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Here is a problem from my homework assignment that I am struggling with:

Consider the differential equation $\frac{d^2\phi}{dx^2}+\lambda\phi=0 $.

Determine the eigenvalues $\lambda$ if $\phi$ satisfies the following boundary conditions:

$\phi(a)=0$
$\phi(b)=0$

I have been able to successfully complete this problem with 6 other sets of boundary conditions, but this one is giving me trouble. The question also states we only need to consider $\lambda>1$. Here is what I have done so far:

Because $\lambda>1$, we get our general solution of: $\phi = C_1cos(\sqrt\lambda x)+C_2sin(\sqrt\lambda x)$.

We can plug in our first boundary condition and get:

$\phi(a) = C_1cos(\sqrt\lambda a)+C_2sin(\sqrt\lambda a)=0$

Usually I would solve for $C_1$ or $C_2$ to plug into the second equation, so here I solve for $C_1$:

$C_1=\frac{-C_2 sin(\sqrt\lambda a)}{cos\sqrt\lambda a)}$

Now I substitute $C_1$ in and plug in the $b$ value to get:

$\phi(a) = \frac{-C_2 sin(\sqrt\lambda a)}{cos\sqrt\lambda a)}cos(\sqrt\lambda b)+C_2sin(\sqrt\lambda b)=0$

Now I will factor our $C_2$ to get:

$\phi(a) = C_2[\frac{sin(\sqrt\lambda a)}{cos\sqrt\lambda a)}cos(\sqrt\lambda b)+sin(\sqrt\lambda b)]=0$

Here we assume that $C_2$ cannot equal zero, or we would also get $C_1$ equal to zero, which would give us the trivial solution. Thus we conclude that:

$\frac{sin(\sqrt\lambda a)}{cos\sqrt\lambda a)}cos(\sqrt\lambda b)+sin(\sqrt\lambda b)=0$

And that is as far as I can get....

The back of the book says the answer is: $\lambda = (\frac{n \pi}{b-a})^2$

Any help would be greatly appreciated!

Thanks.

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  • $\begingroup$ Try writing in terms of the tangent function $$\tan x=\frac{\sin x}{\cos x}$$ $\endgroup$ – ClassicStyle Sep 19 '14 at 0:22
  • $\begingroup$ Essentially what i would get is $tan(\sqrt\lambda a)=-tan(\sqrt\lambda b)$.... Which would mean that $a=-b$. You suggest i should substitute that back in? $\endgroup$ – Riley Sep 19 '14 at 0:27
  • $\begingroup$ Try graphing your equation. There are several solutions. $\endgroup$ – ClassicStyle Sep 19 '14 at 0:33
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Up to a scalar multiple, the eigenfunction $\phi$ satisfying $\phi(a)=0$, $\phi'(a)\ne 0$ is $$ \sin(\sqrt{\lambda}(x-a)). $$ This is an actual eigenfunction iff $\sin(\sqrt{\lambda}(b-a))=0$. Equivalently, $$ \sqrt{\lambda}(b-a)=\pm\pi,\pm 2\pi,\pm 3\pi,\cdots,\\ \lambda = \frac{n^{2}\pi^{2}}{(b-a)^{2}},\;\;\; n=1,2,3,\cdots. $$ $\lambda = 0$ cannot be considered using this method for $\lambda=0$ because $\sin(\sqrt{\lambda}(x-a))\equiv 0$ in that case. The correct solution for $\lambda=0$ is $C(x-a)$, but this function does cannot vanish at $b$. So $\lambda=0$ is not an eigenvalue.

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  • $\begingroup$ Nice and simple, just as it should be. Thanks! $\endgroup$ – Riley Sep 21 '14 at 22:30
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Write $$\lambda = \mu^{2},\ \ \ \mu > 0$$

Then as you found the general solution would be $\phi = c_{1}\ cos(\mu x) + c_{2} \ sin(\mu x)$.

As for you final answer you may write in terms of $tan\ (\mu x) =\frac{sin\ (\mu x)}{cos \ (\mu x)} $.

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It is also possible to solve this with your original method. From

$$-\frac{\sin(\sqrt{λ}a)}{\cos(\sqrt{λ}a)}\cos(\sqrt{λ}b)+\sin(\sqrt{λ}b)=0$$

(beware, you forgot a minus there) you obtain (by division with $\cos(\sqrt{λ}b)$)

$$-\tan(\sqrt{λ}a)+\tan(\sqrt{λ}b)=\frac{\sin(\sqrt{λ}(b-a))}{\cos(\sqrt{λ}a)\cos(\sqrt{λ}b)}= 0\ .$$

And from $\sin(\sqrt{λ}(b-a))=0$ you get desired solution.

This works under conditions $$\lambda_n\neq\frac{\pi(2n+1)}{2a}\ ,\quad \lambda_n\neq\frac{\pi(2n+1)}{2b}\ ,$$

which you must take into account when dealing with special intervals e.g. $b=-a$. It results in different eigenfunctions for even and odd $n$.

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