Unit - 2
Derivation of Heat equation
Derivation one dimensional heat equation-
Consider a homogeneous bar of uniform cross-section 
square cm.
Suppose that the sides are covered with a material impervious so that the stream lines of heat flow are all parallel and perpendicular to the area 
.
Take one end of the bar as the region and the directions of flow as the positive x-axis.
Suppose the density is 
, h is the specific heat and k is the thermal conductivity.
Let u(x, -t) is the temperature at a distance x from O. If 
is the temperature change in a slab of 
of the bar.
Then the quantity of hear flow in slab = 
.
Hence the rate of increase of heat in this slab,
Where 
and 
are respectively the rate of inflow and outflow of heat.
Now
Now
Which means
Writing 
, which is also called the diffusivity of the substance, and taking the limit as 
, we get
This is called the one-dimensional heat flow equation.
Example 1. A rod of length 1 with insulated sides is initially at a uniform temperature u. Its ends are suddenly cooled to 0° Celsius and are kept at that temperature. Prove that the temperature function u (x, t) is given by
Where 
is determined from the equation.
Solution. Let the equation for the conduction of heat be
Let us assume that u = XT, where X is a function of x alone and T that of t alone.
Substituting these values in (1) we get 
i.e. 
Let each side be equal to a constant 
And 
Solving (3) and (4) we have
Putting x = 0, u = 0 in (5), we get
(5) becomes 
Again putting x = l, u =0 in (6), we get
Hence (6) becomes 
This equation satisfies the given conditions for all integral values n. Hence taking n = 1, 2, 3,…, the most general solution is
By initial conditions 
Example 3. The ends A and B of a rod 20 cm long having the temperature at 30 degree Celsius and at 80 degree Celsius until steady state prevails. The temperature of the ends are changed to 40 degree Celsius and 60 degree Celsius respectively. Find the temperature distribution in the rod at time t.
Solution. The initial temperature distribution in the rod is
And the final distribution (i.e. steady state) is
To get u in the intermediate period, reckoning time from the instant when the end temperature were changed we assumed
Where 
is the steady state temperature distribution in the rod (i.e. temperature after a sufficiently long time) and 
is the transient temperature distribution which tends to zero as t increases.
Thus, 
Now 
satisfies the one dimensional heat flow equation
Hence u is of the form
Since 
Hence 
Using the initial condition i.e.
Putting this value of 
n (1), we get
Two dimensional heat flow-
Let us consider the heat flow in a metal plate of uniform thickness, in the direction parallel to length and breadth of the plate.
Let u(x, y) be the temperature at any point (x, y) of the plate at time t is given as-
In the steady state, u doesn’t change with t,
Equation (1) becomes-
Which is known as Laplace’s equation in two dimensions.
Derivation one dimensional wave equation-
The classical one dimensional wave equation is hyperbolic, which we study in transverse vibrations of longitudinal vibrations of a rod.
Suppose there is an elastic string, stretched to its length L and placed along the x-axis, with its two ends x = 0 and x = L fixed.
Suppose 
is the constant density of the string.
Let the function u(x, t) denote the displacement of string at any point x and at any time t > 0 from the equilibrium position (x-axis). When the string is distorted, then it vibrates. The small transverse vibrations of such a vibrating string are modelled by one-dimensional wave equation.
Let us consider a tightly stretched elastic string of length L.
Now using Newton’s second law of motion for a small portion of the string between x and x + 
x, [See fig]
Suppose 
and 
be tension at the end points P and Q of this portion of the string.
Assume that the points on the string move only in the vertical direction, there is no motion in the horizontal direction. Thus the sum of the forces in the horizontal direction must be zero.
Which means
Or
Neglect the gravitational force on the string, the only two forces acting on the string are the vertical
Components of tension 
at P and 
at Q with upward direction takes as positive.
By Newton’s second law-
Divide (2) by (1), we obtain
Putting
And
Then
Taking the limit as 
Thus
Here 
.
The above equation (4) is known as one-dimensional wave equation.
D Alembert’s solution of the one dimensional wave equation
The method of d' Alembert provides a solution to the one-dimensional wave equation
That models vibrations of a string.
