant is negligible (since the eigenfunction can be scaled up), therefore $$ y(t) = \sin(\lambda t) $$

The second and final B.C then gives $$ y'(2\pi) = \lambda \cos(2\pi \lambda) = 0 $$

This is achieved if $$ 2\pi \lambda = (2n+1)\frac{\pi}{2} $$

Then $$ \lambda_n = \frac{2n+1}{4} $$

And the final solution is $$ y_n(t) = y_n(x-\pi) = \sin\left[\frac{2n+1}{4}(x-\pi)\right] $$

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EDIT: If you don't like this solution, you can do it the "normal" way. But first note that your second boundary equation is wrong, since $$ y'(x) = \lambda\big[{-A}\sin(\lambda x) + B\cos(\lambda x)\big] $$

Then $$ y'(3\pi) = 0 \implies -A\sin(3\pi\lambda) + B\cos(3\pi\lambda) = 0 $$

The system can be expressed as a matrix equation $$ \left[\begin{matrix} \cos(\pi\lambda) & \sin(\pi\lambda) \\ -\sin(3\pi\lambda) & \cos(3\pi\lambda) \end{matrix}\right] \left[\begin{matrix} A \\ B \end{matrix}\right] = M\left[\begin{matrix} A \\ B \end{matrix}\right] = 0 $$

If $\det(M) \ne 0$, then the only solution is $A=B=0$, which we don't want, so it must follow that $ \det(M) = 0 $ which gives

$$ \cos(\pi\lambda)\cos(3\pi\lambda) + \sin(\pi\lambda)\sin(3\pi\lambda) = \cos(2\pi\lambda) = 0 $$

This gives the same eigenvalues as above $$ \lambda_n = \frac{2n+1}{4} $$

You can then assign an arbitrary solution for the pair $(A,B)$ such that they satisfy one of the equations. Since $\det(M) = 0$, the remaining equation will automatically hold. So let's put $$ A = -\sin(\pi\lambda_n), B = \cos(\pi\lambda_n) $$

This gives

$$ y(x) = -\sin(\pi\lambda_n)\cos(\lambda x) + \cos(\pi\lambda_n)\sin(\lambda_n x) = \sin(\lambda_n(x-\pi)) $$

which shows why shifting the solution from the beginning is much better.

Answer 2

One of the issues with eigenvalue problems is that there is a scale factor that creates ambiguity. Any non-trivial solution of the eigenvalue problem is also a non-trivial solution if you scale by any non-zero scale factor. So, to eliminate this ambiguity, solve the problem $$ y''+\lambda^2y=0,\;\;\; y(\pi)=0,\; y'(\pi)=1,\; y'(3\pi)=0. $$ Any non-trivial solution of the original problem can be scaled to satisfy $y'(\pi)=1$ because it cannot satisfy $y(\pi)=0,\; y'(\pi)=0$ or it would be identically $0$ by uniqueness of such a solution. The solution satisfying $y(\pi)=0$, $y'(\pi)=1$ is $$ y(x)=\frac{\sin(\lambda(x-\pi))}{\lambda}. $$ This works even for $\lambda=0$, if you take the limit as $\lambda\rightarrow 0$ to obtain $y(x)=x-\pi$.

So the eigenvalue equation is $$ 0 = y'(3\pi)=\cos(\lambda(x-\pi))|_{x=3\pi}=\cos(2\pi\lambda). $$ The permissible $\lambda$ are the $\lambda_n$ satisfying $$ 2\pi \lambda_n = (n+\frac{1}{2})\pi,\\ \lambda_n = \frac{2n+1}{4}. $$ The negative values of $\lambda_n$ are redundant because of the $\lambda^2$ in the original equation.