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23MA401- Numerical Methods
Boundary Value Problems
By
Dr. S.Shyamala AP/Maths
Ms.G. Nandhini AP/Maths
Excel Engineering College (Autonomous)
INTRODUCTION
 The basic partial differential equations of heat conduction, wave
propagation, and potential theory that we discuss in this chapter are
associated with three distinct types of physical phenomena: diffusive
processes, oscillatory processes, and time-independent or steady processes.
 Consequently, they are of fundamental importance in many branches of
physics, and are significant mathematically.
 The partial differential equations whose theory is best developed and
whose applications are most significant and varied are the linear
equations of second order.
 All such equations can be classified as one of three types:
The heat equation, the wave equation, and the potential equation, are
prototypes of these categories.
HEAT CONDUCTION IN A ROD: ASSUMPTIONS
 Consider a heat equation conduction problem for a straight bar of
uniform cross section and homogeneous material.
 Let the x-axis be chosen to lie along the axis of the bar, and let x
= 0 and x = L denote the ends of the bar. See figure below.
 Suppose that the sides of the bar are perfectly insulated so that
no heat passes through them.
 Assume the cross-sectional dimensions are so small that the
temperature u can be considered constant on the cross sections.
 Then u is a function only of the axial coordinate x and time t.
HEAT CONDUCTION:
INITIAL AND BOUNDARY CONDITIONS
 In addition, we assume that the initial temperature
distribution in the bar is given, and hence
where f is a given function.
 Finally, we assume that the ends of the bar are held at
fixed temperatures: the temperature T1 at x = 0 and T2 at
x = L.
 However, as will be shown in Section 10.6, we need only
consider T1 = T2 = 0.
 Thus we have the boundary conditions
L
x
x
f
x
u 

 0
),
(
)
0
,
(
0
,
0
)
,
(
,
0
)
,
0
( 

 t
t
L
u
t
u
HEAT CONDUCTION PROBLEM
 Thus the fundamental problem of heat conduction is to find u(x,t)
satisfying
 With respect to the time variable t, this is an initial value problem;
an initial condition is given and the differential equation governs
what happens later.
 With respect to the spatial variable x, it is a boundary value
problem; boundary conditions are imposed at each end of the bar
and the differential equation describes the evolution of the
temperature in the interval between them.
L
x
x
f
x
u
t
t
L
u
t
u
t
L
x
u
u t
xx










0
),
(
)
0
,
(
0
,
0
)
,
(
,
0
)
,
0
(
0
,
0
,
2

HEAT CONDUCTION: BOUNDARY
PROBLEM
 Alternatively, we can consider the problem as a boundary
value problem in the xt-plane, see figure below.
 The solution u(x,t) satisfying the heat conduction problem
is sought in the semi-definite strip 0 < x < L, t > 0, subject
to the requirement that u(x,t) must assume a prescribed
value at each point on the boundary of this strip.
L
x
x
f
x
u
t
t
L
u
t
u
t
L
x
u
u t
xx










0
),
(
)
0
,
(
0
,
0
)
,
(
,
0
)
,
0
(
0
,
0
,
2

HEAT CONDUCTION:
LINEAR HOMOGENEOUS EQUATION
 The heat conduction problem
is linear since u appears only to the first power
throughout.
 The differential equation and boundary conditions are
also homogeneous.
 This suggests that we might approach the problem by
seeking solutions of the differential equation and
boundary conditions, and then superposing them to
satisfy the initial condition.
 We next describe how this plan can be implemented.
L
x
x
f
x
u
t
t
L
u
t
u
t
L
x
u
u t
xx










0
),
(
)
0
,
(
0
,
0
)
,
(
,
0
)
,
0
(
0
,
0
,
2

SEPARATION OF VARIABLES
METHOD
 Our goal is to find nontrivial solutions to the differential
equation and boundary conditions.
 We begin by assuming that the solution u(x,t) has the
form
 Substituting this into our differential equation
we obtain
or
)
(
)
(
)
,
( t
T
x
X
t
x
u 
t
xx u
u 
2

T
X
T
X 



2

T
T
X
X 



2
1

ORDINARY DIFFERENTIAL EQUATIONS
 We have
 Note left side depends only on x and right side only on t.
Thus in order for this equation to be valid for 0 < x < L, t >
0, it is necessary for both sides of this equation to equal
the same constant, call it -. Then
 Thus the partial differential equation is replaced by two
ordinary differential equations.
T
T
X
X 