The general solution can be obtained by introducing new variables 
and 
, and applying the chain rule to obtain
Using (4) and (5) to compute the left and right sides of (3) then gives
Respectively, so plugging in and expanding then gives
This partial differential equation has general solution
Where f and g are arbitrary functions, with f representing a right-traveling wave and g a left-traveling wave.
The initial value problem for a string located at position 
as a function of distance along the string x and vertical speed 
can be found as follows. From the initial condition and (12),
Taking the derivative with respect to t then gives
And integrating gives
Solving (13) and (16) simultaneously for f and g immediately gives
So plugging these into (13) then gives the solution to the wave equation with specified initial conditions as
Example. Find the deflection of a vibrating string of unit length having fixed ends with initial velocity zero and initial deflection f (x)=k (sinx –sin2x).
Solution. By d’Alembert’s method, the solution is
i.e., the given boundary corrections are satisfied.
Wave equation in two dimensions-
The equation
Is the wave equation in two dimensions.
Solution of wave equation in two dimensions (rectangular membrane)-
Let us assume that the solution of the above equation (1) is of the form-
Put these values in the equation and dividing by XYT, we get-
If each member is a constant then this can hold good.
Now choosing the constants suitably, we g
Et-
Hence the solution of equation (1) is-
Now let the membrane is rectangular and it is stretched between the lines x = 0, x = a, y = 0, y = b.
Then the condition u = 0 when x = 0 gives-
Then putting 
in (2) and applying the condition u = 0, when x = a,we get-
Now applying the conditions u = , when y = 0 and y = b, we get
Therefore the solution (2) becomes-
Where 
Choosing the constant 
so that 
We can write the general solution of the equation-1 as-
If the membrane starts from rest from the initial position u = f(x,y)
Which means-
Then equation 3 gives-
Also using the condition u = f(x, y) when t = 0, we get-
Which is the double Fourier series.
Now multiply the both sides by 
and integrating from x = 0 to x = a and y = 0 to y = b,
Every term on the right except one, become 0, therefore we get-
This is called the generalised Euler’s formula.
Example: Find the deflection u(x,y,t) of the square membrane with a = b = 1 and c = 1. If the initial velocity is zero and the initial deflection is f(x, y) = 
.
Sol.
Here taking a = b = 1 and f(x, y) = 
in equation above (5)-
We get-
Also from equation (3) above-
Therefore from equation (4),
The solution will be-
Equation of vibrating string
Consider an elastic string tightly stretched between two points O and A. Let O be the origin and OA as x axis. On giving a small displacement to the string perpendicular to its length (parallel to the y axis). Let y be the displacement at the point P (x, y) at any time. The wave equation
Example. Obtain the solution of the wave equation
Using the method of separation of variables.
Solution. 
Let y = XT where X is a function of x only and T is a function of t only.
Since T and X are functions of a single variable only.
Substituting these values in the given equation we get
By separating the variables we get
(Each side is constant since the variables x and y are independent)
Auxiliary equations are
Case 1. If k>0
Case 2. If k<0
Case 3. If k =0
These are the three cases depending upon the particular problems. Hare we are dealing with wave motion (k<0)
Example. Find the solution of the wave equation
Such that 
is a constant) when x = 1 and y = 0 when x =0
Solution:
Put y = 0, when x = 0
(2) is reduced to
Put 
when x=1
Equating the coefficient of sin and cos on both sides
Example: The vibrations of an elastic string is governed by the partial differential equation
The length of the string is π and the ends are fixed. The initial velocity is zero and the initial deflection is u (x, 0)=2 (sinx + sin3x). Find the deflection u (x, t) of the vibrating string for t ≥ 0.
Solution:
On putting x =0, u = 0 in (1) we get
On putting 
in (1) we get
On putting 
and u =0 in (2) we have
On substituting the value of p in (2) we get
On differentiating (3(,w.r.t t we get
On putting 
in (4) we have
On putting 
Given u (x, 0) = 2 (sin x+ sin3x)
On putting t = 0 in (5) we have
On substituting the value of 
Duhamel's principle for one dimensional wave equation
One dimensional heat flow
Let heat flow along a bar of uniform cross section in the direction perpendicular to the cross section. Take one end of the bar as origin and the direction of the heat flow is along x axis.
