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Suppose I have the parabolic system $$u_t=\nabla\cdot(k(x)\nabla u)+f,\quad (x,t)\in\Omega\times I$$ with dirichlet boundary conditions $$u=g, \quad x\in\partial\Omega$$ and initial condition $$u(x,t)= h,\quad t=0.$$

Often times in engineering, we are more interested in the asymptotic (steady state) behavior of this pde rather than the transient behavior. So, we sometimes neglect the time derivative term and solve the elliptic system $$-\nabla\cdot(k(x)\nabla u)=f,\quad (x,t)\in\Omega\times I$$ instead. The assumption is that over infinite time, $$lim_{t\rightarrow \infty}u_{parabolic}(x,t)= u_{elliptic}(x,t)$$.

I've observed that when $f\equiv 0$, this limit is true, but I'm not sure if this is the case for arbitrary $f$ or if there are other necessary conditions to guarantee this limit to be true. Do the boundary conditions have to converge asymptotically to a constant value in order for the parabolic solution to converge to the elliptic solution?

Though I'm my question is formulated in the continuous case, I'm also curious if the same conditions are true for the discrete case. That is, assuming I use a stable and consistent finite difference scheme to approximate $u_{parabolic}$ and $u_{elliptic}$, should I expect $$lim_{t\rightarrow \infty}u_{parabolic}^{fdm}(x,t)= u_{elliptic}^{fdm}(x,t)$$ if $u_{elliptic}^{fdm}$ and $u_{parabolic}^{fdm}$ are discretized on the same spatial grid and $\Delta t\rightarrow\infty$?

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1 Answer 1

up vote 2 down vote accepted

Yes, with Dirichlet boundary conditions you always have exponential convergence to the staedy state. Any PDE book will have a proof. For a nice explanation from a numerical perspective, see chapter 2 of LeVeque's FDM book.

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The same should be true if there were mixed conditions (neumann and dirichlet), right? –  Paul Jan 24 '13 at 15:51
@Paul Yes, the only sticky case in 1D is when both boundaries are Neumann. –  David Ketcheson Jan 27 '13 at 8:37

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