Problem 4 from our heat flow handout

Determine the values of $\lambda$ for which the boundary value problem
$$-u''=\lambda u, \: \: u(0)=u(1)=0$$
has a non-trivial solution.

Using the motivation that Mark will undoubtedly brag on the amount of activity that is happening on discourse,

We can start by finding the general solution to the information $-u''={\lambda}u$ to get,

$$u(t)=a\sin(\sqrt{\lambda}*x)+b\cos(\sqrt{\lambda}*x) $$

Then using the information $u(0)=u(1)=0$ to solve for $a$ and $b$,

$$u(0)=a\sin(0)+b\cos(0)=0$$
$$u(0)=b*(1)=0$$

Therefore $b=0$, now for $u(1)=0$ with our new information,

$$u(1)=a\sin(\sqrt{\lambda})=0$$

So we have to solve for when $\sin(\theta)=0$ which is when $\theta=n\pi$. so therefore,

$$\sqrt{\lambda}=n\pi$$
$$\lambda=(n\pi)^2$$

So we get infinitely many values of $\lambda$ given by $\lambda=(n\pi)^2$.