Summary of Separation of Variables — Question 5

Question 5 – Solution

Strategy. A repeated eigenvalue allows a common time factor without forcing the spatial combination to be a single coordinate product.

Step 1: Identify the repeated eigenvalue. Zero edges give $\sin(mx)\sin(ny)$ with eigenvalues $m^2+n^2$. The pairs $(1,2)$ and $(2,1)$ both give five. Therefore $$\boxed{U=e^{-5t}\bigl[\sin x\sin(2y)+2\sin(2x)\sin y\bigr].}$$ Every edge and the initial field are correct, and applying the Laplacian multiplies the bracket by $-5$.

Step 2: Distinguish two meanings of separation. The displayed formula is exactly one product $T(t)F(x,y)$. It is not one product $T(t)X(x)Y(y)$. Indeed, the spatial slice at $y=\pi/2$ is $2\sin(2x)$, whereas at $y=\pi/4$ it is $\sin x+\sqrt 2\sin(2x)$. These two nonzero slices are not proportional by sine orthogonality, as they would have to be for $X(x)Y(y)$.

Step 3: Determine the nodal geometry. Factor the bracket as $2\sin x\sin y\,(\cos y+2\cos x)$. The first two factors are positive inside the square, so the interior zero set is $$\boxed{\cos y=-2\cos x,\qquad \pi/3<x<2\pi/3.}$$ Its limiting endpoints are $(\pi/3,\pi)$ and $(2\pi/3,0)$. At fixed $y$, the field is positive for $x<\arccos(-\cos y/2)$ and negative to its right. The zero curve is unchanged by the positive decay factor.

Step 4: Resolve the equal-rate inverse problem. At any one sensor, the entire trace has the form $e^{-5t}(A\alpha+B\beta)$ with fixed spatial factors $\alpha,\beta$; it supplies only one scalar equation. At the two specified sensors the unwanted product vanishes and the wanted product equals one. Hence $$\boxed{A=e^{5t}U(\pi/2,\pi/4,t),\qquad B=e^{5t}U(\pi/4,\pi/2,t).}$$ Distinct time samples at one point cannot separate equal decay rates; spatially independent observations can.