Let the temperature of the bar at any time t at a point x distance from the origin be u(x,t). Then the equation of one dimensional heat flow is 
Example-1: A rod of length 1 with insulated sides is initially at a uniform temperature u. Its ends are suddenly cooled to 0° Celsius and are kept at that temperature. Prove that the temperature function u (x, t) is given by
Where 
is determined from the equation.
Solution: Let the equation for the conduction of heat be
Let us assume that u = XT, where X is a function of x alone and T that of t alone.
Substituting these values in (1) we get 
i.e. 
Let each side be equal to a constant 
And 
Solving (3) and (4) we have
Putting x = 0, u = 0 in (5), we get
(5) becomes 
Again putting x = l, u =0 in (6), we get
Hence (6) becomes 
This equation satisfies the given conditions for all integral values n. Hence taking n = 1, 2, 3,…, the most general solution is
By initial conditions 
Example-2: Find the solution of
For which u ( 0, t) = u (l.t) =0 =sin 
bi method of variable separable.
Solution:
In example 10 the given equation was
On comparing (1) and (2) we get 
Thus solution of (1) is
On putting x =0
u =0 in (3) we get
(3) reduced to
On putting x = l and u =0 in (4) we get
Now (4) is reduced to
On putting t = 0, u = 
This equation will be satisfied if
On putting the values of 
and n in (5) we have
Example-3: The ends A and B of a rod 20 cm long having the temperature at 30 degree Celsius and at 80 degree Celsius until steady state prevails. The temperature of the ends are changed to 40 degree Celsius and 60 degree Celsius respectively. Find the temperature distribution in the rod at time t.
Solution:
The initial temperature distribution in the rod is
And the final distribution (i.e. steady state) is
To get u in the intermediate period, reckoning time from the instant when the end temperature were changed we assumed
Where 
is the steady state temperature distribution in the rod (i.e. temperature after a sufficiently long time) and 
is the transient temperature distribution which tends to zero as t increases.
Thus, 
Now 
satisfies the one dimensional heat flow equation
Hence u is of the form
Since 
Hence 
Using the initial condition i.e.
Putting this value of 
n (1), we get
Derivation of Laplace equation-
The equation-
Is known as the Laplace’s equation in two dimensions.
Solution of Laplace’s equation-
Let
Put the value in (1), we get-
Separating the variables-
Since x and y are the independent variables, equation (2) can hold good only if each side of (2) is equal to a constant (k),
Then (2) leads the ordinary differential equation-
On solving these equations, we get-
- When k is positive and it is equals to
, say
2. When k is negative and it is equals to 
, say
3. When k is zero-
Example: Solve the Laplace’s equation 
subject to the conditions u(0, y) = u(l, y) = u(x, 0) = 0 and u(x, a) = sin n
Sol.
The three possible solutions of Laplace’s equation-
Are-
We need to solve equation (1) satisfying the following boundary conditions-
u(0, y) ........... (5)
u(l, y) = 0........(6)
u(x, 0) = 0 ..........(7)
And u(x, a) = sin n
...... (8)
Using (5), (6) and (2), we get-
Solving these equations, we get-
Which leads to trivial solution.
Similarly we get a trivial solution by using (5), (6) and (4).
Hence the solution for the present problem is solution (3).
Now using (5) in (3), we get-
Therefore, equation (3) becomes-
Using (6), we get-
Therefore either-
If we take 
then we get a trivial solution.
Thus sin pl = 0 whence 
Equation (9) becomes-
Using (6), we have 0 = 
i.e.
Thus the solution suitable for this problem is-
Now using the condition (8)-
We get-
Hence the required solution is-
Method of separation of variables
In this method, we assume that the dependent variable is the product of two functions, each of which involves only one of the independent variables. So to ordinary differential equations are formed.
Example 1. Using the method of separation of variables, solve
Solution. 
Let, u = X(x). T (t). (2)
Where X is a function of x only and T is a function of t only.
Putting the value of u in (1), we get
(a) 
On integration log X = cx + log a = log 
(b)
On integration 
Putting the value of X and T in (2) we have
But, 
i.e. 
