Eigenvalues and Eigenfunctions — Question 10

Question 10 – Solution

Strategy. Match displacement and stiffness-weighted slope, then preserve every branch of the determinant equation.

Step 1: Exclude nonpositive eigenvalues. Adding the two integrations by parts gives $\lambda\int_0^1y^2dx=\int_0^1p(y')^2dx$: endpoint values vanish and the interface terms cancel because both $y$ and $py'$ match. The right side is positive for a nonzero admissible function; equality would make it constant on both sides, hence identically zero. Therefore $\lambda>0$.

Step 2: Form the interface system. The endpoint-compatible pieces are $A\sin(kx)$ on the left and $B\sin(k(1-x)/2)$ on the right. With $t=k/4$, matching gives $$A\sin 2t=B\sin t,\qquad A\cos 2t=-2B\cos t.$$ The second equation uses flux, not ordinary derivative. The determinant vanishes exactly when $2\sin 2t\cos t+\cos 2t\sin t=0$.

Step 3: Preserve both spectral branches. Factoring yields $\sin t(6\cos^2t-1)=0$. Thus the complete spectrum is $$\boxed{\lambda=16t^2,\qquad t>0,\quad t=m\pi\ (m\geq 1)\ \text{or}\ \cos^2t=1/6.}$$ At $t=m\pi$, the first matching equation is zero and the second gives $B=(-1)^{m+1}A/2$. On the other branch, $\sin t\ne 0$ and $B=2A\cos t$. Each branch gives a nonzero solution and a one-dimensional eigenspace. The apparent root $t=0$ is excluded by Step 1.

Step 4: Construct the first mode and its corner. The smallest positive root is $t_0=\arccos(1/\sqrt 6)\in(0,\pi/2)$. Hence $\boxed{\lambda_1=16t_0^2}$ and, taking $A=1$, $$\boxed{y=\begin{cases}\sin(4t_0x),&0\leq x\leq 1/2,\\ \sqrt{2/3}\sin(2t_0(1-x)),&1/2\leq x\leq 1.\end{cases}}$$ At the join both values equal $\sqrt 5/3$. The one-sided slopes are $-8t_0/3$ and $-2t_0/3$, so $y'(1/2-)=4y'(1/2+)$ as required by flux continuity. The slope change is physical; smoothing it would violate the model.