Unit - 3

The Cauchy problem

Q1) What is Cauchy’s problem?

A1)

Let

Where A(x, y), B(x, y) and C(x, y) are functions of x and y and be the curve in x, y plane.

The problem of finding the solution u(x, y) and the PDE (1) in the neighbourhood of satisfying the following conditions

On is called a Cauchy problem and the above two conditions are called Cauchy conditions.

Consider the Cauchy problem in the case of Laplace’s equation

With the following data prescribed on the x-axis,

U(x, 0) = 0

The intital curve is here is the x-axis,

It is easy to verify that

Is the solution of the above problem.

Observe that when n tends to infinity the function uniformly.

However does not become small an n tends to infinity for any non-zero y.

So that the solution is not stable.

We will now show that the solution to the Dirichlet’s problem is stable.

Suppose and are the solutions of

And

Let v = (, then

By the maximum and minimum principle, the harmonic function v attains its maximum and minimum on B which is nothing but the maximum and minimum of .

Q2) What is wave equation(non-homogenous)?

A2)

Let us consider the non-homogenous wave equation

With homogeneous initial conditions,

We suppose the function v(x, t;r) which satisfies the equation given below with respect to x and t for t < r,

And the following conditions at t = r

v(x, t;r) = 0,   

By using D’ Alembert’s solution, the solution of this problem is-

Consider

Here we will show that u is the solution,

Since

Therefore

We observe that u(x, t) satisfies the condition (2),

If (2) replaced by

Then the solution of equation (1) is obtained by superposing u from equation (6) on D’Alembert’s solution.

Q3) Give the method of separation of variables.

A3)

Let us consider the following problem

So f and g are the intital displacement and velocity respectively.

Let us assume the solution of equation (1)

Then

Here the RHS is a function of t while the LHS is function of x, here each of them must be constant and equal to , therefore

Now from equation (4)

We have

Since T(t) is non-zero, we get X(0) = 0

Similarly y(l, t) = 0 implies that X(l) = 0

Hence we have

Which is an Eigenvalue problem

Case-1: , the solution of the above eigen value is

Where A and B are the arbitrary constants,

To satisfy the boundary condition

Possibility is A = B = 0,

Hence there is no eigen value .

Case-2: , the solution of the eigenvalue problem is

The boundary condition implies that A = 0 and A + Bl = 0

Therefore A = B = 0, hence is not an eigen value.

Case-3: , in this case the solution is

The condition X(0) = 0 implies that A = 0 and X(l) = 0 implies that

As B = 0 gives the trivial solution, we must have for a non-trivial solution.

Hence

is called the eigen value and the functions are the corresponding eigen functions.

Therefore

For each , we have

Where are arbitrary constants, hence

Which is the solution of the equation (1) and satisfies the boundary condition (4)

Q4) What is heat conduction problem.

A4)

Let us consider the following heat conduction problem in an infinite rod with the following conditions

1. The rod position coincides with the x- axis and the rod is homogenous.
2. The heat is uniformly distributed over its cross sections at a time t. 
3. The surface is isolated to prevent heat loss.
4. u(x, t) is the temperature at the point x at time t.

Then the problem we need to solve is

Suppose the Fourier transform of u(x, t) is

Taking the Fourier transform of equations (1) and assuming that u, as , we get

Its solution is given by

Where is an arbitrary function to be determined from the initial condition as follows

Hence

By the convolution theorem

Equations (3) gives the solution of (1) and (2)

Now consider the case k = 1 and

Put

Where erf(x) is the error functions.

Q5) What is initial boundary condition.

A5)

PDE’s are usually specified through a set of boundary or initial conditions. A boundary condition expresses the behavior of a function on the boundary (border) of its area of definition. An initial condition is like a boundary condition, but then for the time-direction. Not all boundary conditions allow for solutions, but usually the physics suggests what makes sense. Let you remind the situation for ordinary differential equations, one you should all be familiar with, a particle under the influence of a constant force.

Q6) What do you understand by semi-infinite String with a Fixed End?

A6)

Let us first consider a semi-infinite vibrating string with a fixed end,

That is,

…..(1)

It is evident here that the boundary condition at x = 0 produces a wave moving to the right with the velocity c. Thus, for x > ct, the solution is the same as that of the infinite string, and the displacement is influenced only by the initial data on the interval [x − ct, x + ct],

When x < ct, the interval [x − ct, x + ct] extends onto the negative x-axis where f and g are not prescribed. But from the d’Alembert formula

…(2)

Where

We see that

If we let α = −ct, then

And hence,

The solution of the initial boundary-value problem, therefore, is given by

In order for this solution to exist, f must be twice continuously differentiable and g must be continuously differentiable, and in addition

f (0) = f’’(0) = g (0) = 0.

Q7) Determine the solution of the initial boundary-value problem

A7)

For x > 2t,

And for x < 2t,

And for x < 2t,

Notice that u (0, t) = 0 is satisfied by u (x, t) for x < 2t (that is, t > 0).

Semi-infinite String with a Free End

We consider a semi-infinite string with a free end at x = 0. We will determine the solution of

As in the case of the fixed end, for x > ct the solution is the same as that of the infinite string. For x < ct, from the d’Alembert solution

We have

Thus

Integration yields

Where K is a constant. Now, if we let α = −ct, we obtain

Replacing α by x − ct, we have

And hence,

The solution of the initial boundary-value problem, therefore, is given by

For x < ct

We note that for this solution to exist, f must be twice continuously differentiable and g must be continuously differentiable, and in addition,

F’ (0) = g’(0) = 0.

Q8) Give the equations with Non-homogeneous Boundary Conditions.

A8)

In the case of the initial boundary-value problems with non-homogeneous boundary conditions, such as

u(0, t) = p(t)    ,

We proceed in a manner similar to the case of homogeneous boundary conditions.

We apply the boundary condition to obtain

If we let = −ct, we have

Replacing by x − ct, the preceding relation becomes

Thus, for 0 ≤ x < ct,

Where

(x + ct = ) , and () is given by

In this case, in addition to the differentiability conditions satisfied by f and g, as in the case of the problem with the homogeneous boundary conditions, p must be twice continuously differentiable in t and

We next consider the initial boundary-value problem

We apply the boundary condition to obtain

Then, by integrating

If we let α = −ct, then

Replacing α by x − ct, we obtain

The solution of the initial boundary-value problem for x < ct, therefore, is

Given by

Here f and g must satisfy the differentiability conditions, as in the case of

The problem with the homogeneous boundary conditions. In addition

f’(0) = q (0), g’(0) = q’(0)

The solution for the initial boundary-value problem involving the boundary condition

Where h = constant

Can also be constructed in a similar manner from the d’Alembert solution.

Q9) What do you understand by vibration of Finite String with Fixed Ends.

A9)

We know that the solution of the wave equation is

Applying the initial conditions, we have

Solving for and , we find

Hence

For 0 ≤ x + ct ≤ l and 0 ≤ x − ct ≤ l. The solution is thus uniquely

Determined by the initial data in the region

For larger times, the solution depends on the boundary conditions. Applying

The boundary conditions, we get

If we set = −ct, equation first equation of above becomes

…..(1)

And if we set = l + ct, the second equation becomes

With = −

……….(2)

From (1) and (2), we get

We see that the range of () is extended to −l ≤ ≤ l.

If we put =

Then, by putting = 2l −