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Comparing the Laplace Transform and Backward-Euler Method

Module by: Doug Daniels, Steven Cox

Summary: This module compares two methods for solving time-dependent systems, the Backward-Euler method and the Laplace transform.

Comparing the two representations, the Laplace:

xt=ℒ-1sI-B-1ℒg+x0 , x t ℒ s I B ℒ g x 0 , (1)
and the Backward Euler:
x ˜ jdt=Idt-B-1 x ˜ j-1 d t dt+gjdt , x ˜ j dt I dt B x ˜ j 1 d t dt g j dt , (2)
we see that they both produce the solution to the general linear system of ordinary equations,
x ′=Bx+g , x B x g , (3)
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 2 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- d t B-1jx0 . x ˜ j dt I d t 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-to-one, yields the exponential xt=ⅇBtx0 , x t B t x 0 , clearly the correct solution to Equation 3. A careful explication of the matrix exponential and its relationship to Equation 1 will have to wait until we have mastered the inverse laplace transform.

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