 Robin diffusion with a growing mode

These Robin boundary conditions give the spatial Sturm-Liouville operator $-d^2/d{x}^2$ one negative eigenvalue, $-1$, with eigenfunction $e^x$. The other eigenvalues are $n^2{\pi }^2$, with eigenfunctions $V_n=n\pi \cos (n\pi x)+\sin (n\pi x)$, $n\ge 1$; there is no zero eigenvalue. Hence the diffusion equation has a growing contribution $c_0{e}^{t+x}$ in addition to decaying modes.

There is no contradiction with ordinary diffusive smoothing. The outward normal derivative is $-u_x$ at the left endpoint and $u_x$ at the right endpoint, so imposing the same coordinate derivative at both ends does not impose dissipative outward-normal data at both ends. The energy identity is $\frac{1}{2}{\partial }_t\int u^2=[u{u}_x{]}_0^1-\int u_x^2$, whose boundary term is $u(1{)}^2-u(0{)}^2$ here and can supply energy. Orthogonal projection of initial data $f$ gives $c_0=2{\int }_0^{1}f{e}^{x}dx/(e^2-1)$; it vanishes precisely when the growing mode is absent.

The reflected boundary conditions $u_x(0,t)=-u(0,t)$ and $u_x(1,t)=-u(1,t)$ have the same spectrum, because reflection $x\mapsto 1-x$ reverses the coordinate derivative. Here the growing eigenfunction is $e^{-x}$ and the decaying functions are $W_n=n\pi \cos (n\pi x)-\sin (n\pi x)$, givingFor square-integrable initial data $f$, orthogonal projection givesThe norms are $(1-e^{-2})/2$ and $(n^2{\pi }^2+1)/2$. The spatial Sturm-Liouville problem has no zero eigenvalue and exactly one negative eigenvalue for $-D^2$, namely $-1$. The energy boundary term now reads $u(0{)}^2-u(1{)}^2$, so the left endpoint can supply energy. The growing mode disappears exactly when ${\int }_0^1{e}^{-x}f(x)dx=0$.