# HeatWave by abhishekgoel137

VIEWS: 3 PAGES: 75

• pg 1
```									PARTIAL DIFFERENTIAL
EQUATIONS AND BOUND
ARY VALUE PROBLEMS
VIBRATING STRINGS AN
D HEAT CONDUCTION
THE VIBRATING STRING:
Suppose that a flexible string is pulled taut on
the x-axis and fastened at two points, which w
e denote by x = 0 and x = L. The string is then
drawn aside into a certain curve y = f(x) in the
xy plane and then released from rest.
Problem: To find the vertical displacement yx
(,t) from the equilibrium position of the string
at the point (distance x from the origin) at time
t.
We make certain assumptions:
The subsequent motion of the string is entir
ely transverse (i.e. along the xy plane). (Thu
s the displacement of the string from the eq
uilibrium position is a function        y=y
(x, t).)

We look at a small piece of the string whic
h in the equilibrium position has length x
. If m is the (uniform) density of
the string, the mass of the piece is m x. By
Newton’s second law of motion, the transver
se force F acting on the piece is given by

¶2 y
F = m Dx 2
¶t

Since the string is flexible, the tension T at
any point is directed along the tangent (see
the figure) and has T sin  as its y- compon
ent.
T sin 
T

We now assume that the motion of the string i
s due strictly to the tension T in it. (That is the
gravitational force is negligible.)
Hence F =  (T sin  )   (T tan  ) (as
                                      ¶y
is small) and so =  (T tan  ) = D (T    )
¶x
Thus we get            ¶y         ¶ y
2
D (T    ) = m Dx 2
¶x         ¶t
Dividing by x and letting x  0, we get

¶     ¶y    ¶ y     2
(T    )=m 2
¶x    ¶x    ¶t
Or as T is constant (the same for all x), we get
¶ y2
¶ y  2
T      =m 2
¶x 2
¶t
T
Putting a =2

M,

We get the one-dimensional wave equation:
¶ y ¶ y
2    2
a2
= 2
¶x 2
¶t
which is usually written as
¶ y
2
1 ¶ y
2
= 2 2
¶x 2
a ¶t
It is proved in Physics that y(x,t) satisfies the ‘o
ne-dimensional wave equation’
2 y 2 y
a 2
 2
x 2
t
where a is a constant (depending on the mass o
f the string and the tension of the string). We th
us want to solve the boundary value-problem

 y  y
2      2
 (1)
a2
 2
x 2
t
subject to the boundary conditions
y (0, t )  y ( L, t )  0 for all t > 0  (2)
and the initial conditions
y
 0 for t = 0              (3)
t
y(x,0) = f(x), 0 ≤ x ≤ L      (4)
We find the solution by the method of ‘sep
aration of variables’.
That is we let y(x,t) = u(x) v(t)
(a function of x  a function of t )
Substituting for y in the equation (1), we get

                        
a u  x ) v (t )  u ( x ) v  t )
2
(                        (

or           
u ( x )       
1 v ( t )
 2                (5)
u ( x)    a v (t )
where the ‘dashes’ denote derivative w.r.t. the c
oncerned variable. The L.H.S of (5) is indepen
dent of t and the R.H.S is independent of x and
hence both are independent
of both x and t and hence equals a pure consta
nt, say, - .
            
u  x ) 1 v t )
(             (
Thus, we get u ( x )  a 2 v (t )  

y (0, t )  y ( L , t )  0   for all t > 0
implies u(0) = u(L) =0.
So we have to solve for u

u   u  0
with boundary conditions u(0) = u(L) =0.
This eigenvalue problem has nonttrivial soluti
ons only when
np 2 2
l = 2 (n =1,2,...)
L
Corresponding solutions are
np x
u n = sin      ( n = 1, 2, ...)
