a²u Solve 0 0 ax? subject to u(0, t) = 0, u(1, t) = 0, t> 0 ди и(х, 0) %3D 0, sin 7x, 0
a²u Solve 0 0 ax? subject to u(0, t) = 0, u(1, t) = 0, t> 0 ди и(х, 0) %3D 0, sin 7x, 0
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Question
I don't understand why up(x,s)=(1/s^2+pi^2)sinpix. Can you please explain it to me. Thank you
![492
CHAPTER 14 INTEGRAL TRANSFORMS
EXAMPLE 2 Using the Laplace Transform to Solve a BVP
a²u
Solve
0 <x< 1,
t> 0
subject to
u(0, г) —D 0,
и(1, г) —D 0, 1>0
ди
u(х, 0) %3 0,
0 <x< 1.
= sin Tx,
at \1=0
SOLUTION The partial differential equation is recognized as the wave equation
with a = 1. From (4) and the given initial conditions the transformed equation is
d²U
-s²U
dx?
= -sin 7x,
(5)
where U(x, s) = L{u(x, t)}. Since the boundary conditions are functions of t, we
must also find their Laplace transforms:
L{u(0, t)}
U(0, s)
= ()
and
L{u(1, t)} = U(1, s) = 0.
(6)
The results in (6) are boundary conditions for the ordinary differential equation (5).
Since (5) is defined over a finite interval, its complementary function is
U.(x, s) = c, cosh sx + c2 sinh sx.
The method of undetermined coefficients yields a particular solution
U,(x, s)
1
sin Tx.
s2 + 7?
U(x, s) = c1 cosh sx + c2 sinh sx +
1
sin TX.
Hence
s² + 7²
But the conditions U(0, s)
= 0 and U(1, s) = 0 yield, in turn, c¡ = 0 and c2 = 0. We
conclude that
1
U(x, s)
sin X
s2 + T
1
sin Tx
1
sin Tx L-1.
TT
L-1
s² + T´
и(х, t)
%3|
s² + 7²]
TT
Therefore
и(х, 1)
1
sin 7x sin t.
TT
EXAMPLE 3 Using the Laplace Transform to Solve a BVP
A very long string is initially at rest on the nonnegative x-axis. The string is secured
at x = 0, and its distant right end slides down a frictionless vertical support. The
string is set in motion by letting it fall under its own weight. Find the displace-
ment u(x, t).
SOLUTION Since the force of gravity is taken into consideration, it can be shown
that the wave equation has the form
q2.
dx
x > 0,
t> 0.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fca708747-373f-4dbe-b127-10492ca0e68f%2F6e746f5f-2aea-466e-a134-044d75515900%2F2u1og6_processed.png&w=3840&q=75)
Transcribed Image Text:492
CHAPTER 14 INTEGRAL TRANSFORMS
EXAMPLE 2 Using the Laplace Transform to Solve a BVP
a²u
Solve
0 <x< 1,
t> 0
subject to
u(0, г) —D 0,
и(1, г) —D 0, 1>0
ди
u(х, 0) %3 0,
0 <x< 1.
= sin Tx,
at \1=0
SOLUTION The partial differential equation is recognized as the wave equation
with a = 1. From (4) and the given initial conditions the transformed equation is
d²U
-s²U
dx?
= -sin 7x,
(5)
where U(x, s) = L{u(x, t)}. Since the boundary conditions are functions of t, we
must also find their Laplace transforms:
L{u(0, t)}
U(0, s)
= ()
and
L{u(1, t)} = U(1, s) = 0.
(6)
The results in (6) are boundary conditions for the ordinary differential equation (5).
Since (5) is defined over a finite interval, its complementary function is
U.(x, s) = c, cosh sx + c2 sinh sx.
The method of undetermined coefficients yields a particular solution
U,(x, s)
1
sin Tx.
s2 + 7?
U(x, s) = c1 cosh sx + c2 sinh sx +
1
sin TX.
Hence
s² + 7²
But the conditions U(0, s)
= 0 and U(1, s) = 0 yield, in turn, c¡ = 0 and c2 = 0. We
conclude that
1
U(x, s)
sin X
s2 + T
1
sin Tx
1
sin Tx L-1.
TT
L-1
s² + T´
и(х, t)
%3|
s² + 7²]
TT
Therefore
и(х, 1)
1
sin 7x sin t.
TT
EXAMPLE 3 Using the Laplace Transform to Solve a BVP
A very long string is initially at rest on the nonnegative x-axis. The string is secured
at x = 0, and its distant right end slides down a frictionless vertical support. The
string is set in motion by letting it fall under its own weight. Find the displace-
ment u(x, t).
SOLUTION Since the force of gravity is taken into consideration, it can be shown
that the wave equation has the form
q2.
dx
x > 0,
t> 0.
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