 Solution

Use separation of variables in the form $u=X(x)T(t)$, with $T^{\prime }/T=X^{\prime \prime }/X=-\lambda$. Then $T=e^{-\lambda t}$ and the spatial Sturm-Liouville problem isThese 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 toThus $\mu =n\pi$, 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 becomesHence $\kappa =1$, giving the single negative eigenvalue $-1$ and spatial eigenfunction $e^x$. Completeness of this regular Sturm-Liouville problem now givesThis 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.