Wave Equation & Heat Equation

# Wave Equation & Heat Equation | Engineering Mathematics - Civil Engineering (CE) PDF Download

``` Page 1

Chapter 03.indd   65 5/31/2017   12:42:43 PM
Page 2

Chapter 03.indd   65 5/31/2017   12:42:43 PM

Laplace Equation
When the temperature in a homogeneous material is in steady
state and the temperature does not vary with time then the
heat conduction equation becomes
?
?
+
?
?
+
?
?
=
2
2
2
2
2
2
0
u
x
u
y
u
z

and this is known as
Laplace’ s equation in cartesian system While solving the
boundary value problems the following results may be used
If u(x, t) is a function of x and t
1. L
u
t
su xs ux
?
?
?
?
?
?
?
?
=- (, )( ,) 0
2. L
u
t
su xs su xu x
t
?
?
?
?
?
?
?
?
=- -
2
2
2
00 (, )( ,) (, )
3. L
u
x
du
dx
?
?
?
?
?
?
?
?
=
4. L
u
x
du
dx
?
?
=
2
2
2
2
where L {u (x, t)} = u (x, s)
Example
Solve the one dimensional heat equation
?
?
=
?
?
u
t
u
x
2
2
2
sat-
isfying the boundary conditions u (0, t) = 0 = u (4 ,t) and u
(x, 0) = 8sin 2p x.
Solution
Taking Laplace transform on both sides of the equation
?
?
=
?
?
u
t
u
x
2
2
2
L
u
t
L
u
x
?
?
?
?
?
?
?
?
=
?
?
?
?
?
?
?
?
2
2
2
su - u (x, 0) =· 2
2
2
du
dx
or
?
?
-
2
2
2
u
x
s
u = -4sin 2 p x  as u (x, 0) = 8sin 2p x
The general solution of the above equation is u
= Ae
SX () /2
+ Be
x
s
SX -
-
--
()
sin
()
/2
2
42
2
2
p
p

or
uAe
SX
=
/2
+ Be
SX - /2
+
82
8
2
sin p
p
x
s +
(1)
But u (0, t) = 0 = u (4, t)
\  u (0, s) = 0, u (4, s) = 0
\ From Eq. (1), we have A + B = 0
and 0 = Ae Be
ss // 22
44
+
-
+
88
8
2
sin p
p + s
? Ae Be
ss // 22
44
+
-
= 0
Solving we get A = B = 0
\ From (1) we have u
x
s
=
+
82
8
2
sin p
p
?=
+
?
?
?
?
?
?
-
yL
s
x
1
2
8
8
2
p
p sin
i.e.,                      ye x
t
=
-
82
8
2
p
p sin.
Example
Solve the wave equation of a stretched string given by
?
?
=
?
?
2
2
2
2
9
u
t
u
x
satisfying the boundary conditions u (x, 0) =
0, u
t
(x, 0) = 0, x > 0 and u (0, t) = F (t), lim (, ), .
x
ux tt
?8
== 00
Solution
Given
?
?
=
?
?
2
2
2
2
9
u
t
u
x
.
Taking Laplace transform on both sides of the equation
with the boundary conditions we have
L
u
t
L
u
x
?
?
?
?
?
?
?
?
=
?
?
?
?
?
?
?
?
2
2
2
2
9
or s
2u
(x, s) - su (x, 0) - u
t
(x, 0) =· -= 9
9
0
2
2
2
2
2
du
dx
du
dx
s
u or
(1)
Also u (0, s) = Ft edtF
st
0
8
-
?
= () and u (x, s) = 0 as x ? 8
\ The general solution Eq. of (1) is u (x, s)
=+
-
ce ce
s
x
s
x
1
3
2
3
and u (x, s) = 0 as x ? 8 ? c
1
= 0
and u (0, s) = F(s) = c
2
Hence, u (x, s) = F (s) e
sx
-
3

?=
?
?
?
?
?
?
?
?
?
?
-
-
ux tL eF s
sx
(, )( )
1
3

=
-
?
?
?
?
?
?
>
<
?
?
?
?
?
?
?
=
-
Ft
x
t
x
t
x
LF sF t
33
0
3
,
,
{( )( )}, as

when expressed in terms of Heaviside’s unit step function.
u(x, t)
=-
?
?
?
?
?
?
·-
?
?
?
?
?
?
Ft
x
Ht
x
33
.
Chapter 03.indd   66 5/31/2017   12:42:48 PM
```

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