PARTIAL DIFFERENTIAL EQUATIONS AND BOUNDARY VALUE PROBLEMS

1
PARTIAL DIFFERENTIAL EQUATIONS AND BOUNDARY VALUE
PROBLEMS VIBRATING STRINGS AND HEAT CONDUCTION
2
THE VIBRATING STRING
Suppose that a flexible string is pulled taut on
the x-axis and fastened at two points, which we
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 y(x,t)
from the equilibrium position of the string at
the point (distance x from the origin) at time t.
3
We make certain assumptions The subsequent
motion of the string is entirely transverse (i.e.
along the xy plane). (Thus the displacement of
the string from the equilibrium position is a
function y y (x, t).)
We look at a small piece of the string which in
the equilibrium position has length ?x. If m is
the (uniform) density of
4
the string, the mass of the piece is m ?x. By
Newtons second law of motion, the transverse
force F acting on the piece is given by
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- component.
5
T sin ?
T
We now assume that the motion of the string is
due strictly to the tension T in it. (That is the
gravitational force is negligible.)
Hence F ? (T sin ? ) ? ? (T tan ? ) (as ?
is small) and so ? (T tan ? )
6
Thus we get
Dividing by ?x and letting ?x ? 0, we get
Or as T is constant (the same for all x), we get
7
Putting
We get the one-dimensional wave equation
which is usually written as
8
It is proved in Physics that y(x,t) satisfies the
one-dimensional wave equation
where a is a constant (depending on the mass of
the string and the tension of the string). We
thus want to solve the boundary value-problem
? (1)
9
subject to the boundary conditions
for all t gt 0 ? (2)
and the initial conditions
for t 0 ? (3)
y(x,0) f(x), 0 x L ? (4)
We find the solution by the method of separation
of variables. That is we let y(x,t) u(x) v(t)
(a function of x ? a function of t )
10
Substituting for y in the equation (1), we get
or
? (5)
where the dashes denote derivative w.r.t. the
concerned variable. The L.H.S of (5) is
independent of t and the R.H.S is independent of
x and hence both are independent
11
of both x and t and hence equals a pure constant,
say, - ?.
Thus, we get
for all t gt 0
implies u(0) u(L) 0.
So we have to solve for u
with boundary conditions u(0) u(L) 0.
12
This eigenvalue problem has nonttrivial solutions
only when
Corresponding solutions are
and non-zero multiples of them.
13
for t 0 implies
Now for each n, v satisfies the d.e.
with the initial condition
Thus
14
Hence
We shall take c11 and denote the corresponding
solution (which depends on n ) by vn . Thus
15
Thus for each n 1,2,3,
is a solution to the given problem satisfying the
first three conditions. So also any finite or
infinite lc of them. But no finite lc of them
would satisfy the initial condition
y(x,0) f(x), 0 x L
So we consider their infinite lc so as to satisfy
the above initial condition.
16
Thus look at the infinite linear combination
Now t 0 implies
Thus bn Fourier sine coefficient of f(x) in 0,
L
17
Thus the solution to the wave equation satisfying
the initial and boundary conditions is
0 x L, t ? 0
where
18
Example Solve the wave equation when
Thus
19
Thus the solution to the given problem is
0 x L, t ? 0
20
Example Solve the wave equation when
Thus
21
(No Transcript)
22
(No Transcript)
23
Hence the solution to the wave equation is
24
Example Solve the wave equation when
Thus
25
(No Transcript)
26
(No Transcript)
27
Thus the solution to the given problem is
28
DAlemberts solution of the wave equation
Consider the wave equation
We change the independent variables x, t to u, v
by substituting u x at , v x - at
29
(assuming that all mixed partial derivates are
continuous). Similarly
30
Hence the wave equation becomes
or
Integrating partially w.r.t u, we get
a constant independent of u
a function of v g(v) say
Again integrating partially w.r.t v, we get
31
(where a is a constant 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 initial
and boundary conditions. This is extremely
difficult. Hence we have to adopt the Fourier
Series method for solving the wave equation.
32
ONE-DIMENSIONAL HEAT EQUATION
33
The heat equation
We consider the flow of heat in a thin
cylindrical rod of cross-sectional area A whose
lateral surface is perfectly insulated so that no
heat flows through it. The word thin means that
the temperature is uniform on any cross- section
and thus is a function of the time t and its
horizontal distance x from one end, w(x,t).
34

• We note the following physical principles
• Heat flows in the direction of decreasing
  temperature, that is from hot regions to cold
  regions
• The rate at which heat flows across an area is
  proportional to the area and to the rate of
  change of temperature with respect to the
  distance in a direction perpendicular to the
  area. (This proportionality factor is denoted by
  k and called the thermal conductivity of the
  substance.)

