Summary of Separation of Variables — Question 2

Question 2 – Solution

Strategy. The spatial constant is an actual eigenfunction for insulated sides; its remaining equation changes with the type of PDE.

Step 1: Retain the zero eigenvalue. The problem $-X''=\lambda X$, $X'(0)=X'(\pi)=0$ has $X_0=1$, $\lambda_0=0$, and $X_n=\cos(nx)$, $\lambda_n=n^2$ for $n\ge 1$. Their squared $L^2(0,\pi)$ norms are $\pi$ and $\pi/2$, respectively. The mean coefficient therefore uses $1/\pi$, not the positive-mode factor $2/\pi$.

Step 2: Solve the zero and first-mode equations. For $\lambda=0$, the remaining equations are $a'=0$, $b''=0$ and $d''=0$. Using all the supplied data gives $$\boxed{\begin{aligned} H(x,t)&=2+e^{-t}\cos x,\\ W(x,t)&=2+3t+(\cos t+2\sin t)\cos x,\\ P(x,y)&=2y+\frac{\sinh y}{\sinh 1}\cos x. \end{aligned}}$$ The cosine derivatives vanish at the side boundaries. The scalar factors verify the PDEs, both wave initial data and the two harmonic edge traces.

Step 3: Compare the three means. Since $\int_0^\pi\cos x\,dx=0$, their horizontal means are $$\boxed{\overline H=2,\qquad \overline W=2+3t,\qquad \overline P=2y.}$$ Omitting the constant spatial eigenfunction would lose the heat initial mean, both the wave initial mean and mean velocity, and the harmonic top mean. The zero mode preserves a heat mean, permits a freely drifting wave mean, and linearly joins two spatial mean values in the harmonic problem.

Step 4: Separate drift energy from modal energy. Put $q=\cos t+2\sin t$. Orthogonality gives $$E=\frac 12\left[9\pi+\frac\pi 2(q'^2+q^2)\right] =\frac{9\pi}{2}+\frac{5\pi}{4}=\boxed{\frac{23\pi}{4}}.$$ The spatially constant displacement $2+3t$ contributes no elastic energy, although its velocity contributes a fixed kinetic energy. Thus $W$ becomes unbounded through its mean while its energy remains constant. A displacement bound from this energy alone would require control of that mean.