2
1

0
0
1
2
2














T
T
X
X
T
T
X
X





BOUNDARY CONDITIONS
 Recall our original problem
 Substituting u(x,t) = X(x)T(t) into boundary condition at x
= 0,
 Since we are interested in nontrivial solutions, we
require X(0) = 0 instead of T(t) = 0 for t > 0. Similarly,
X(L) = 0.
 We therefore have the following boundary value problem
L
x
x
f
x
u
t
t
L
u
t
u
t
L
x
u
u t
xx










0
),
(
)
0
,
(
0
,
0
)
,
(
,
0
)
,
0
(
0
,
0
,
2

0
)
(
)
0
(
)
,
0
( 
 t
T
X
t
u
0
)
(
)
0
(
,
0 




 L
X
X
X
X 
EIGENVALUES AND EIGEN FUNCTIONS
 Thus
 From Section 10.1, the only nontrivial solutions to this
boundary value problem are the eigen functions
associated with the eigen values
 With these values for , the solution to the first order
equation
is
0
)
(
)
0
(
,
0 




 L
X
X
X
X 
  
,
3
,
2
,
1
,
/
sin
)
( 
 n
L
x
n
x
Xn 

,
3
,
2
,
1
,
/ 2
2
2

 n
L
n
n 

0
2


 T
T 

constant.
,
2
)
/
(
n
t
L
n
n
n k
e
k
T 


FUNDAMENTAL SOLUTIONS
 Thus our fundamental solutions have the form
where we neglect arbitrary constants of proportionality.
 The functions un are sometimes called fundamental solutions
of the heat conduction problem.
 It remains only to satisfy the initial condition
 Recall that we have often solved initial value problems by
forming linear combinations of fundamental solutions and then
choosing the coefficients to satisfy the initial conditions.
 Here, we have infinitely many fundamental solutions.
  ,
,
3
,
2
,
1
,
/
sin
)
,
(
2
)
/
(


 
n
L
x
n
e
t
x
u t
L
n
n 

L
x
x
f
x
u 

 0
),
(
)
0
,
(
FOURIER COEFFICIENTS
 Our fundamental solutions are
 Recall the initial condition
 We therefore assume that
where the cn are chosen so that the initial condition is
satisfied:
 Thus we choose the coefficients cn for a Fourier sine
series.
  ,
,
3
,
2
,
1
,
/
sin
)
,
(
2
)
/
(


 
n
L
x
n
e
t
x
u t
L
n
n 

L
x
x
f
x
un 

 0
),
(
)
0
,
(
 









1
)
/
(
1
/
sin
)
,
(
)
,
(
2
n
t
L
n
n
n
n
n L
x
n
e
c
t
x
u
c
t
x
u 

 





1
/
sin
)
(
)
0
,
(
n
n L
x
n
c
x
f
x
u 
SOLUTION
 Therefore the solution to the heat conduction problem
is given by
where
 





1
)
/
(
/
sin
)
,
(
2
n
t
L
n
n L
x
n
e
c
t
x
u 

 


L
n dx
L
x
n
x
f
L
c
0
/
sin
)
(
2

L
x
x
f
x
u
t
t
L
u
t
u
t
L
x
u
u t
xx










0
),
(
)
0
,
(
0
,
0
)
,
(
,
0
)
,
0
(
0
,
0
,
2

EXAMPLE 1: HEAT
CONDUCTION PROBLEM
 Find the temperature u(x,t) at any time in a metal rod 50
cm long, insulated on the sides, which initially has a
uniform temperature of 20° C throughout and whose ends
are maintained at 0° C for all t > 0.
 This heat conduction problem has the form
50
0
,
20
)
0
,
(
0
,
0
)
,
50
(
,
0
)
,
0
(
0
,
50
0
,
2










x
x
u
t
t
u
t
u
t
x
u
u t
xx

EXAMPLE 1: SOLUTION
 The solution to our heat conduction problem is
where
 Thus
 





1
)
50
/
(
50
/
sin
)
,
(
2
n
t
n
n x
n
e
c
t
x
u 

   
   












even
,
0
odd
,
/
80
cos
1
40
50
/
sin
5
4
50
/
sin
20
50
2
/
sin
)
(
2
50
0
50
0
0
n
n
n
n
n
dx
x
n
dx
x
n
dx
L
x
n
x
f
L
c
L
n






