Laplace transform

### 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) dz

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 20.4.3). The Laplace transform has the following properties:

L(y′)(x) | =−y(0)+xL(y)(x) |  |  |  |  |  |  |  |  |
L(y′′)(x) | =−y′(0)+xL(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)

sint |  |  |  |  |  |  |  |  |  |

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
|  | ⎛ ⎝ | x3−18 x c0+6 x c1+6 c0 | ⎞ ⎠ | e3 x

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

LsL(I)(s)+RL(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+L2 ω2

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 20.1.1):

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.