Eigenvalues and Eigenfunctions — Question 1

Question 1 – Solution

Strategy. Enforce both homogeneous endpoint conditions in each sign case before normalizing a nonzero solution.

Step 1: Classify all real parameters. For $\lambda=-\mu^2<0$, $y(0)=0$ gives $y=A\sinh(\mu x)$; since $\sinh(\mu L)>0$, the right condition forces $A=0$. For $\lambda=0$, the left condition gives $y=Ax$ and the right gives $A=0$. For $\lambda=k^2>0$, $y=A\sin(kx)$, so a nonzero $A$ requires $kL=n\pi$. Thus $$\boxed{\lambda_n=(n\pi/L)^2,\quad E_n=\operatorname{span}\{\sin(n\pi x/L)\}, \quad n=1,2,\ldots.}$$

Step 2: Fix amplitude and sign. The identity $\sin^2t=(1-\cos 2t)/2$ gives $\int_0^L\sin^2(n\pi x/L)dx=L/2$. The unit norm requires $|A|=\sqrt{2/L}$, and $y'(0)=An\pi/L>0$ selects the positive sign: $$\boxed{\phi_n(x)=\sqrt{2/L}\sin(n\pi x/L).}$$

Step 3: Count the interior zeros. The zeros are $x=jL/n$, $j=0,\ldots,n$, of which exactly $n-1$ lie inside. They are simple because the derivative there is nonzero. The first three modes on the unit interval therefore have zero, one and two interior zeros.

Step 4: Rescale the interval. Replacing $L$ by $2L$ divides each eigenvalue by four and the normalization factor by $\sqrt 2$. For example, $\cos(kx)$ solves the differential equation with $\lambda=k^2$ but fails the left endpoint condition. The equation, both boundary conditions and nontriviality are all part of the eigenproblem.