Comparing the Laplace Transform and the Backward-Euler Method 2.3 2001/07/06 2002/07/12 Doug Daniels rainking@alumni.rice.edu Steven Cox cox@rice.edu Doug Daniels rainking@alumni.rice.edu Steven Cox cox@rice.edu backward-euler method inverse laplace transform laplace transform matrix exponential This module compares two methods for solving time-dependent systems, the Backward-Euler method and the Laplace transform. Comparing the two representations, the Laplace: x t ℒ s I B ℒ g x 0 , and the Backward Euler: x ˜ j dt I dt B x ˜ j 1 d t dt g j dt , we see that they both produce the solution to the general linear system of ordinary equations, x B x g , 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 d t 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-to-one, yields the exponential x t B t x 0 , clearly the correct solution to . A careful explication of the matrix exponential and its relationship to will have to wait until we have mastered the inverse laplace transform.