= Solution
Use <separation of variables> in the form $u=X(x)T(t)$, with $T'/T=X''/X=-\lambda$. Then $T=e^{-\lambda t}$ and the spatial <Sturm-Liouville problem> is
$$
-X''=\lambda X,\qquad X'(0)=X(0),\qquad X'(1)=X(1).
$$
These real separated <Robin boundary conditions> give a <self-adjoint differential operator>, so the <eigenvalues> are real. All three sign cases must be considered.
If $\lambda=\mu^2>0$, write $X=A\cos\mu x+B\sin\mu x$. The first condition gives $B=A/\mu$, while the second reduces to
$$
-A\left(\mu+\frac1\mu\right)\sin\mu=0.
$$
Thus $\mu=n\pi$, and scaling the <eigenfunction> gives $V_n=n\pi\cos(n\pi x)+\sin(n\pi x)$.
If $\lambda=0$, then $X=A+Bx$. The first condition gives $B=A$, and the second gives $B=A+B$, so $A=B=0$; there is no zero <eigenvalue>.
If $\lambda=-\kappa^2<0$, write $X=A\cosh\kappa x+B\sinh\kappa x$. The first condition gives $B=A/\kappa$, and the second becomes
$$
A\left(\kappa-\frac1\kappa\right)\sinh\kappa=0.
$$
Hence $\kappa=1$, giving the single negative <eigenvalue> $-1$ and spatial <eigenfunction> $e^x$. Completeness of this regular <Sturm-Liouville problem> now gives
$$
\boxed{u(x,t)=c_0e^{t+x}+\sum_{n=1}^{\infty}c_ne^{-n^2\pi^2t}\left[n\pi\cos(n\pi x)+\sin(n\pi x)\right].}
$$
This is <Robin diffusion with a growing mode>. The growth is compatible with the stated conditions: they prescribe the same coordinate <derivative> at both ends, not identical dissipative outward-normal conditions.
Back to article page