L
and non-zero multiples of them.
y
 0 for t = 0 implies v  0 )  0
(
t
Now for each n, v satisfies the d.e.
n p a
2   2   2
v ¢¢ +     2
v = 0
L
with the initial condition v  0 )  0
(

np at           np at
Thus v = c1 cos       + c 2 sin
L               L
np a        np at      np a     np at
v¢ = -      c1 sin       + c2      cos
L            L         L         L
v0)  0  c2  0
(

Hence               np a
v = c1 cos      t
L
We shall take c1=1 and denote the correspo
nding solution (which depends on n ) by vn
. Thus                     np a
v n = cos       t
L
Thus for each n =1,2,3,…
np x     np a t
y n = u n v n = s in      cos
L         L
is a solution to the given problem satisfying th
e first three conditions. So also any finite or in
finite lc of them. But no finite lc of them wou
ld satisfy the initial condition
y(x,0) = f(x), 0 ≤ x ≤ L
So we consider their infinite lc so as to satisfy t
he above initial condition.
Thus look at the infinite linear combination
¥
np x     np a t
y =   å   yn =      å
n =1
b n s in
L
cos
L

Now t = 0 implies
¥
np x
f ( x ) = å bn sin      , 0£ x£ L
n =1      L
Thus bn= Fourier sine coefficient of f(x) in [0, L]
n x
L
2

L   f ( x ) sin L dx
0
Thus the solution to the wave equation satisf
ying the initial and boundary conditions is

¥
np x     np a t
y( x,t) =   ån =1
b n s in
L
cos
L
0 ≤ x ≤ L, t  0
where
L
2                   np x
bn =
L   ò
0
f ( x ) sin
L
d x , n = 1, 2, ...
Example: Solve the wave equation when

f ( x ) = x ( L - x ), 0 £ x £ L
Thus             2
L
np x
bn = ò x ( L - x )sin      dx
L0                 L
L
é           æ    np x ö              æ     np x ö     æ     np x öù
2ê           ç cos L ÷                ç sin      ÷     ç cos      ÷ú
= ê x( L - x) ç -         - ( L - 2 x) ç - 2 L ÷ + (-2) ç 3 L ÷ú
Lê               np ÷                 ç   n p2 ÷       ç   n p 3 ÷ú
ç         ÷              ç          ÷     ç          ÷ú
ê
ë           è    L ø                 è    L ø
2
è L   3
øû 0
ì 8L2
4L 2
ï 3 3 n = 1,3,..
= 3 3 (1 - cos np )   = ín p
np                     ï0    n = 2,4,..
î
Thus the solution to the given problem is

y(x,t)
8L 2 ¥
1        (2n -1)p x    (2n -1)p at
= 3
p
å (2n -1)3 sin L cos L
n=1
0 ≤ x ≤ L, t  0
Example: Solve the wave equation when

             L
 x      0x
f ( x)               2
L
L  x     xL
        2
Thus
L                                  
nx                    nx 
2              L
2
bn   sin
x       dx   L  x ) sin
(              dx 
L 0      L                      L


L
2


éì    æ        np x   ö       np x   ü
L/2

êï
2 êï    ç - cos L       ÷ s in L       ï
ï
=    íx   ç               ÷+             ý
L êï          np             n p
2   2
ï
ç               ÷
êï    è      L        ø              ï0
ëî                                   þ
2
L

ì        æ        np x   ö      np x       ü
L
ù
ï
ï        ç - co s L      ÷ sin L           ï
ï
ú
+ í(L - x) ç               ÷-                ý    ú
ï              np            n 2p 2        ï    ú
ç               ÷
ï
î        è      L        ø     L 2
ïL/2 ú
þ    û
2 é    L  2
np     L     2
np
=   ê - 2 np c o s 2 + n 2p        2
s in
L ë                                            2

L 2
np x   L        2
np ù
+      co s      + 2                sin    ú
2 np       L    n p           2
2 û

ì 0              n = 2, 4, 6, ...
