Separation of Variables — Question 3

Question 3 – Solution

Strategy. The zero eigenvalue carries a conserved quantity and must be analyzed before any division by the separation constant.

Step 1: Retain the constant spatial mode. Integration by parts gives $\lambda\int X^2=\int(X')^2\ge 0$ because both endpoint derivatives vanish. Thus negative eigenvalues are excluded. For $\lambda=0$, $X=ax+b$ and the endpoints force $a=0$; nonzero constants are allowed. For $\lambda=k^2>0$, the left derivative condition removes the sine term and the right requires $\sin(kL)=0$. Consequently $$\boxed{\lambda_0=0,\ X_0=1;\qquad \lambda_n=(n\pi/L)^2,\ X_n=\cos(n\pi x/L),\ n\ge 1.}$$

Step 2: Track the conserved mean. For $m(t)=L^{-1}\int_0^L u(x,t)\,dx$, $m'(t)=\kappa[u_x]_0^L/L=0$. Integrating $X''+\lambda X=0$ gives $\lambda\int_0^L X\,dx=-[X']_0^L=0$. Thus every $\lambda>0$ factor has zero mean. The zero mode has a constant time factor and can carry a nonzero mean.

Step 3: Verify the two-mode construction. Let $\omega=\kappa(\pi/L)^2$. Then $$\boxed{u(x,t)=2+3e^{-\omega t}\cos(\pi x/L).}$$ Its time derivative is $-3\omega e^{-\omega t}\cos(\pi x/L)$, equal to $\kappa u_{xx}$. Its endpoint derivatives vanish, and its initial trace is the prescribed function. Its mean is two for every time. Since the decaying term has absolute value at most $3e^{-\omega t}$, it tends uniformly to $u=2$. No infinite-series argument is required for this finite construction.

Step 4: Diagnose the missing mode. Dividing by $\lambda$ loses the legitimate constant eigenfunction and the corresponding stationary PDE solutions. Every finite linear combination of positive-eigenvalue modes has zero spatial mean at each time, while the specified initial data have mean two. Such a combination cannot reproduce these data. This obstruction follows from the integral, not from an inability to find enough trigonometric coefficients.