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The Backward-Euler Method

Module by: Doug Daniels, Steven Cox

Summary: This module examines the Backward-Euler method, which uses a finite difference quotient to approximate the solution of a time-dependent system of equations. This module is a continuation of the nerve fiber example introduced in the dynamic Strang quartet module.

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Where in the Inverse Laplace Transform module we tackled the derivative in

x ′=Bx+g , x B x g , (1)
via an integral transform we pursue in this section a much simpler strategy, namely, replace the derivative with a finite difference quotient. That is, one chooses a small d t d t and 'replaces' Equation 1 with
x ˜ t− x ˜ t− d t d t =B x ˜ t+gt . x ˜ t x ˜ t d t d t B x ˜ t g t . (2)
The utility of Equation 2 is that it gives a means of solving for x ˜ x ˜ at the present time, tt, from the knowledge of x ˜ x ˜ in the immediate past, t− d t t d t . For example, as x ˜ 0=x0 x ˜ 0 x 0 is supposed known we write Equation 2 as I d t −B x ˜ dt=x0dt+gdt . I d t B x ˜ dt x 0 dt g dt . Solving this for x ˜ dt x ˜ dt we return to Equation 2 and find Idt−B x ˜ 2dt=xdtdt+g2dt . I dt B x ˜ 2 dt x dt dt g 2 dt . and solve for x ˜ 2dt x ˜ 2 dt . The general step from past to present,
x ˜ jdt=Idt−B-1 x ˜ j−1 d t d t +gjdt x ˜ j dt I dt B x ˜ j 1 d t d t g j dt (3)
is repeated until some desired final time, Tdt T dt , is reached. This equation has been implemented in fib3.m with dt=1 dt 1 and BB and gg as in the dynamic Strang module. The resulting x ˜ x ˜ ( run fib3.m yourself!) is indistinguishable from the plot we obtained in the Inverse Laplace module.

Comparing the two representations, this equation and Equation 3, we see that they both produce the solution to the general linear system of ordinary equations, see this eqn, by simply inverting a shifted copy of BB. The former representation is hard but exact while the latter is easy but approximate. Of course we should expect the approximate solution, x ˜ x ˜ , to approach the exact solution, xx, as the time step, dt dt , approaches zero. To see this let us return to Equation 3 and assume, for now, that g≡0 g 0 . In this case, one can reverse the above steps and arrive at the representation

x ˜ jdt=I− dt B-1jx0 x ˜ j dt I dt B j x 0 (4)
Now, for a fixed time tt we suppose that dt =tj dt t j and ask whether xt=limj→∞I−tjB-1jx0 x t j I t j B j x 0 This limit, at least when BB is one-by-one, yields the exponential xt=ⅇBtx0 x t B t x 0 clearly the correct solution to this equation. A careful explication of the matrix exponential and its relationship to this equation will have to wait until we have mastered the inverse laplace transform.

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