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Please show me how to calculate the eigenvalues for the following boundary value problem:

$$x''+\lambda x=0\\x(0)=0\\x(\pi)=0\\x'(\pi)=0$$

This is what I did: let $\lambda=\mu^2$

$$X(x)=A\cos\mu x+B\sin \mu x$$

$$X^{'}(x)=-A\mu \sin\mu x+B\mu \cos \mu x$$

Now applying the boundary conditions we get $$A=0$$

and using last two boundary conditions we get

$$A\cos\mu \pi+B\sin\mu\pi=0$$

$$-A\mu \cos\mu \pi+B\mu \sin \mu \pi=0$$

How to solve it further? Here the eigenvalue is $$\lambda=\mu^2=n^2$$

Does the eigenvalue satisfy this equation:

$$\sqrt{\lambda}+\tan \sqrt{\lambda}\pi=0 \textrm{ ?}$$

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  • 3
    $\begingroup$ From your equations you would find that $A=B=0$, except that you mixed up $\sin$, $\cos$ in the second system. That's because your problem is not well posed, it has too many boundary conditions. If you dropped one of the conditions at $x=\pi$, it would go smoothly. $\endgroup$ – user138530 Sep 1 '14 at 20:27
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Hint:you have $Acos\mu \pi+Bsin\mu\pi=0$

$-A\mu cos\mu \pi+B\mu sin \mu \pi=0$. From these equation you have $2\mu Bsin\mu\pi=0$. Now when $\mu$ is not equal 0 then $sin\mu\pi=0$.As B can not be 0,otherwise X(x) will become zero. so you get $\mu\pi=n\pi$ where n is non-zero integer.That imply $\mu=n$. Now substitute this value you will get eigen value.

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Look at the following solution :

Case 1: $\lambda=0$

In this case, we have $x''=0\implies x=Ct+D$. Now $x(0)=0\implies D=0$, so
$x=Ct$. Again $x(\pi)=0\implies C=0$.

Hence for $\lambda=0$, there are no eigen values.

Case 2: $\lambda >0$

Then, we let $\lambda=\alpha^2$. $ $$0\ne\alpha\in \mathbb{R}$.

So we have $x''+\lambda x=x''+\alpha^2 x=0. $ $\to(1)$
Letting $x=e^{mt}$ and considering the auxiliary equation the solution for $(1)$ is

$x=A\cos \alpha t+B\sin \alpha t.$ Now by the boundary condition $x(0)=0,$ we get
$0=A$, so $x=B\sin \alpha t$.

By the second boundary condition $x(\pi)=0$, we see that :
$0=B\sin \alpha \pi$.
For non trivial eigen values we must have $\sin \alpha \pi=0\implies \alpha\pi=n\pi\implies\alpha=n, n\in \mathbb{N}$

Thus, $\lambda=n^2$ are the eigen values and correspondingly

$x=B\sin nt$ are the eigen vectors.

Case 3: $\lambda <0$

Assume $\lambda=-\beta^2, 0\ne \beta\in \mathbb{R}$ and proceed similarly to show that there are no real eigen values. $\square$

Please Note : $x'(\pi)=0$ is a redundant boundary condition, which doesn't do any good. Further it makes the problem look ill-posed.

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