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:


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!


  • $\begingroup$ Try writing in terms of the tangent function $$\tan x=\frac{\sin x}{\cos x}$$ $\endgroup$ 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. $\endgroup$ Sep 19 '14 at 0:33

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.

  • $\begingroup$ Nice and simple, just as it should be. Thanks! $\endgroup$
    – Riley
    Sep 21 '14 at 22:30

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)} $.


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


(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 $n$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.