 laplace - Maple Help

MTM

 laplace
 Laplace integral transform Calling Sequence laplace(M) laplace(M,s) laplace(M,t, s) Parameters

 M - array or expression t - variable s - variable Description

 • L = laplace(F) is the Laplace transform of the scalar F with default independent variable t.  If F is not a function of t, then F is  assumed to be a function of the independent variable returned by findsym(F,1).The default return is a function of s.
 • If F = F(s), then laplace returns a function of t.
 • By definition,

$L\left(s\right)={{\int }}_{0}^{\mathrm{\infty }}F\left(t\right){ⅇ}^{-ts}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}t$

where the integration above proceeds with respect to t.

 • laplace(F,x) makes L a function of the variable x instead of the default s.
 • laplace(F,z,x) makes L a function of x instead of the default s. The integration is then with respect to z.
 • The laplace(M) function computes the element-wise Laplace transform of M.  The result, R, is formed as R[i,j] = laplace(M[i,j]). Examples

 > $\mathrm{with}\left(\mathrm{MTM}\right):$
 > $\mathrm{laplace}\left(\mathrm{BesselI}\left(0,2t\right)\right)$
 $\frac{{1}}{\sqrt{{{s}}^{{2}}{-}{4}}}$ (1)
 > $\mathrm{laplace}\left(\mathrm{BesselI}\left(0,2s\right)\right)$
 $\frac{{1}}{\sqrt{{{t}}^{{2}}{-}{4}}}$ (2)
 > $\mathrm{laplace}\left(\mathrm{BesselI}\left(0,2t\right),x\right)$
 $\frac{{1}}{\sqrt{{{x}}^{{2}}{-}{4}}}$ (3)
 > $\mathrm{laplace}\left(\mathrm{BesselI}\left(0,z\cdot 2t\right),z,x\right)$
 $\frac{{1}}{\sqrt{{-}{4}{}{{t}}^{{2}}{+}{{x}}^{{2}}}}$ (4)
 > $M≔\mathrm{Matrix}\left(\left[\mathrm{BesselI}\left(0,2t\right),\mathrm{BesselI}\left(0,z\cdot 2t\right)\right]\right):$
 > $\mathrm{laplace}\left(M\right)$
 $\left[\begin{array}{cc}\frac{{1}}{\sqrt{{{s}}^{{2}}{-}{4}}}& \frac{{1}}{\sqrt{{{s}}^{{2}}{-}{4}{}{{z}}^{{2}}}}\end{array}\right]$ (5)