Separation of Variables — Question 10

Question 10 – Solution

Strategy. Separation reveals individual growth rates; superposition shows how a small unstable component can dominate a decaying one.

Step 1: Construct and verify the finite sum. For $X_n=\sin nx$, $X_n''=-n^2X_n$, so the factor equation is $T_n'=(r-n^2)T_n$. Consequently $$\boxed{u=e^{(r-4)t}\sin 2x+\varepsilon e^{(r-1)t}\sin x.}$$ Each time derivative has coefficient $r-n^2$, matching $u_{xx}+ru$. Both sine factors vanish at $0,\pi$, and their initial coefficients are $1,\varepsilon$. Linearity justifies adding these two verified modes.

Step 2: Compare exact cancellation with small contamination. For $\varepsilon=0$, the solution decays uniformly because $r-4<0$. For fixed $\varepsilon>0$, the first mode grows because $r-1>0$. Moreover $$e^{-(r-1)t}u=\varepsilon\sin x+e^{-3t}\sin 2x, \qquad\|e^{-(r-1)t}u-\varepsilon\sin x\|_\infty=e^{-3t}.$$ This proves uniform convergence of the normalized shape. The unnormalized solution grows without bound in sup norm, as is also seen at $x=\pi/2$.

Step 3: Compare modal norms exactly. Since $\int_0^\pi\sin^2(nx)\,dx=\pi/2$, the squared component norms are $$N_2^2=\frac\pi 2 e^{2(r-4)t},\qquad N_1^2=\frac\pi 2\varepsilon^2e^{2(r-1)t}.$$ Their equality is $\varepsilon^2e^{6t}=1$, giving $\boxed{t_c=\tfrac 13\log(1/\varepsilon)}$. The growing component is larger for $t>t_c$. The crossing time is independent of $r$ in the stated range because the difference of modal rates is always three. Orthogonality also gives $\|u\|_2^2=N_1^2+N_2^2$.

Step 4: Interpret the crossing and the threshold. For $r=2$, $\varepsilon=1/8$, the amplitudes are $a_1=e^t/8$ and $a_2=e^{-2t}$. They cross at $t_c=\log 2$, with common value $1/4$. A decaying mode only describes its own initial shape. A nonzero first-mode component decays for $r<1$, is stationary for $r=1$, and grows for $r>1$. Thus one decaying example cannot establish decay for every admissible field.