The Backward-Euler Method 2.5 2001/07/06 2002/07/12 00:00:00.005 GMT-5 Doug Daniels rainking@alumni.rice.edu Steven Cox cox@rice.edu Doug Daniels rainking@alumni.rice.edu Steven Cox cox@rice.edu Jacob Grabczewski jgrab@owlnet.rice.edu backward-euler method 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. Where in the Inverse Laplace Transform module we tackled the derivative in x B x g , 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 and 'replaces' with x ˜ t x ˜ t d t d t B x ˜ t g t . The utility of is that it gives a means of solving for x ˜ at the present time, t, from the knowledge of x ˜ in the immediate past, t d t . For example, as x ˜ 0 x 0 is supposed known we write as I d t B x ˜ dt x 0 dt g dt . Solving this for x ˜ dt we return to and find I dt B x ˜ 2 dt x dt dt g 2 dt . and solve for x ˜ 2 dt . The general step from past to present, x ˜ j dt I dt B x ˜ j 1 d t d t g j dt is repeated until some desired final time, T dt , is reached. This equation has been implemented in fib3.m with dt 1 and B and g as in the dynamic Strang module. The resulting x ˜ ( run fib3.m yourself!) is indistinguishable from the plot we obtained in the Inverse Laplace module. Comparing the two representations, this equation and , 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 B. The former representation is hard but exact while the latter is easy but approximate. Of course we should expect the approximate solution, x ˜ , to approach the exact solution, x, as the time step, dt , approaches zero. To see this let us return to and assume, for now, that g 0 . In this case, one can reverse the above steps and arrive at the representation x ˜ j dt I dt B j x 0 Now, for a fixed time t we suppose that dt t j and ask whether x t j I t j B j x 0 This limit, at least when B is one-by-one, yields the exponential 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.