Putting the value of a b and c in (3) we have
Which is the required solution.
Example 2. Use the method of separation of variables to solve equation
Given that v = 0 when t→
as well as v =0 at x = 0 and x = 1.
Solution. 
Let us assume that v = XT where X is a function of x only and T that of t only
Substituting these values in (1), we get
Let each side of (2) equal to a constant 
Solving (3) and (4) we have
Putting x = 0, v = 0 in (5) we get
On putting the value of 
in (5) we get
Again putting x = l, v= 0 in (6) we get
Since 
cannot be zero.
Inputting the value of p in (6) it becomes
Hence, 
This equation satisfies the given condition for all integral values of n. Hence taking n = 1, 2, 3,… the most general solution is
Example. Using the method of separation of variables, solve 
Where 
Solution. Assume the given solution 
Substituting in the given equation, we have
Solving (i) 
From (ii) 
Thus 
Now, 
Substituting these values in (iii) we get
Which is the required solution.
Key takeaways
- One-dimensional heat flow equation-
2. Laplace’s equation in two dimensions-
3. One-dimensional wave equation-
4. Wave equation in two dimensions-
The second order PDE in two independent variables of the form
Here a,b,c,d,e,f and g are the functions of the independent variables x and y.
The principal part of the operator L, can be given as-
The equation is classified as-
Hyperbolic if- 
Parabolic if- 
Elliptic if- 
Here 
is the discriminant of the operator L.
Example: Classify the following PDEs into hyperbolic, parabolic or elliptic.
Sol. In the first PDE, a = 1, b = 0 and c = 
So that-
Thus we can say that the given PDE is hyperbolic.
Now in second PDE,
A = 1, b = 0 and c = 
So that-
Therefore the second PDE is elliptic.
Key takeaways
- Hyperbolic if-
- Parabolic if-
- Elliptic if-
Suppose the equation is-
Where A,B,C,D.E and F are the real constants.
Then the equation-
1. Elliptical- if 
2. Parabolic- if 
3. Hyperbolic- if if 
The equation (1) can be reduced to more simple form by the change of the independent variable.
This form is said to be canonical form of equation (1),
Lets introduce the new independent variables- 
, by means of the transformation
Now we compute derivatives of u, regarding 
as intermediate variables.
So that,
Are given by (1), we first find
And
Now using (2), we determine 
Since 
And
In finding 
and 
, the situation somewhat more difficult and we need to be very careful to remember what is involved.
We are regarding u as a function of 
, where 
are themselves the functions of x and y.
That is
where 
and 
.
Thus 
and 
are regarded as the function of 
, where 
are themselves the functions of x and y.
That is
And
We compute
And since
We get
Same as, we get
Put this value in (4), we get
Assume that 
has second continuous derivative with respect to 
.
The cross-derivatives are equal and we have
Same as we find
… (6)
And
And
Now we put these values in PDE, we get
By rearranging terms, this becomes
Thus using the transformation, equation (1) becomes
Where
And
Canonical form of hyperbolic, parabolic and elliptic equations
Type of equation | Canonical form |
Hyperbolic  |  |
Parabolic  |  |
Elliptic  |  |
Example: Let us consider the equation
Here we see that the equation is hyperbolic as 
.
Since A is non-zero,
Let us consider the transformation,
Here 
and 
are the roots of quadratic equation 
.
We find that 
The transformation (2) becomes
Apply (3) to (1), we get
Divide by -36 and transposing terms, we get the canonical form,
Example: Consider an equation
Here we see that the equation is elliptic as 
.
Since A is non-zero,
Let us consider the transformation,
Where 
are the conjugate complex roots of the quadratic equation 
.
We find that 
The transformation (2) becomes
Apply (3) to (1), we get
Divide by 4 and transposing terms, we get the canonical form,
Key takeaways
- Hyperbolic-
- Parabolic-
3. Elliptic-
References:
1. Erwin Kreyszig, Advanced Engineering Mathematics, 9thEdition, John Wiley & Sons, 2006.
2. Advanced engineering mathematics, by HK Dass
3. Differential equations, Shepley L. Ross, Willey India.
4. Partial differential equations, Phoolan Prasad, Renuka ravindran, John Willey & Sons
