Unit – 6
Applications of Partial Differential Equations (PDE)
Differential equation in which partial derivatives one involved, are called partial differential equation (PDE).
The order of the PDE is the highest order of partial derivatives presenting it.
We know that,
Partial differential equation of order one obtained by eliminating one arbitrary function 
.
Having two order obtained by eliminating two arbitrary function 
and .
While deriving a differential equation corresponding to a string problem we assume the following assumptions.
The string is perfectly plastic and doesn’t offer any resistance to bending.
The mass of the string per unit length is constant.
The tension is so large.
Consider the forces acting on a small portion of string. Let T1 and T2 be the tensions at end P and Q.
The horizontal components of T1 and T2 must be constant.
(1)
and vertical components of 
and 
are 
and 
By using (1), we can divide this by
We know that, 
and 
As 
Let 
This is called one-dimensional wave equation.
Solution of Wave Equation by method of separation of variable:
The vibration of an elastic string are governed by are dimensional wave equation
where
is the deflection of the sting.
Since the string is fixed at the ends 
and 
.
We have two boundary conditions,
and 
for all t
We have two initial conditions:
1. If 
represents the vibrations of a string of length 
fixed at both ends. Find the solution with boundary conditions.
and initial conditions.
Solution:
Given 
The most general solutions is given by
Applying conditions (1), 
To apply conditions (iii) 
we first obtain 
Here, 
The most general solution will be
(1)
Applying condition (ii) 
and 
Solution (1) becomes
Combining all these solution we get
Applying condition (iv)
This is Fourier Half Range she series
For
This is Fourier Half Range she series
For 
Substituting in equation (2) we get
2. A tightly stretched string with fixed end points 
and 
is initially in a position given by 
If it is released from rest from this positon. Find the displacement y at any distance x from one end at any time t.
Solution:
The differential equation satisfied by 
is 
The initial and boundary conditions are given by
(i) 
(ii) 
(iii) 
(iv) 
The most general solution is given by
From condition (i) we get, 
From condition (iii) we get, 
The most general solution will become
(1)
From Condition (ii)
Equation (1) becomes
Combining all these solutions we get,
(2)
Applying condition (iv)
By using 
Comparing we get,
Substitute in equation (2) we get,
One dimensional Heat flow by method of separation of variables
We have to obtain solution of P.D.E.
The most general solution is
1. Solve: 
subject to condition
(i) 
(ii) 
(iii) 
is odd
(iv) 
for 
Solution:
We have
Step 1: The most general Heat equation is
(1)
Step 2: Use first B.C
Step 3: Equation (1) becomes
(2)
Step 4: Use 2nd B.C 
, 
Step 5: Equation (2) becomes
Step 6: Take addition of all solutions and
(3)
Step 7: Use I.C 
Step 8: To find 
we have,
Put in eqn (3) we get,
2. Solve : 
subject to :
Solution:
We have
Step 1: The most general heart equation is
….. (1)
Step 2: Use B.C 
Step 3: Equation (1) becomes
(2)
Step 4: Use 2nd B.C 
, 
Step 5: Equation (2) becomes
Step 6: Take addition of all solutions and
(3)
Step 7: Use I.C. 
Step 8: To find 
Put in Equation (3) we get,
Two dimensional Heat flow by method of separation of variables:
Two dimensional Heat flow also called Laplace’s equation in two dimensions.
The equation is 
The most general solution is given by
Use this Equation 
when 
1. Solve the equation 
with conditions
(i) 
when 
for all 
.
(ii) 
when 
for all values of 
.
(iii) 
when 
for all values of 
.
(iv) 
when 
for 
Solution:
Here in condition (i), 
when 
for all values of 
, for all z is given
Step 1:
The most general solution is
(1)
Step 2:
Now condition (i) that 
must remain finite as 
, this is possible only if 
Step 3:
Apply condition (ii) 
Step 4: The equation (1) becomes
Step 5:
By condition (iii)
Step 6:
Combining all these solution we get
Applying condition (iv) we have
Which is half range since series
The complete solution is
2. A rectangular plate with insulated surfaces is 10 cm wide and so long compared to its width that may be considered infinite in length without introducing an appreciable error. If the temperature along short edge 
is given 
while the two long 
and 
as well as the other short edge are kept at 0
C. Find steady state temperature 
Solution:
We have to solve the P.D.E
Subject to the conditions
(i) 
(ii) 
(iii) 
(iv) 
Step 1: The most general solution is
(1)
Step 2:
By (i) condition, 
Step 3:
By (ii) condition 
The equation (1) becomes
Step 4: By (iii) condition
Step 5:
Applying condition (iv), we have
By comparing we get,
, 
The answer is
1. Solve by method separation of variables
where 
Solution:
Given 
Let the solution of this equation be S
Solving
Integrate both sides
By solving
Integrating both sides,
When 
,
(given)
By comparing
Also
Extra Practice Problems:
1. An infinity long uniform metal plate is enclosed between lines 
and 
for 
. The temperature is zero along the edges 
and at infinity. If the edge 
is kept at a constant temperature 
. Find the temperature distribution 
.
Solution:
We have to solve P.D.E.
Subject to the boundary condition.
In condition (iii), 
when 
is given
(i) 
(ii) 
(iii) 
(iv) 
In condition (iii), 
when 
is given
The most general solution is given by
Now condition (iii) u(x , y) must remain as 
By condition (i) 
By condition (ii)
Apply condition (iv) we have
Which is represented by half range Fourier fine series
For 
in 
The complete solution is
2. Solve the equation where satisfies the following conditions
(i) 
(ii) 
(iii) 
(iv) 
is finite
Ans: 
we have
Solution:
We have
Step 1: The most general solution is
Step 2: Use 1st B.C 
step 3: Equation (1) becomes
Step 4: Use 2nd B.C 
,
Step 5: Equation (2) becomes
Step 6: Take addition of all solutions and
Step 7: Use I.C 
Step 8: To find 
we have
3. A string is stretched and fastened to two points 
apart. Motion is started by displacing by the string in form from which it is released at time t=0 find the displacement 
from one end.
Solution:
We have,
Subject to the condition:
(i) 
(ii) 
(iii) 
(iv) 
The general solution is
By condition (i) 
By condition (iii) 
Eq (1) becomes
Applying condition (ii) we get
Substituting in equation (2) we get
Combining these solutions we get
Applying condition (iv) we get
By comparing
Final Answer is