EXAMPLE 1: RAPID
CONVERGENCE
 Thus the temperature along the rod is given by
 The negative exponential factor in each term cause the
series to converge rapidly, except for small values of t or
2
.
 Therefore accurate results can usually be obtained by
using only a few terms of the series.
 In order to display quantitative results, let t be measured
in seconds; then 2
has the units cm2
/sec.
 If we choose 2
= 1 for convenience, then the rod is of a
material whose properties are somewhere between copper
and aluminum.
THANK YOU

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23MA401 Numerical Methods Boundary value problems

  • 1. 23MA401- Numerical Methods Boundary Value Problems By Dr. S.Shyamala AP/Maths Ms.G. Nandhini AP/Maths Excel Engineering College (Autonomous)
  • 2. INTRODUCTION  The basic partial differential equations of heat conduction, wave propagation, and potential theory that we discuss in this chapter are associated with three distinct types of physical phenomena: diffusive processes, oscillatory processes, and time-independent or steady processes.  Consequently, they are of fundamental importance in many branches of physics, and are significant mathematically.  The partial differential equations whose theory is best developed and whose applications are most significant and varied are the linear equations of second order.  All such equations can be classified as one of three types: The heat equation, the wave equation, and the potential equation, are prototypes of these categories.
  • 3. HEAT CONDUCTION IN A ROD: ASSUMPTIONS  Consider a heat equation conduction problem for a straight bar of uniform cross section and homogeneous material.  Let the x-axis be chosen to lie along the axis of the bar, and let x = 0 and x = L denote the ends of the bar. See figure below.  Suppose that the sides of the bar are perfectly insulated so that no heat passes through them.  Assume the cross-sectional dimensions are so small that the temperature u can be considered constant on the cross sections.  Then u is a function only of the axial coordinate x and time t.
  • 4. HEAT CONDUCTION: INITIAL AND BOUNDARY CONDITIONS  In addition, we assume that the initial temperature distribution in the bar is given, and hence where f is a given function.  Finally, we assume that the ends of the bar are held at fixed temperatures: the temperature T1 at x = 0 and T2 at x = L.  However, as will be shown in Section 10.6, we need only consider T1 = T2 = 0.  Thus we have the boundary conditions L x x f x u    0 ), ( ) 0 , ( 0 , 0 ) , ( , 0 ) , 0 (    t t L u t u
  • 5. HEAT CONDUCTION PROBLEM  Thus the fundamental problem of heat conduction is to find u(x,t) satisfying  With respect to the time variable t, this is an initial value problem; an initial condition is given and the differential equation governs what happens later.  With respect to the spatial variable x, it is a boundary value problem; boundary conditions are imposed at each end of the bar and the differential equation describes the evolution of the temperature in the interval between them. L x x f x u t t L u t u t L x u u t xx           0 ), ( ) 0 , ( 0 , 0 ) , ( , 0 ) , 0 ( 0 , 0 , 2 
  • 6. HEAT CONDUCTION: BOUNDARY PROBLEM  Alternatively, we can consider the problem as a boundary value problem in the xt-plane, see figure below.  The solution u(x,t) satisfying the heat conduction problem is sought in the semi-definite strip 0 < x < L, t > 0, subject to the requirement that u(x,t) must assume a prescribed value at each point on the boundary of this strip. L x x f x u t t L u t u t L x u u t xx           0 ), ( ) 0 , ( 0 , 0 ) , ( , 0 ) , 0 ( 0 , 0 , 2 
  • 7. HEAT CONDUCTION: LINEAR HOMOGENEOUS EQUATION  The heat conduction problem is linear since u appears only to the first power throughout.  The differential equation and boundary conditions are also homogeneous.  This suggests that we might approach the problem by seeking solutions of the differential equation and boundary conditions, and then superposing them to satisfy the initial condition.  We next describe how this plan can be implemented. L x x f x u t t L u t u t L x u u t xx           0 ), ( ) 0 , ( 0 , 0 ) , ( , 0 ) , 0 ( 0 , 0 , 2 
