Robin boundary condition

A Robin boundary condition prescribes a linear combination of a function and its outward normal derivative on the boundary.

On a bounded region, two sufficiently regular solutions of the Poisson equation with the same forcing and boundary condition $\alpha {\partial }_n\varphi +\varphi =\gamma$, where $\alpha \ge 0$, have harmonic difference $w$. Green's first identity gives $\int |\nabla w{|}^2={\int }_{\partial V}w{\partial }_{n}w=-{\int }_{\partial V}\alpha ({\partial }_{n}w{)}^2\le 0$. Hence $w$ is constant on each component, and the boundary condition makes that constant zero. The proof never divides by $\alpha$, so it includes variable coefficients and Dirichlet portions where $\alpha =0$.

For a decaying harmonic function on a semi-infinite strip with equal outward Robin boundary conditions and reflection-symmetric end data, the top and bottom value traces agree. The global relation and its reflected spectral companion express their transforms through $D(k)$. With $H(k)=\frac{1}{2}{\int }_0^{\ell }{e}^{ky}g(y)dy$, the relation is $\psi (-ik)+\psi (ik)=2H(k)/D(k)$. Real zeros of $D$ require compatible data; unrestricted negative Robin parameters do not guarantee a decaying solution.

Let ${\lambda }_n$ be the transverse eigenvalues of a Sturm-Liouville problem for the two Robin boundary conditions. For a harmonic strip solution decaying in the longitudinal direction, projection of the prescribed end Neumann boundary condition onto every nonpositive eigenvalue must vanish. Positive modes give $q_n(x)=-g_n{e}^{-\sqrt{{\lambda }_n}x}/\sqrt{{\lambda }_n}$. Zero modes are affine and negative modes are oscillatory, so neither can contribute to a decaying solution. The condition also establishes uniqueness within the decay class.

For $Lu=-{\partial }_j(a^{ij}{\partial }_{i}u)+b^i{\partial }_{i}u+cu$ with smooth coefficients on a bounded smooth domain, the weak formulation of $(L+\lambda )u=f$ and conormal Robin boundary condition $a^{ij}{\partial }_{i}u{\nu }_j+\beta u=0$ isThe bilinear form is bounded by the Sobolev trace theorem. If $a^{ij}$ satisfy the positive uniformly elliptic operator inequality, the multiplicative trace inequality on a bounded smooth domain and the Young inequality show that the form is coercive when $\lambda$ is sufficiently large, even if $\beta$ and $c$ have negative parts. The Lax-Milgram theorem then gives a unique weak solution for every $f\in L^2(U)$.