4L      np   ï
= 2 2 sin    = í 4L / n p 2   2
n = 1, 5, 9, ...
n p       2   ï - 4 L / n 2p 2 n = 3, 7 ,1 1, ...
î
Hence the solution to the wave equatio
n is
y( x,t)

4L ¥            1          (2n -1)p x     (2n -1)p at
= 2 å(-1) n+1
sin            cos
p n=1        (2n -1) 2
L               L
Example: Solve the wave equation when

ì x,        0£ x £p /4
ï
f ( x ) = í p / 4, p / 4 £ x £ 3p / 4
ï p - x, 3p / 4 £ x £ p
î
p
Thus         2
bn =
p   ò f ( x) sin nx dx
0
2é
p /4                 3p / 4
p
= ê ò x sin nx dx + ò sin nx dx
pë0               p /4
4
p
+     ò     (p - x)sin nx dx]
p
3 /4
2 éì
p /4          3p / 4
cos nx sin nx ü p ì cos nx ü
= ê í x( -       ) + 2 ý + í-         ý
p êî
ë         n       n þ0  4î    n þp / 4

cos nx sin nx ü ù
p
ì
+ í(p - x)(-       )- 2 ý ú
î            n      n þ3p / 4 ú
û
2      np       3np
= 2 (sin    + sin     )
pn       4        4
Thus the solution to the given problem is

y( x,t)

2   ¥
1  np     3np
= å 2 (sin + sin     )sin nx cos nat
p n=1 n   4      4
D’Alembert’s solution of the wave equation:

Consider the wave equation  y  1  y
2         2

x 2
a t
2   2

We change the independent variables x, t to u,
v by substituting u = x + at , v = x - at
y   y u   y v   y   y
                  
x   u x   v x   u   v
 y
2
   y    y 
     u    x   v 
x 2
x             
 y u  y v  y u  y v
2            2        2         2
 2                  2
u x vu x uv x v x
y2
u  y
2       2
 2 2      2
u    uv v
(assuming that all mixed partial derivates are
continuous). Similarly
y  2
2 y   u  y
2   2
 a  2 2
2
u        2
t 2
     uv v 

Hence the wave equation becomes
 y
2
 y
2
or    y
2
2        2                  0
u v      u v       uv

Integrating partially w.r.t u, we get
y    = a constant independent of u
v
= a function of v = g(v) say
Again integrating partially w.r.t v, we get
y   (v ) dv  a
g                (where a is a constant
v
independent of v)
= G (v) + F (u)(say)

= F (x + at) + G (x - at)
We have to chose F,G so as to satisfy the initi
al and boundary conditions. This is extremel
y difficult. Hence we have to adopt the Fouri
er Series method for solving the wave equati
on.
ONE-DIMENSIONAL
HEAT EQUATION
The heat equation:

We consider the flow of heat in a thin cylin
drical rod of cross-sectional area A whose l
ateral surface is perfectly insulated so that n
o heat flows through it. The word ‘thin’ me
ans that the temperature is uniform on any
cross- section and thus is a function of the t
ime t and its horizontal distance x from one
end, w(x,t).
We note the following physical principles:
• Heat flows in the direction of decreasing te
mperature, that is from hot regions to cold re
gions
• The rate at which heat flows across an area i
s proportional to the area and to the rate of ch
ange of temperature with respect to the dista
nce in a direction perpendicular to the area. (
This proportionality factor is denoted by k an
d called the thermal conductivity of the subst
ance.)
c) The quantity of heat gained or lost by a b
ody when its temperature changes is propo
rtional to the mass of the body and to the c
hange of temperature. (This proportionalit
y factor is denoted by c and called the spe
cific heat of the substance.)
Using the above three principles, we can deri
ve the heat equation describing the temperatu
re w(x,t) as
2 w    1 w
 2
x 2
a t
Here the constant a2 depends on three things:
,density of the rod, c, specific heat of the su
bstance and k, thermal conductivity of the subs
tance.
k
In fact a =
2

cr
We assume the boundary conditions
w (0, t )  w ( L , t )  0
This means that the ends are maintained at zer
o temperature at all times.