  • 8. SEPARATION OF VARIABLES METHOD  Our goal is to find nontrivial solutions to the differential equation and boundary conditions.  We begin by assuming that the solution u(x,t) has the form  Substituting this into our differential equation we obtain or ) ( ) ( ) , ( t T x X t x u  t xx u u  2  T X T X     2  T T X X     2 1 
  • 9. ORDINARY DIFFERENTIAL EQUATIONS  We have  Note left side depends only on x and right side only on t. Thus in order for this equation to be valid for 0 < x < L, t > 0, it is necessary for both sides of this equation to equal the same constant, call it -. Then  Thus the partial differential equation is replaced by two ordinary differential equations. T T X X     2 1  0 0 1 2 2               T T X X T T X X     
  • 10. BOUNDARY CONDITIONS  Recall our original problem  Substituting u(x,t) = X(x)T(t) into boundary condition at x = 0,  Since we are interested in nontrivial solutions, we require X(0) = 0 instead of T(t) = 0 for t > 0. Similarly, X(L) = 0.  We therefore have the following boundary value problem L x x f x u t t L u t u t L x u u t xx           0 ), ( ) 0 , ( 0 , 0 ) , ( , 0 ) , 0 ( 0 , 0 , 2  0 ) ( ) 0 ( ) , 0 (   t T X t u 0 ) ( ) 0 ( , 0       L X X X X 
  • 11. EIGENVALUES AND EIGEN FUNCTIONS  Thus  From Section 10.1, the only nontrivial solutions to this boundary value problem are the eigen functions associated with the eigen values  With these values for , the solution to the first order equation is 0 ) ( ) 0 ( , 0       L X X X X     , 3 , 2 , 1 , / sin ) (   n L x n x Xn   , 3 , 2 , 1 , / 2 2 2   n L n n   0 2    T T   constant. , 2 ) / ( n t L n n n k e k T   
  • 12. FUNDAMENTAL SOLUTIONS  Thus our fundamental solutions have the form where we neglect arbitrary constants of proportionality.  The functions un are sometimes called fundamental solutions of the heat conduction problem.  It remains only to satisfy the initial condition  Recall that we have often solved initial value problems by forming linear combinations of fundamental solutions and then choosing the coefficients to satisfy the initial conditions.  Here, we have infinitely many fundamental solutions.   , , 3 , 2 , 1 , / sin ) , ( 2 ) / (     n L x n e t x u t L n n   L x x f x u    0 ), ( ) 0 , (
  • 13. FOURIER COEFFICIENTS  Our fundamental solutions are  Recall the initial condition  We therefore assume that where the cn are chosen so that the initial condition is satisfied:  Thus we choose the coefficients cn for a Fourier sine series.   , , 3 , 2 , 1 , / sin ) , ( 2 ) / (     n L x n e t x u t L n n   L x x f x un    0 ), ( ) 0 , (            1 ) / ( 1 / sin ) , ( ) , ( 2 n t L n n n n n L x n e c t x u c t x u          1 / sin ) ( ) 0 , ( n n L x n c x f x u 
  • 14. SOLUTION  Therefore the solution to the heat conduction problem is given by where        1 ) / ( / sin ) , ( 2 n t L n n L x n e c t x u       L n dx L x n x f L c 0 / sin ) ( 2  L x x f x u t t L u t u t L x u u t xx           0 ), ( ) 0 , ( 0 , 0 ) , ( , 0 ) , 0 ( 0 , 0 , 2 
  • 15. EXAMPLE 1: HEAT CONDUCTION PROBLEM  Find the temperature u(x,t) at any time in a metal rod 50 cm long, insulated on the sides, which initially has a uniform temperature of 20° C throughout and whose ends are maintained at 0° C for all t > 0.  This heat conduction problem has the form 50 0 , 20 ) 0 , ( 0 , 0 ) , 50 ( , 0 ) , 0 ( 0 , 50 0 , 2           x x u t t u t u t x u u t xx 
  • 16. EXAMPLE 1: SOLUTION  The solution to our heat conduction problem is where  Thus        1 ) 50 / ( 50 / sin ) , ( 2 n t n n x n e c t x u                       even , 0 odd , / 80 cos 1 40 50 / sin 5 4 50 / sin 20 50 2 / sin ) ( 2 50 0 50 0 0 n n n n n dx x n dx x n dx L x n x f L c L n      
  • 17. EXAMPLE 1: RAPID CONVERGENCE  Thus the temperature along the rod is given by  The negative exponential factor in each term cause the series to converge rapidly, except for small values of t or 2 .  Therefore accurate results can usually be obtained by using only a few terms of the series.  In order to display quantitative results, let t be measured in seconds; then 2 has the units cm2 /sec.  If we choose 2 = 1 for convenience, then the rod is of a material whose properties are somewhere between copper and aluminum.