Separation of Variables — Question 2

Question 2 – Solution

Strategy. Let the actual boundary operators select the admissible separation constants.

Step 1: Prove positivity before solving the ODE. At a time where $T\ne 0$, the endpoint conditions give $X(0)=0$, $X'(L)=0$. Multiplication by $X$ and integration yield $$\lambda\int_0^L X^2\,dx =\int_0^L (X')^2\,dx-[XX']_0^L=\int_0^L(X')^2\,dx.$$ If $\lambda=0$, then $X'$ vanishes identically; $X(0)=0$ forces $X=0$. Thus a nonzero factor has $\boxed{\lambda>0}$; negative values are impossible.

Step 2: Apply the two different endpoint conditions. Put $k=\sqrt\lambda$. The general solution is $A\cos(kx)+B\sin(kx)$. The left condition gives $A=0$, and the right gives $Bk\cos(kL)=0$. For $B\ne 0$, $$\boxed{k_n=\frac{(n+1/2)\pi}{L},\quad \lambda_n=k_n^2,\quad X_n(x)=\sin(k_nx),\quad n=0,1,\ldots.}$$ The right condition is on the derivative, so it selects zeros of cosine. Using $\sin(kL)=0$ would impose an unrequested Dirichlet condition there.

Step 3: Compare decay measures. The modes are $C X_n(x)e^{-\kappa\lambda_n t}$. Their amplitude half-lives are $\tau_n=\log 2/(\kappa\lambda_n)$, so $\tau_1=\tau_0/9$. Squared amplitude decays as $e^{-2\kappa\lambda_n t}$ and has half-life $\tau_n/2$. Confusing those two quantities introduces a factor of two.

Step 4: Check the plotted nodes and slopes. For $n=0,1,2$, the normalized factors are $\sin((n+1/2)\pi\xi)$. There are respectively zero, one and two interior zeros: none; $2/3$; and $2/5,4/5$. Every right slope is zero because $\cos((n+1/2)\pi)=0$. The right values alternate $1,-1,1$ and need not vanish.