We also assume that the rod has an initial temp
erature distribution f(x) at time t = 0
i.e. w(x,0) = f(x)    0<x<L
Again ‘separating’ the variables, we find that
w=uv
where u satisfies the eigenvalue problem

u   u  0
u(0)  u(L)  0        (1)
and that v satisfies first order d.e
v    a 2 v        (2)
Thus for nontrivial solutions,
n 2p 2
ln = 2 ( n = 1, 2,...)
L
Corresponding eigenfunctions are
np x
u n = sin
L
n 2p 2 a 2 t
-
For each n=1,2,3,…    vn = e                       L2

np       2       2
is a solution of (2) with ln = 2
a
Hence for each n = 1,2,3,…
n 2p 2 a 2 t
np x     -
wn = u n vn = sin      e         L2
L
is a solution of the heat equation.
Again we assume that the infinite linear combin
ion
n 2p 2 a 2 t
¥             ¥
np x       -
w = å b n w n = å b n s in      e           L2

n =1        n =1        L
is also a solution.             
Putting t = 0, we get f ( x)   b sin nx
n
n 1     L
Hence bn = nth Fourier sine coefficient of f(x)
n x
L
2
in [0, L]  f ( x ) sin        dx
L 0              L
Problem

Solve the heat equation given that
w (0, t ) = w ( L , t ) = 0;
ì x    0 < x £ L/2
w ( x , 0) = f ( x ) = í
îL - x L / 2 £ x £ L
Solution

The solution to the heat equation subject to the
given boundary and initial conditions are:
n 2p 2 a 2 t
¥               ¥
np x       -
w = å b n w n = å b n s in      e           L2

n =1        n =1        L
where bn = nth Fourier sine coefficient of f(x)
n x
L
2
in [0, L] 
L   f ( x ) sin L dx
0
L                                 
nx                    nx 
L
2 2
bn   sin
x      dx   L  x ) sin
(              dx 
L 0      L                      L


L
2



Continued on the next
slide
éì    æ        np x   ö       np x   ü
L/2

êï
2 êï    ç - cos L       ÷ s in L       ï
ï
=    íx   ç               ÷+             ý
L êï          np             n p
2   2
ï
ç               ÷
êï    è      L        ø              ï0
ëî                                   þ
2
L

ì        æ        np x   ö      np x       ü
L
ù
ï
ï        ç - co s L      ÷ sin L           ï
ï
ú
+ í(L - x) ç               ÷-                ý    ú
ï              np            n 2p 2        ï    ú
ç               ÷
ï
î        è      L        ø     L 2
ïL/2 ú
þ    û
2 é    L  2
np     L     2
np
=   ê - 2 np c o s 2 + n 2p        2
s in
L ë                                        2

L    2
np   L     2
np ù
+      co s    + 2             sin    ú
2 np       2  n p        2
2 û

ì 0              n = 2, 4, 6, ...
4L      np   ï
= 2 2 sin    = í 4L / n p 2   2
n = 1, 5, 9, ...
n p       2   ï - 4 L / n 2p 2 n = 3, 7 ,1 1, ...
î
The solution to the heat equation subject to the
given boundary and initial conditions are:

w( x , t ) =
( 2 n -1) 2 p 2 a 2t
4L    ¥
(2 n - 1)p x     -
2 å
n +1
( -1) sin              e             L2
p n =1                L
We assume that when t is large, w a function i
ndependent of t, w(x) say. This is called the ste
ady state temperature. And noting that in stead
y state       w
0
t
w d w
2      2
we get           2 0
x 2
dx
Therefore w = c1 x + c 2
Suppose that the ends are maintained at temper
ature w1,w2 at all times. Hence
x  0  w  w1  c2  w1
1
x = L Þ w = w2 Þ c1 = ( w2 - w1 )
L
Hence the steady state temperature is
1
w  w1  ( w2  w1 ) x   0xL
L
Problem
Find the solution of the one-dimensional heat e
quation satisfying the boundary and initial cond
itions
w ( 0 , t )  100 , w ( L , t )  0 , for all t.

