= Robin diffusion with a growing mode
{c}
{title2=$u_t=u_{xx},\quad u_x(0)=u(0),\quad u_x(1)=u(1)$}
These <Robin boundary conditions> give the spatial <Sturm-Liouville operator> $-d^2/dx^2$ one negative <eigenvalue>, $-1$, with <eigenfunction> $e^x$. The other eigenvalues are $n^2\pi^2$, with eigenfunctions $V_n=n\pi\cos(n\pi x)+\sin(n\pi x)$, $n\geq1$; there is no zero eigenvalue. Hence the <diffusion equation> has a growing contribution $c_0e^{t+x}$ 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 $\tfrac12\partial_t\int u^2=[uu_x]_0^1-\int u_x^2$, whose boundary term is $u(1)^2-u(0)^2$ here and can supply energy. Orthogonal projection of initial data $f$ gives $c_0=2\int_0^1fe^x\,dx/(e^2-1)$; 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 $x\mapsto1-x$ reverses the coordinate derivative. Here the growing eigenfunction is $e^{-x}$ and the decaying functions are $W_n=n\pi\cos(n\pi x)-\sin(n\pi x)$, giving
$$
u(x,t)=c_0e^{t-x}+\sum_{n\geq1}c_ne^{-n^2\pi^2t}W_n(x).
$$
For square-integrable initial data $f$, <orthogonal projection> gives
$$
c_0=\frac{2}{1-e^{-2}}\int_0^1e^{-x}f(x)dx,\qquad c_n=\frac{2}{n^2\pi^2+1}\int_0^1W_n(x)f(x)dx.
$$
The 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 $\int_0^1e^{-x}f(x)dx=0$.
Back to article page
