### 13.4.2 Laplace transform

Denoting by L the Laplace transform, you get the following:

L(y)(x)

=

∫

+∞
0

e−x uy(u)du

L−1(g)(x)

=

1
2iπ

∫

C

ez xg(z)d z

where C is a closed contour enclosing the poles of g.

The laplace command finds the Laplace transform of a function.

- laplace takes one mandatory argument and two optional arguments:
  - expr, an expression involving a variable.
  - Optionally, x, the variable name (by default x).
  - Optionally, s, a variable for the output (by default x).
- laplace(expr⟨,x⟩) returns the Laplace transform of expr.

The ilaplace or invlaplace command finds the inverse Laplace transform of a function.

- ilaplace takes one mandatory argument and two optional arguments:
  - expr, an expression involving a variable.
  - Optionally, x, the variable name (by default x).
  - Optionally, s, a variable for the output (by default x).
- ilaplace(expr⟨,x⟩) returns the inverse Laplace transform of expr.

You also can use the addtable command Laplacians of unspecified functions (see Section 21.4.3). The Laplace transform has the following properties:

L(y′)(x) | =−y(0)+x L(y)(x) |  |  |  |  |  |  |  |  |
L(y′′)(x) | =−y′(0)+x L(y′)(x) |  |  |  |  |  |  |  |  |
| =−y′(0)−xy(0)+x2L(y)(x) |  |  |  |  |  |  |  |  |

These properties make the Laplace transform and inverse Laplace transform useful for solving linear differential equations with constant coefficients. For example, suppose you have

| y′′ +p y′ +q y=f(x) |  |  |  |  |  |  |  |  |
| y(0)=a,y′(0)=b |  |  |  |  |  |  |  |  |

then

L(f)(x) | =L(y′′+py′+qy)(x) |  |  |  |  |  |  |  |  |
| =−y′(0)−x y(0)+x2 L(y)(x)−p y(0)+p x L(y)(x))+q L(y)(x) |  |  |  |  |  |  |  |  |
| =(x2+p x+q)L(y)(x)−y′(0)−(x+p)y(0) |  |  |  |  |  |  |  |  |

Therefore, if a=y(0) and b=y′(0), you get

L(f)(x)=(x2+p x+q)L(y)(x)−(x+p)a−b

and the solution of the differential equation is:

y(x)=L−1

⎛
⎜
⎜
⎝

L(f)(x)+(x+p)a+b
x2+p x+q

⎞
⎟
⎟
⎠

.

#### Examples

laplace(sin(x))

1
x2+1

With t as the original variable:

laplace(sin(t),t)

1
t2+1

With t as the original variable and s as the transform variable:

L:=laplace(sin(t),t,s)

1
s2+1

Apply the inverse transform to obtain the original function:

ilaplace(L,s,t)

sin t |  |  |  |  |  |  |  |  |  |

Solve y′′ −6 y′+9 y=x e3 x, y(0)=c0, y′(0)=c1. Here, p=−6 and q=9.

laplace(x*exp(3*x))

1
x2−6 x+9

ilaplace((1/(x^2-6*x+9)+(x-6)*c_0+c_1)/(x^2-6*x+9))

1
6

⎛
⎝

x3−18x c0+6x c1+6c0

⎞
⎠

e3x

Note that this equation could be solved directly:

desolve(y''-6*y'+9*y=x*exp(3*x),y)

e3 x

⎛
⎝

c0 x+c1

⎞
⎠

+

1
6

x3 e3 x

#### Applications

Laplace transform is useful in analysis of eletric circuits. A simple RL circuit is shown in Figure 13.1. Given the input voltage Vin, the problem is to find the current I(t). From Kirchhoff laws:

VL+VR=LI′(t)+RI(t)=Vin.

By transforming both sides of the above equation and assuming I(0)=0, you get

Ls L(I)(s)+R L(I)(s)=L(Vin)(s),

from which follows

I(t)=L−1

⎛
⎜
⎜
⎝

L(Vin)(s)
Ls+R

⎞
⎟
⎟
⎠

.

> Figure 13.1: RL circuit (source)

Let Vin(t)=V0sin(ω t). You find I(t) by entering:

assume(omega>0):; I:=ilaplace(laplace(V0*sin(omega*t),t,s)/(L*s+R),s,t):; factor(I)

⎛
⎜
⎝

−ω L cos

⎛
⎝

ω t

⎞
⎠

+ω L e

−

R t
L

+R sin

⎛
⎝

ω t

⎞
⎠

⎞
⎟
⎠

V0

R2+L2ω2

Let now Vin be a single rectangular pulse Vin={

V0, | 0≤t≤ 1,
0, | t>1.

. It is easily defined by using the boxcar command (see Section 21.1.1):

Vin:=V0*boxcar(0,1,t):; I:=ilaplace(laplace(Vin,t,s)/(L*s+R),s,t):; normal(piecewise(I))

⎧
⎪
⎪
⎪
⎨
⎪
⎪
⎪
⎩

−V0 e

−

R t
L

+V0

R

,

1>t

−V0 e

−

R t
L

+V0 e

−

R t−R
L

R

,

otherwise

Laplace transform can be used with arbitrary periodic functions. For example, Let Vin=V0(1−(t−1)2) for t∈[0,2] define a periodic function with period T=2.

Vin:=V0*periodic(1-(t-1)^2,t=0..2):; I:=ilaplace(laplace(Vin,t,s)/(L*s+R),s,t)

⎛
⎜
⎜
⎝

2 L

⎛
⎝

R+L

⎞
⎠

e

−

R

⎛
⎜
⎜
⎝

t−2

⎢
⎢
⎢
⎣

t
2

⎥
⎥
⎥
⎦

⎞
⎟
⎟
⎠

L

+4 R2

⎢
⎢
⎢
⎣

t
2

⎥
⎥
⎥
⎦

⎛
⎜
⎜
⎝

t−

⎢
⎢
⎢
⎣

t
2

⎥
⎥
⎥
⎦

−1

⎞
⎟
⎟
⎠

−t

⎛
⎝

t−2

⎞
⎠

R2−4 R

⎢
⎢
⎢
⎣

t
2

⎥
⎥
⎥
⎦

L+2 L

⎛
⎝

R t−R−L

⎞
⎠

⎞
⎟
⎟
⎠

V0

R3

Subexpressions in the above result are factored manually for a more readable presentation. Xcas computes this result by using symbolic periodic summation.

Applying the above technique to more complex circuits involving resistors, capacitors and inductors is straightforward; contour analysis by Kirchhoff laws produces a system of linear differential equations which is transformed to a system of linear equations by Laplace transform. From there, it is easy to find transforms of contour currents.