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The Old Laplace Transform

Module by: Doug Daniels, Steven Cox

Summary: This module examines the Laplace Transform, an analytical tool that produces exact solutions for small, closed-form, tractable systems. We use the Laplace transform to move toward a solution for the nerve fiber potentials modeled by the dynamic Strang Quartet in the earlier module of the same name.

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The Laplace Transform is typically credited with taking dynamical problems into static problems. Recall that the Laplace Transform of the function hh is ℒhs≡∫0∞ⅇ-sthtdt . ℒ h s t 0 s t h t . MATLAB is very adept at such things. For example:

Example 1: The Laplace Transform in MATLAB


	>> syms t

	>> laplace(exp(t))

	ans = 1/(s-1)

	>> laplace(t*(exp(-t))

	ans = 1/(s+1)^2

The Laplace Transform of a matrix of functions is simply the matrix of Laplace transforms of the individual elements.

Example 2: Laplace Transform of a matrix of functions

ℒⅇttⅇ-t=1s−11s+12 ℒ t t t 1 s 1 1 s 1 2

Now, in preparing to apply the Laplace transform to our equation from the dynamic strang quartet module: x ′=Bx+g , x B x g , we write it as

ℒddtx=ℒBx+g ℒ t x ℒ B x g (1)
and so must determine how ℒℒ acts on derivatives and sums. With respect to the latter it follows directly from the definition that
ℒBx+g=ℒBx+ℒg=Bℒx+ℒg . ℒ B x g ℒ B x ℒ g B ℒ x ℒ g . (2)
Regarding its effect on the derivative we find, on integrating by parts, that ℒddtx=∫0∞ⅇ-stddtxtdt=xtⅇ-st|0∞+s∫0∞ⅇ-stxtdt . ℒ t x t 0 s t t x t 0 x t s t s t 0 s t x t . Supposing that xx and ss are such that xtⅇ-st→0 x t s t 0 as t→∞ t we arrive at
ℒddtx=sℒx−x0 . ℒ t x s ℒ x x 0 . (3)
Now, upon substituting Equation 2 and Equation 3 into Equation 1 we find sℒx−x0=Bℒx+ℒg , s ℒ x x 0 B ℒ x ℒ g , which is easily recognized to be a linear system for ℒx ℒ x , namely
sI−Bℒx=ℒg+x0 . s I B ℒ x ℒ g x 0 . (4)
The only thing that distinguishes this system from those encountered since our first brush with these systems is the presence of the complex variable ss. This complicates the mechanical steps of Gaussian Elimination or the Gauss-Jordan Method but the methods indeed apply without change. Taking up the latter method, we write ℒx=sI−B-1ℒg+x0 . ℒ x s I B ℒ g x 0 . The matrix sI−B-1 s I B is typically called the transfer function or resolvent, associated with BB, at ss. We turn to MATLAB for its symbolic calculation. (for more information, see the tutorial on MATLAB's symbolic toolbox). For example,

Example 3


	>> B = [2 -1; -1 2]
	
	>> R = inv(s*eye(2)-B)
	
	R =
	
	[ (s-2)/(s*s-4*s+3), -1/(s*s-4*s+3)]
	
	[ -1/(s*s-4*s+3), (s-2)/(s*s-4*s+3)]

We note that sI−B-1 s I B is well defined except at the roots of the quadratic, s2−4s+3 s 2 4 s 3 . This quadratic is the determinant of sI−B s I B and is often referred to as the characteristic polynomial of BB. Its roots are called the eigenvalues of BB.

Example 4

As a second example let us take the BB matrix of the dynamic Strang quartet module with the parameter choices specified in fib3.m, namely

B=-0.1350.12500.5-1.010.500.5-0.51 B -0.1350.1250 0.5-1.010.5 00.5-0.51 (5)
The associated sI−B-1 s I B is a bit bulky (please run fib3.m) so we display here only the denominator of each term, i.e.,
s3+1.655s2+0.4078s+0.0039 . s 3 1.655 s 2 0.4078 s 0.0039 . (6)
Assuming a current stimulus of the form i 0 t=t3ⅇ-t610000 i 0 t t 3 t 6 10000 and E m =0 E m 0 brings ℒgs=0.191s+16400 ℒ g s 0.191 s 1 6 4 0 0 and so Equation 6 persists in ℒx=sI−B-1ℒg=0.191s+164s3+1.655s2+0.4078s+0.0039s2+1.5s+0.270.5s+0.260.2497 ℒ x s I B ℒ g 0.191 s 1 6 4 s 3 1.655 s 2 0.4078 s 0.0039 s 2 1.5 s 0.27 0.5 s 0.26 0.2497

Now comes the rub. A simple linear solve (or inversion) has left us with the Laplace transform of xx. The accursed

Theorem 1: No Free Lunch Theorem

We shall have to do some work in order to recover xx from ℒx ℒ x .

confronts us. We shall face it down in the Inverse Laplace module.

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