Unit - 2
Derivation of Heat equation
Q1) Derive one dimensional heat equations.
A1)
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.
Q2) 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.
A2)
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
Q3) What do you understand heat flow?
A3)
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.
Q4) What is one dimensional wave equation?
A4)
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.
Q5) Give D Alembert’s solution of the one dimensional wave equation
A5)
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
Q6) 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).
A6)
By d’Alembert’s method, the solution is
i.e., the given boundary corrections are satisfied.
Q7) Give the solution for wave equation in two dimensions.
A7)
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.
Q8) Obtain the solution of the wave equation
Using the method of separation of variables.
A8)
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)
Q9) Find the solution of the wave equation
Such that is a constant) when x = 1 and y = 0 when x =0
A9)
Put y = 0, when x = 0
(2) is reduced to
Put when x=1
Equating the coefficient of sin and cos on both sides
Q10) 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.
A10)
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
Q11) 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.
A11)
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
Q12) Give the solution for Laplace’s equation.
A12)
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-
Q13) Using the method of separation of variables, solve
Q14)
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.
Q15) Using the method of separation of variables, solve Where
A14)
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.
Q16) Determine the canonical form of the following equation.
A15)
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,
Q17) Determine the canonical form of the following equation.
A16)
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,