Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2012 / Paper 2 / 2

Use separation of variables with $u=W(x)T(t)$ and choose $W^{\prime \prime }/W=T^{\prime }/T=k$. Then $T=e^{kt}$ and $W^{\prime }(0)=-W(0)$, $W^{\prime }(1)=-W(1)$.

If $k=0$, write $W=A+Bx$. The first condition gives $B=-A$, and the second gives $A+2B=0$. Together they force $A=B=0$, so this case contributes only the zero solution.

If $k={\alpha }^2>0$, write $W=A\cosh (\alpha x)+B\sinh (\alpha x)$. The first condition gives $B=-A/\alpha$. The second becomesA nonzero solution therefore requires $\alpha =1$, giving $W\propto e^{-x}$ and time factor $e^t$.

If $k=-{\beta }^2<0$, write $W=A\cos (\beta x)+B\sin (\beta x)$. Now $B=-A/\beta$, and the remaining condition is $-A(\beta +{\beta }^{-1})\sin \beta =0$. Thus $\beta =n\pi$, $n\ge 1$. Choosing a convenient amplitude gives $W_n=n\pi \cos (n\pi x)-\sin (n\pi x)$. The complete Sturm-Liouville eigenfunction expansion is consequentlyThese are self-adjoint separated Robin boundary conditions: the boundary form $[f^{\ast }{g}^{\prime }-f^{\prime \ast }g{]}_0^1$ vanishes because both functions satisfy $y^{\prime }=-y$ at each end. The regular interval problem has a complete orthogonal eigenbasis; the calculation includes its one negative eigenvalue for $-D^2$, not only its positive ones.

To project the initial data, calculate the squared norms:The second identity uses the half-unit integrals of ${\cos }^2(n\pi x)$ and ${\sin }^2(n\pi x)$ and the zero integral of their product. By orthogonal projection,For square-integrable $f$, the initial expansion is understood in that norm. Classical agreement of boundary derivatives at $t=0$ requires compatible smoother data.

This is Robin diffusion with a growing mode, in the reflected version of its positive-coordinate-derivative example. Growth is not a contradiction: the outward derivative is $-u_x$ at the left endpoint, so these conditions are not dissipative outward-normal conditions at both ends. Indeedand the boundary can supply energy. The growing contribution is absent exactly when the displayed integral defining $c_0$ vanishes.