Separation of Variables — Question 6

Question 6 – Solution

Strategy. The boundary conditions select the spatial eigenvalue, while the PDE determines the sign in the other factor equation.

Step 1: Separate with the correct opposite signs. Substitution yields $X''Y+XY''=0$. Nonzero anchors, as in the previous separation arguments, give $X''=-\lambda X$, $Y''=\lambda Y$ globally. The vertical sides impose $X(0)=X(\pi)=0$. Integration by parts excludes $\lambda\le 0$ for nonzero $X$; solving the positive case gives $\lambda=n^2$ and $X=\sin(nx)$, with the spatial constant absorbed into $Y$.

Step 2: Match the one-mode top trace. The top data are nonzero, so $Y(H)\ne 0$ and $X$ must be proportional to $\sin 2x$. Thus $n=2$. The remaining conditions are $Y(0)=0,Y(H)=1$, which determine $$\boxed{u(x,y)=\sin(2x)\frac{\sinh(2y)}{\sinh(2H)}.}$$ Its second $x$ derivative is $-4u$ and its second $y$ derivative is $4u$. The sum is zero. The sine factor vanishes at the sides, $\sinh 0=0$ gives the bottom, and the ratio equals one at the top. The denominator is nonzero since $H>0$. This uniquely determines the field within the stipulated product class.

Step 3: Reject a plausible sign error. For $\widetilde u=\sin 2x\sin 2y$, both second derivatives equal $-4\widetilde u$, so $\Delta\widetilde u=-8\widetilde u$, not identically zero. Rescaling a nonzero field to repair top values would not remove this interior residual. The required equation $Y''-4Y=0$ has hyperbolic, rather than sinusoidal, solutions because the Laplace equation sums the second derivatives.

Step 4: Interpret growth in a bounded spatial direction. For $a(y)=\sinh(2y)/\sinh(2H)$, one has $a(0)=0$, $a(H)=1$ and $a'(y)=2\cosh(2y)/\sinh(2H)>0$. Hence $0\le a\le 1$ and $|u|\le 1$. For the same $X''=-\lambda X$ with $\lambda>0$, the remaining equations are $$\begin{array}{ll} \text{heat:}&T'+\kappa\lambda T=0,\\ \text{wave:}&T''+c^2\lambda T=0,\\ \text{Laplace:}&Y''-\lambda Y=0. \end{array}$$ Spatial hyperbolic growth in the last equation is compatible with bounded boundary data on a finite rectangle; it is not a statement about time instability.