COMPLEX ANALYSIS: Maple Worksheets, 2001
(c) John H. Mathews Russell W. Howell
mathews@fullerton.edu howell@westmont.edu

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COMPLEX ANALYSIS: for Mathematics & Engineering, 4th Ed, 2001, ISBN: 0-7637-1425-9
Jones and Bartlett Publishers, Inc., 40 Tall Pine Drive, Sudbury, MA 01776
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CHAPTER 11 FOURIER SERIES and the LAPLACE TRANSFORM

Section 11.10 Convolution

Let and denote the transforms of and , respectively. Then the inverse of the product is given by the function and is called the convolution of and and can be regarded as a generalized product of and . Convolution will assist us in solving integral equations.

Theorem 10.24 (Convolution Theorem)

Let and denote the Laplace transforms of and , respectively. Then

h(t) = ,

or

h(t) = .

Load Maple's "Laplace transform" procedures.
Make sure this is done only ONCE during a Maple session.

Then activate Maple's Laplace transform package.

> with(inttrans):
laplace(t,t,s):

Example 11.29, Page 545. Use convolution to find the inverse Laplace transform of
.
Solution: The function is the product of and ,
which arethe Laplace transfoms of and , respectively.

> f:='f': g:='g': h:='h': H:='H':
r:='r': s:='s': t:='t': v:='v':
f0 := t -> sin(t):
g0 := t -> 2*cos(t):
fg := proc(r,t) sin(t-r)*2*cos(r) end:
v := T -> int(fg(r,T), r=0..T):
h := subs(T=t,v):
`f(t) ` = f0(t);
`g(t) ` = g0(t);
`f(t-r)g(r) ` = fg(r,t);
int(f(t-r)*g(r), r) = int(fg(r,t), r);
`h(t) = `, int(f(t-r)*g(r), r=0..t) = h(t);

We can check this with Maple's result using the inverse Laplace transform subroutine.

> `H(s) ` = 2*s/(s^2 + 1)^2;
`h(t) ` = invlaplace(2*s/(s^2 + 1)^2, s, t);

Example 11.30, Page 545. Use the convolution theorem to solve the integral equation
.

Solution: Take the Laplace transform of each of the terms and get:

> F:='F': s:='s': S:='S':
eqn := F(S) = 2*S/(S^2+ 1) - 1/S^2*F(S):
subs(S=s,eqn);
`Solve for F(s):`;
sol := solve(eqn, F(S)):
`F(s) ` = subs(S=s, sol);
F := s -> simplify(subs(S=s, sol)):
`F(s) ` = F(s);
`F(s) ` = convert(F(s),parfrac,s);

Using a table of Laplace transforms, the solution is .

> F1 := 2*s/(s^2+1)^2:
F2 := s/(s^2+1):
f1 := invlaplace(F1, s, t):
f2 := invlaplace(F2, s, t):
`F1(s) ` = F1, ` F2(s) ` = F2;
`f2(t) ` = F2, ` f2(t) ` = f2;
`f(t) ` = - f1 + 2*f2;

We can check this with Maple's result using the inverse Laplace transform subroutine.

> `F(s) ` = F(s);
`f(t) ` = invlaplace(F(s), s, t);

Example 11.31, Page 546. Find the Laplace transform of the Dirac delta function .
Solution: For any we define on the interval [ , ] as follows:

> a:='a': A:='A': d:='d': F:='F': s:='s': S:='S': t:='t': v:='v':
d := proc(A,T) 1/A end:
v := int(d(A,T)*exp(-S*T), T=0..A):
F := proc(a,s) subs(A=a,S=s,v) end:
`On [0,a] we define: `, delta(a,t) = d(a,t);
`F(a,s) ` = F(a,s);

Find to obtain the Laplace transform of .

> `F(a,s) ` = F(a,s);
`F(a,s) ` = simplify(F(a,s));
L := limit(F(a,s), a=0):
Limit(`F(a,s)`, a=0), ` = F(s) ` = L;

We can check this with Maple's result using the Laplace transform subroutine.

> `f(t) ` = Dirac(t);
`F(s) ` = laplace(Dirac(t), t, s);

Example 11.32, Page 547. Solve the initial value problem
with and .
Solution: Taking transforms results in the equation:

> Y:='Y': s:='s': S:='S':
eqn := (S^2 + 4*S + 13)*Y(S) = 3:
subs(S=s,eqn);
sol := solve(eqn, Y(S)):
Y := s -> simplify(subs(S=s, sol)):
`Y(s) ` = Y(s);

Using a table of Laplace transforms, the solution is .
We can check this with Maple's result using the inverse Laplace transform subroutine.

> `Y(s) ` = Y(s);
`y(t) ` = invlaplace(Y(s), s, t);

We can also solve the D.E. with Maple's using the dsolve subroutine.
= with and = .
Notice that we have changed the initial condition to = , because
Maple uses the right hand derivative at .

> DE := diff(y(t),t$2)+4*diff(y(t),t)+13*y(t)=3*Dirac(t):
IC := y(0)=0, D(y)(0)=3:
`D. E. ` = DE;
`I. C. ` = IC;
dsolve({DE}, y(t));
dsolve({DE,IC}, y(t));

Remark. For , this reduces to .

Eample 11.33, Page 549. Use convolution to solve the initial value problem
= with and = .
Solution. First solve the homogeneous equation
= with and = .
The solution to this equation is easy to obtain .

> F:='F': s:='s': S:='S': t:='t': u:='u': U:='U':
u0 := 1:
u1 := 2:
F := 0:
`u''(t) + u(t) = 0`;
`u(0) ` = u0, ` u'(0) ` = u1;
eqn := S^2*U(S) - S*u0 - u1 + U(S) = F:
subs(S=s,eqn);
sol := solve(eqn, U(S)):
U := s -> subs(S=s,simplify(sol)):
`U(s) ` = U(s);
ut := invlaplace(U(s), s, T):
u := t -> subs(T=t,ut):
`u(t) ` = u(t);
cos(t) + 2*sin(t);

Second, solve the following equation for the particular soluton
= with and = .

> g:='g': h:='h': H:='H': v:='v': V:='V': s:='s': t:='t':
`v''(t) + v(t) = tan(t)`;
`v(0) ` = 0, ` v'(0) ` = 0;
H := s -> 1/(s^2 + 1):
`H(s) ` = H(s);
`h(t) ` = invlaplace(H(s), s, t);
h0 := t -> sin(t):
g0 := t -> tan(t):
hg := proc(r,t) sin(t-r)*tan(r) end:
V := int(hg(r,T), r=0..T):
v := t -> subs(T=t,V):
`g(t) ` = g0(t);
`h(t-r)g(r) ` = hg(r,t);
int(h(t-r)*g(r), r) = int(hg(r,t), r);
`v(t) = `, int(h(t-r)*g(r), r=0..t) = v(t);
`v(t) ` = simplify(v(t));
v := t -> cos(t)*ln(cos(t)/(1+sin(t))) + sin(t):
`v(t) ` = v(t);

Error, (in csc) numeric exception: division by zero

Third, the solution is the sum of and .

> y := t -> subs(T=t,u(T)+v(T)):
`y(t) ` = y(t);

End of Section 11.10.