Demonstrate, by running a program, that you can take one large time step with the Backward Euler scheme and compute the solution of (9.38).

1.Many diffusion problems reach a stationary time-independent solution as t → ∞. The model problem from Sect. 9.2.4 is one example where u(x, t) = s(t) = const for t → ∞. When u does not depend on time, the diffusion equation reduces to −βu(x) = f (x), in one dimension, and −β∇2u = f (x), in 2D and 3D. This is the famous Poisson equation, or if f = 0, it is known as the Laplace equation. In this limit t → ∞, there is no need for an initial condition, but the boundary conditions are the same as for the diffusion equation. We now consider a one-dimensional problem − u(x) = 0, x ∈ (0, L), u(0) = C, u (L) = 0, (9.38) which is known as a two-point boundary value problem. This is nothing but the stationary limit of the diffusion problem in Sect. 9.2.4. How can we solve such a stationary problem (9.38)? The simplest strategy, when we already have a solver for the corresponding time-dependent problem, is to use that solver and simulate until t → ∞, which in practice means that u(x, t) no longer changes in time (within some tolerance). A nice feature of implicit methods like the Backward Euler scheme is that one can take one very long time step to “infinity” and produce the solution of (9.38). a) Let (9.38) be valid at mesh points xi in space, discretize u by a finite difference, and set up a system of equations for the point values ui,i = 0,…,N, where ui is the approximation at mesh point xi. b) Show that if Δt → ∞ in (9.16)–(9.18), it leads to the same equations as in a). c) Demonstrate, by running a program, that you can take one large time step with the Backward Euler scheme and compute the solution of (9.38). The solution is very boring since it is constant: u(x) = C.

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