px
w ( x , 0) = 100 cos       0xL
2L
Solution

We write the temperature distribution as
w( x, t ) = wS ( x) + wT ( x, t )

= steady state temperature + Transient tempe
rature.
We impose the boundary conditions              wS(0
) = 100, wS(L) = 0 on wS(x).
Thus
1                 x
ws ( x)  100  (0 100) x  100   0  x  L
1
L                 L

We impose on wT(x,t) the boundary conditions
wT(0,t) = wT(L,t) = 0 for all t
and the initial condition
px
wT ( x, 0) = 100 cos    - wS ( x)   0£ x£ L
2L
That is : wT(x, t) satisfies the boundary conditio
ns
wT(0,t)=wT(L,t) = 0 for all t

and the initial condition
px            x
wT ( x, 0) = 100 cos    - 100(1 - )   0£ x£ L
2L          L
n 2p 2 a 2 t
Thus
¥
np x -
w T ( x , t ) = å b n sin     e        L2

n =1       L
where
2
L
px               x       np x
bn =
L   ò [1 0 0 co s 2 L - 1 0 0 (1 - L )] sin L d x
0

L
2              px     np x
=   ò 1 0 0 co s 2 L sin L d x
L 0
L
2            x      np x
- ò 1 0 0 (1 - ) sin      dx
L 0          L       L
Now
2
L
px     np x
ò 1 0 0 co s 2 L sin L d x
L 0

 ( 2 n  1)x ( 2 n  1)x 
L
100

L      sin 2 L  sin
0 
L      

dx
L
100  2L           (2n 1)x 2L        (2n 1)x 
     (2n 1) cos 2L  (2n 1) cos 2L 
L                                              x0
                                    
200   cos( 2 n  1) 2 1  cos( 2 n  1) 2 
1
                                           
      ( 2 n  1)          ( 2 n  1)     
                                       

200 é 1             1 ù
=     ê (2n + 1) + (2n - 1) ú
p ë                      û
800 n
          ( n  1,2,3,..)
 4n 1
2
np x
L
And    2              x
ò 1 0 0 (1 - L ) sin L d x
L 0

L
ì                np x          np x   ü
ï           co s          sin         ï
200 ï     x            L )- 1         L   ï
=     í (1 - )( -                           ý
L ï      L        np       L n 2p 2      ï
ï
î                L            L 2
ï x=0
þ

200 L   200
=       =
L np   np
Thus
w( x, t )
n2p 2a2
x 200 æ 4n¥
1 ö np x  - 2 t
= 100(1- ) +
L
åç 4n2 -1 - n ÷ sin L e
p n=1 è         ø
L
Problem
The temperatures at the ends x = 0 and x = 100
of a rod 100 cm in length with insulated sides a
re held at 0 and 100 degree centigrade respectiv
ely, until steady-state conditions prevail. Then a
t the instant t = 0, the temperatures of the two e
nds are interchanged. Find the resultant tempera
ture distribution of the rod as a function of x an
d t.
Solution

We note that the steady state temperature (w
hen the ends x = 0 and x = 100 of the rod 10
0 cm in length are held at 0 and 100 degree

f ( x ) = x , 0 < x < 100
Now, the temperatures of the two ends are i
nterchanged.
And we have to find the new temperature di
stribution in the rod.

Hence we have to solve the heat equation wi
th the boundary and initial conditions

w (0, t ) = 100, w (100, t ) = 0, for all t.

