1540470959-Boundary_Value_Problems_and_Partial_Differential_Equations__Powers

374 Chapter 6 Laplace Transform

The transformed equation is readily solved, yielding

U(s)=su^0 +u^1

s^2 −a

+^1

s^2 −a

F(s).

Because 1/(s^2 −a)is the transform of sinh(√at)/√a, we easily determine that
uis

u(t)=u 0 cosh

(√

at

)

+

u 1

√asinh

(√

at

)

+

∫t

0

sinh(√a(t−t′))

√a f

(

t′

)

dt′. (5)

A slightly different problem occurs if the mass in a spring–mass system is
struck while the system is in motion. The mathematical model of the system
might be

u′′+ω^2 u=f(t), u( 0 )=u 0 , u′( 0 )=u 1 ,

wheref(t)=F 0 fort 0 <t<t 1 andf(t)=0 for other values. The transform of
uis

U(s)=su^0 +u^1

s^2 +ω^2

+^1

s^2 +ω^2

F(s).

The inverse transform ofU(s)is then

u(t)=u 0 cos(ωt)+u 1 sin(ωt)

ω

+

∫t

0

sin(ω(t−t′))

ω

f

(

t′

)

dt′.

The convolution in this case is easy to calculate:

∫t

0

sin(ω(t−t′))

ω f

(

t′

)

dt′

=











0 , t<t 0 ,

F 01 −cos(ω(ω 2 t−t^0 )), t 0 <t<t 1 ,

F 0 cos(ω(t−t^1 ))ω− 2 cos(ω(t−t^0 )), t 1 <t.

EXERCISES

1.Solve these initial value problems.

a. u′− 2 u=0, u( 0 )=1;

b. u′+ 2 u=0, u( 0 )=1;

c. u′′+ 4 u′+ 3 u=0, u( 0 )=1, u′( 0 )=0;

d. u′′+ 9 u=0, u( 0 )=0, u′( 0 )=1.