Module 3D

Partial Differential Equation- Higher Order

3.1 Solution to homogenous and non- homogenous linear partial differential equation second and higher order by CF and PI

Rule for finding the CF

Let f(D,D’)z=F(x)

Consider the equation

Step I: Put D=m and D’=1

This is the auxiliary equation .

Step 2 : Solve the AE

Case I- If the roots of AE are real and different say

Case 2 :If roots are equal say m

Solve:

Ans.

Solve :

Or

Solve.

Ans. Given

Its AE is

Complete solution is

Solve.

Ans. Its AE, put D = m, D’ = 1

Complete solution is

Rules for finding the PI

Given f(D D’)z=F(x y)

Case I : When

Put D=a and D’=b

Solve 

Ans.

(

Solve.

Ans. Given

AE is

Expand in ascending power of D or D’ and operate on term by term.

Solve.

Ans. Put,

AE is

Solve.

Ans. Given equation in symbolic form

Its AE is

Complete solution is

Solve.

Ans. Its auxiliary equation is

Put, D = m, D’ = 1

Complete solution is

Solve.

Ans. Put D = m, D’ = 1

The AE is

m2-m=0

m=0,1

in the first integral and in the second integral

Solve. [D2-6DD'+9D'2]z=(6x+2y)+tan(3x+y)+e3x+2y

For complementary function

PI for

PI for

Complete solution CF + PI

Solve

Ans. Put, D = m, D’ = 1

AE is

The complete solution is

Solve.

Ans. Put D = m, D’ = 1

Auxiliary equation is

Complete solution is

3.2 Flows, Vibrations, Diffusion:

Flows :- Flow formulizes the idea of the motion of particle in a fluid.

Vibration :- Vibration are oscillation in mechanical dynamic system.

Diffusion:- The diffusion equation is a parabolic particle differential equation.

3.3 Second order linear equation and their classification

Charpits Method

f(xyz . P q)=0

Dz=Pdx + qdy

Subsidiary equation are

Solve

Ans.

Charpits Equation are

Taking two members

On integrating

Putting for p and q we get

On integrating

Solve.

Ans. Let,

Then charpits Equation are

From first and last equation we get

On integrating

⇒q=xa

Then the equation

So that the relation leads to

On integration

Which is a complete y integral

Solve.

Ans.

Charpits auxiliary equations are

From the first and last members we get

On integrating

3.4 Initial and Boundary condition

Initial value problem is 4(a)=P y’(a)=q

Boundary value problem is 4(a)=Py(b)=q

Solve. Given, if u (x, 0) =0, u (0, t) = t

Ans. LET, u (x, s) = L [u (x, t)(s) for the Laplace transform of u

Then applying the Laplace transform to this equation

This is a constant coefficient first order differential equation

Integrating both sides we get

Using boundary conditions

Taking the inverse Laplace transform we have

Solve. Given,

Ans.

using boundary and initial condition

Now, such that

Solve. Given,

Let,

Ans. where

Also,

After differentiation of above equation with respect to x we get

Where C is integration constant

Using implicit method

Let,

For equation g (k)

This is implicit scheme.

3.5 Alembert’s solution of Wave Equation

Let us introduce that two independent variables =
 

So that y becomes a function of u and v

Putting the value of

From (2) ,(3) and (1) we get

Integrating (4) with respect to v we get

Where is constant in respect to v

Again integrating equation (5)

Where is constant in respect to u.

3.6 Duhamel’s Principle for One Dimensional Wace Equation

Duhamel’s principle is solution to this problem is

Where is the solution of the problem

Example.

Where,

3.7 Method of separation of variables

Using the method of separation of variable

Solve. where

Ans.      (1)

Let, u=X(x)T(t)           (2)

Using (2) in (1) we get

On integrating

On integrating

Putting the value of X and T in (2) we get

Putting the value of a, b, and c we get

3.8 Laplacian in plane, Cylindrical and Spherical polar coordinates:

Laplace equation in polar co-ordinates

Laplace equation in spherical co-ordinates

Method of separation of variables

The equation for variation of the string is

The solution of equation (1) is

Let l be the length of the string

Equation of OB is

Equation of BC is

Equation of CA is

Hence the boundary conditions are

y (0, t) = 0

y (l, t) = 0

when t = 0

And,   y (x, 0) =

From (2)

From (2)

From (6)

From (7)

The most general solution is

Where n = 2m

Which is required expression for y (x, t)

Putting,   in equation (6) we get

Hence mid point of string is always at rest.

Solve.

Ans. Given,

Multiply both sides of the equation by cos x

Let, u (x) = y (x) Cos x         (2)

Find

Thus from last equality we get

Substitution of the (3) in (1) we get

For complementary solution

Put,

The obtained roots give

The general solution is the sum of

Apply Euler’s identity

Redefine as and as

Since these are arbitrary constant

Determine the particular solution to by method of undetermined coefficients

The particular solution to  is of the form

Substitute the particular solution into the differential equation

The general solution is

TEXTBOOKS/REFERENCES:

1. S. J. FARLOW, PARTIAL DIFFERENTIAL EQUATIONS FOR SCIENTISTS AND ENGINEERS, DOVER PUBLICATIONS, 1993.
2. R. HABERMAN, ELEMENTARY APPLIED PARTIAL DIFFERENTIAL EQUATIONS WITH FOURIER SERIES AND BOUNDARY VALUE PROBLEM, 4TH ED., PRENTICE HALL, 1998.
3. IAN SNEDDON, ELEMENTS OF PARTIAL DIFFERENTIAL EQUATIONS, MCGRAW HILL, 1964.
4. MANISH GOYAL AND N.P. BALI, TRANSFORMS AND PARTIAL DIFFERENTIAL EQUATIONS, UNIVERSITY SCIENCE PRESS, SECOND EDITION, 2010.