w ( x , 0) = x ,   0xL
We write the temperature distribution as
w( x, t ) = wS ( x) + wT ( x, t )

= (new) steady state temperature + Transient
temperature.
We impose the boundary conditions         wS(0
) = 100, wS(100) = 0 on wS(x).
Thus
1
ws ( x) = 100 +     (0 - 100) x = 100 - x 0 £ x £ 100
100

We impose on wT(x,t) the boundary conditions
wT(0,t) = wT(100,t) = 0 for all t
and the initial condition
wT ( x,0) = x - wS ( x)   0 £ x £ 100
i.e.    wT ( x,0) = 2 x - 100    0 £ x £ 100
That is : wT(x, t) satisfies the boundary conditio
ns
wT(0,t)=wT(100,t) = 0 for all t

and the initial condition
wT ( x,0) = 2 x - 100   0 £ x £ 100
n 2p 2 a 2 t
Thus
¥
np x     -
w T ( x , t ) = å b n s in      e       1002

n =1       100
100
where b n = 2                              np x
100   ò0
[ 2 x - 1 0 0 ] sin
100
dx

é            æ      np x           ö
2  ê            ç co s 1 0 0          ÷
=     ê(2 x - 100) ç -                   ÷
100 ê            ç np / 1 0 0          ÷
ê
ë            è                     ø
100
æ      np x          öù
ç  sin               ÷ú
+ 2 ç 2 2 100 2          ÷ú
ç n p /100           ÷ú
è                    øú0
û
200                              400
=     (1 - co s np )             =     , n odd
np                               np

Hence the desired temperature distribution is:
w( x, t ) = wS ( x) + wT ( x, t )

= 100 - x
(2 n-1)2 p 2a2t
400 ¥   1       (2n -1)p x              -
+    å (2n -1) sin 100 e
p n=1
1002
END OF WAVE AND HE
AT EQUATIONS
TWO-DIMENSIONAL H
EAT EQUATION
- LAPLACE’S EQUATIO
N
The two-dimensional heat equation:
A thin rectangular plate has its boundary kep
t at temperatures 0,0,0 and f(x) at all times. F
ind the steady temperature distribution in the
plate.
Thus we have to solve the Laplace’s equation
 w  w
2       2
 2 0
x 2
y
subject to the boundary conditions
w ( 0, y )  w ( L , y )  0   0 yM

w( x, M ) = 0, w( x,0) = f ( x) 0 £ x £ L
We again solve by the method of separation of
variables. We assume w(x,y) = X(x) Y(y)
Substituting and separating the variables we get
X   Y
      (a constant)
X     Y
Thus X satisfies the boundary value problem
X ¢¢ + l X = 0, X(0) = X(L) = 0
We know nontrivial solutions exist when
n p2   2
l = ln =   2
( n = 1, 2, 3, ...)
L
and the corresponding eigenfunction is
nx
X n  sin
L
For each n, Y satisfies the BV problem

Y  n Y  0 , Y ( M )  0
ny          ny
Now Y  B sinh      C cosh
L            L
np M          np M
Y ( M ) = 0 Þ B sinh      + C cosh      =0
L             L
np M              np M
Taking    B = - cosh      , C = + sinh
L                 L

we find              np ( M - y )
Y = sinh
L
Thus for each n,
np x      np ( M - y )
wn = bn sin      sinh
L             L
is a solution satisfying the first three boundary
conditions. Hence
¥
np x      np ( M - y )
w = å wn = å bn sin      sinh
n =1      L             L
is also a solution.
Putting y = 0, we get

nM     nx
f ( x)   bn sinh     sin
n 1       L       L
nM
Hence bn sinh
L
= nth Fourier sine coefficient of f(x) in [0, L]
L
2              np x
= ò f ( x ) sin      dx ( n = 1, 2,3,...)
L0              L

Thus we have found out the steady state temp
erature distribution.

THE END
BEST OF LUCK IN T
HE COMPRE

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