Lecture 18

In this Lecture we discuss how to use Mathematica in our studies of the expansion (to use in the solu-

tion of differential equations) of non-integrable functions. The tool to be used here is the Laplace

transform. This is effectively a Fourier transform (and inverse transform) continued into the complex

plane, i.e., the path of the underlying integral is parallel to the imaginary axis instead of the real axis.

Recall that we apply this transform to functions that are zero for arguments < 0, by assumption. We

can start by considering some examples (See Eq. (18.4))

In[6]:= f1@t_D := UnitStep@tD

In[7]:= LaplaceTransform@f1@tD, t, pD

Out[7]=

1

p

Note that Mathematica does not remind us of the constraints on the value of p, e.g., here that Re[p] > 0.

Next consider powers

In[8]:= f2@t_D := t^n UnitStep@tD

In[9]:= LaplaceTransform@f2@tD, t, pD

Out[9]= p-1-n

Gamma@1 + nDwhere again we must really require that Re[p] > 0. Now exponentials

In[10]:= f3@t_D := Exp@-a tD UnitStep@tD

In[11]:= LaplaceTransform@f3@tD, t, pD

Out[11]=

1

a + p

where Re[p+a] > 0 is required to make sense.

In[12]:= f4@t_D := Sin@Β tD UnitStep@tD

In[13]:= LaplaceTransform@f4@tD, t, pD

Out[13]=

Β

p2 + Β2

In[14]:= f5@t_D := Cos@Β tD UnitStep@tD

In[15]:= LaplaceTransform@f5@tD, t, pD

Out[15]=

p

p2 + Β2

In[16]:= f6@t_D := Sinh@Γ tD UnitStep@tD

In[17]:= LaplaceTransform@f6@tD, t, pD

Out[17]=

Γ

p2 - Γ2

In[18]:= f7@t_D := Cosh@Γ tD UnitStep@tD

In[19]:= LaplaceTransform@f7@tD, t, pD

Out[19]=

p

p2 - Γ2

Then we have the inverse transform (a defined function to use instead of the contour integrals dis-

cussed in the Lecture)

In[20]:= InverseLaplaceTransformB1

p

, p, tF

Out[20]= 1

In[21]:= InverseLaplaceTransformAp-1-n Gamma@1 + nD, p, tEOut[21]= t

n

Note that Mathematica does not explicitly return the implied theta function, UnitStep[t], which is explicit

in the analytic approach.

In[22]:= InverseLaplaceTransformB1

a + p, p, tF

Out[22]= ã-a t

In[23]:= InverseLaplaceTransformBΒ

p2 + Β2, p, tF

Out[23]= Sin@t ΒD

In[24]:= InverseLaplaceTransformBp

p2 + Β2, p, tF

Out[24]= Cos@t ΒD

In[25]:= InverseLaplaceTransformBΓ

p2 - Γ2, p, tF

Out[25]=

1

2

ã-t Γ I-1 + ã2 t ΓM

Also known as Sinh[Γ t]

In[26]:= InverseLaplaceTransformBp

p2 - Γ2, p, tF

Out[26]=

1

2

ã-t Γ I1 + ã2 t ΓM

Or Cosh[Γ t].

2 Lec18_228_09.nb

For fun we can also try the inverse transform as the explicit (Bromwich) integral

In[27]:=

1

2 Π äIntegrateB

p

p2 - Γ2Exp@p tD, 8p, Γ + 1 - ä ¥, Γ + 1 + ä ¥

In[34]:= LaplaceTransform@x''@tD + 2 x'@tD + x@tD, t, pDOut[34]= LaplaceTransform@x@tD, t, pD +

2 p LaplaceTransform@x@tD, t, pD + p2 LaplaceTransform@x@tD, t, pDThen we can transform the differential equation and solve for the transform of the solution, which we

identify in the function call

In[35]:= Solve@LaplaceTransform@x''@tD + 2 x'@tD + x@tD, t, pD LaplaceTransform@t, t, pD,LaplaceTransform@x@tD, t, pDD

Out[35]= ::LaplaceTransform@x@tD, t, pD ® 1p2 H1 + pL2 >>

Thus we are back to where we were above and the transform back

In[36]:= InverseLaplaceTransform@%, p, tD

Out[36]= 99x@tD ® -2 + 2 ã-t + t + ã-t t==Alternatively we could include the initial conditions inside Solve

In[37]:= Clear@xD

In[38]:= Solve@8LaplaceTransform@x''@tD + 2 x'@tD + x@tD, t, pD LaplaceTransform@t, t, pD,x@0D == 0, x'@0D == 0

In[40]:= PlotA-2 + 2 ã-t + t + ã-t t, 8t, -1, 10