35
c) The quantity of heat gained or lost by a body
when its temperature changes is proportional to
the mass of the body and to the change of
temperature. (This proportionality factor is
denoted by c and called the specific heat of the
substance.)
36
Using the above three principles, we can derive
the heat equation describing the temperature
w(x,t) as
Here the constant a2 depends on three things
?,density of the rod, c, specific heat of the
substance and k, thermal conductivity of the
substance.
In fact
37
We assume the boundary conditions
This means that the ends are maintained at zero
temperature at all times. We also assume that the
rod has an initial temperature distribution f(x)
at time t 0 i.e. w(x,0) f(x) 0 lt x lt
L Again separating the variables, we find
that w u v where u satisfies the eigenvalue
problem
38
(1)
and that v satisfies first order d.e
(2)
Thus for nontrivial solutions,
Corresponding eigenfunctions are
39
For each n1,2,3,
is a solution of (2) with
Hence for each n 1,2,3,
is a solution of the heat equation.
40
Again we assume that the infinite linear
combination
is also a solution. Putting t 0, we get
Hence bn nth Fourier sine coefficient of f(x)
in 0, L
41
Problem
Solve the heat equation given that
42
Solution
The solution to the heat equation subject to the
given boundary and initial conditions are
where bn nth Fourier sine coefficient of f(x)
in 0, L
43
Continued on the next slide
44
(No Transcript)
45
(No Transcript)
46
The solution to the heat equation subject to the
given boundary and initial conditions are
47
Steady state temperature We assume that when t is
large, w? a function independent of t, w(x) say.
This is called the steady state temperature. And
noting that in steady state
we get
Therefore
48
Suppose that the ends are maintained at
temperature w1,w2 at all times. Hence
Hence the steady state temperature is
49
Problem
Find the solution of the one-dimensional heat
equation satisfying the boundary and initial
conditions
for all t.
0 ? x ? L
50
Solution
We write the temperature distribution as
steady state temperature Transient
temperature.
We impose the boundary conditions wS(0)
100, wS(L) 0 on wS(x).
51
Thus
We impose on wT(x,t) the boundary conditions
wT(0,t) wT(L,t) 0 for all t
and the initial condition
52
That is wT(x, t) satisfies the boundary
conditions
wT(0,t)wT(L,t) 0 for all t
and the initial condition
Thus
53
where
54
Now
55
(No Transcript)
56
And
57
Thus
58
Problem
The temperatures at the ends x 0 and x 100 of
a rod 100 cm in length with insulated sides are
held at 0 and 100 degree centigrade respectively,
until steady-state conditions prevail. Then at
the instant t 0, the temperatures of the two
ends are interchanged. Find the resultant
temperature distribution of the rod as a function
of x and t.
59
Solution
We note that the steady state temperature (when
the ends x 0 and x 100 of the rod 100 cm in
length are held at 0 and 100 degree centigrade
respectively) is
Now, the temperatures of the two ends are
interchanged.
60
And we have to find the new temperature
distribution in the rod.
Hence we have to solve the heat equation with the
boundary and initial conditions
for all t.
0 ? x ? L
61
We write the temperature distribution as
(new) steady state temperature Transient
temperature.
We impose the boundary conditions wS(0)
100, wS(100) 0 on wS(x).
62
Thus
We impose on wT(x,t) the boundary conditions
wT(0,t) wT(100,t) 0 for all t
and the initial condition
i.e.
63
That is wT(x, t) satisfies the boundary
conditions
wT(0,t)wT(100,t) 0 for all t
and the initial condition
Thus
64
where
65
Hence the desired temperature distribution is
66
END OF WAVE AND HEAT EQUATIONS
67
TWO-DIMENSIONAL HEAT EQUATION - LAPLACES
EQUATION
68
The two-dimensional heat equation A thin
rectangular plate has its boundary kept at
temperatures 0,0,0 and f(x) at all times. Find
the steady temperature distribution in the plate.
69
Thus we have to solve the Laplaces equation
subject to the boundary conditions
We again solve by the method of separation of
variables. We assume w(x,y) X(x) Y(y)
70
Substituting and separating the variables we get
(a constant)
Thus X satisfies the boundary value problem
X(0) X(L) 0
We know nontrivial solutions exist when
71
and the corresponding eigenfunction is
For each n, Y satisfies the BV problem
Now
72
Taking
we find
Thus for each n,
is a solution satisfying the first three boundary
conditions. Hence
73
is also a solution. Putting y 0, we get
Hence
nth Fourier sine coefficient of f(x) in 0, L
74
Thus we have found out the steady state
temperature distribution.
THE END
75
BEST OF LUCK IN THE COMPRE