Separation of Variables — Question 4

Question 4 – Solution

Strategy. Test separation with an invariant determinant, then interpret the changing spatial shape.

Step 1: Establish an exact product test. For a product, both terms of $D$ equal $X(x_1)X(x_2)T(t_1)T(t_2)$. Conversely, use the nonzero anchor to write $u(x,t)u(x_0,t_0)=u(x,t_0)u(x_0,t)$. Hence $$\boxed{X(x)=u(x,t_0),\qquad T(t)=u(x_0,t)/u(x_0,t_0)}$$ gives a product everywhere, including zeros. The nonzero-anchor restriction excludes only the identically zero field, which is already a trivial product.

Step 2: Verify the solution and reject one-product form. Each term has time derivative equal to its second spatial derivative; the sum solves the PDE and vanishes at $x=0,\pi$. At the stated points, $$D=\frac{\sqrt 3}{2}\left(\frac 1{16}-\frac 12\right) =\boxed{-\frac{7\sqrt 3}{32}\ne 0}.$$ Thus superposition preserves the linear PDE but does not preserve the class of single products. The two spatial modes decay at different rates.

Step 3: Track the moving interior node. Factoring gives $u=e^{-t}\sin x[1+2e^{-3t}\cos x]$. On $0<x<\pi$, $\sin x>0$, so the only possible interior zero is $$x_*(t)=\arccos(-e^{3t}/2),\qquad 0\le t<t_*:=\frac{\log 2}{3}.$$ It moves from $2\pi/3$ toward $\pi$. At $t=t_*$ it is at the endpoint, not an interior zero. For $t\ge t_*$ the field is strictly positive in the interior. Before then it is positive to the left of the node and negative to its right.

Step 4: Interpret the normalized plots. The three normalized fields are $\sin x+\sin 2x$, $\sin x+2^{-1/2}\sin 2x$, and $\sin x+\tfrac 12\sin 2x$. Their changing zeros reveal changing shape. If $e^t u$ were a product, multiplying its time factor by $e^{-t}$ would make $u$ a product too, contradicting the determinant. Normalization changes amplitudes but cannot repair this failure of separation.