Module 3

Partial differential equations

A differential equation involving partial derivatives with respect to more than one independent variable is called a partial differential equation.

The independent variables will be denoted by x and y and the dependent variable by z. The partial differential coefficients are denoted as-

ORDER of a partial differential equation is the same as that of the order of the highest differential coefficient in it.

Classification of partial differential equation-

Suppose the equation is-

Here A, B, C are the constants of x and y, then the equation-

1. Elliptical- if

2. Parabolic- if

3. Hyperbolic- if if

Formation of partial differential equation-

Method of elimination of arbitrary constants-

Example: Form a partial differential equation from-

Sol.

Here we have-

It contains two arbitrary constants a and c

Differentiate the equation with respect to p, we get-

Or

Now differentiate the equation with respect to q, we get-

Now eliminate ‘c’,

We get

Now put z-c in (1), we get-

Or

The second method we use is method of elimination of arbitrary function.

Solution of partial differential equation by direct partial Integration-

Example: Solve-

Sol.

Here we have-

Integrate w.r.t. x, we get-

Integrate w.r.t. x, we get-

Integrate w.r.t. y, we get-

Example: Solve the differential equation-

Given the boundary condition that-

At x = 0,

Sol.

Here we have-

On integrating partially with respect to x, we get-

Here f(y) is an arbitrary constant.

Now form the boundary condition-

When x = 0,

Hence-

On integrating partially w.r.t.x, we get-

Example: Solve the differential equation-

Given that when y = 0, and u = when x = 0.

Sol.

We have-

Integrating partially w.r.t. y, we get-

Now from the boundary conditions,

Then-

From which,

It means,

On integrating partially w.r.t. x givens-

From the boundary conditions, u = when x = 0

From which-

Therefore the solution of the given equation is-

Lagrange’s linear equation in an equation of the type-

Here P, Q and R are the functions of x, y, z and p = and q =

Working steps to solve-

Step-1: Write down the auxiliary equation-

Step-2: Solve the auxiliary equations-

Suppose the two solutions are- u = and v =

Step-3: Then f(u, v) = 0 or u = ∅(v) is the required solution of

Method of multipliers-

Let the auxiliary equation be

L, m, n may be the constants of x, y, z then we have-

L, m, n are selected in a such a way that-

Thus

On solving this differential equation, if the solution is- u =

Similarly, choose another set of multipliers and if the second solution is v =

So that the required solution is f(u, v) = 0.

Example: Solve-

Sol.

We have-

Then the auxiliary equations are-

Consider first two equations only-

On integrating

…….. (2)

Now consider last two equations-

On integrating we get-

…………… (3)

From equation (2) and (3)-

Example: Find the general solution of-

Sol. The auxiliary simultaneous equations are-

……….. (1)

Using multipliers x, y, z we get-

Each term of (1) is equals to-

Xdx + ydy + zdz=0

On integrating-

………… (2)

Again equation (1) can be written as-

Or

………….. (3)

From (2) and (3), the general solution is-

Non-linear partial differential equations-

Type-1: Equation of the type f(p, q) = 0

Method-

Let the required solution is-

Z = ax + by + c …….. (1)

So that-

On putting these values in f(p, q) = 0

We get-

f(a, b) = 0

So from this, find the value of b in terms of a and put the value of b in (1). It will be the required solution.

Type-2: Equation of the type-

Z = px + qy + f(p, q)

Its solution will be-

Z = ax + by + f(a, b)

Type-3: Equation of the type f(z, p, q) = 0

Type-4: Equation of the type-

Method-

Let-

Example: Solve-

Sol.

This equation can be transformed as-

………. (1)

Let

Equation (1) can be written as-

………… (2)

Let the required solution be-

From (2) we have-

Example: Solve-

Sol.

Let u = x + by

So that-

Put these values of p and q in the given equation, we get-

Example: Solve-

Sol.

Let-

That means-

Put these values of p and q in

An equation of the type-

Is called a homogeneous linear partial differential equation of n’th order with constant coefficient.

Rules for finding the complementary functions

Let us consider the equation-

Or

Step-1: Put D = m and D’ = 1

This is the auxiliary equation.

Step-2: Solve the auxiliary equation.

Case-1: If roots of the auxiliary equation are real and different, say

Then C.F.-

Case-2: If roots are equal-

Then C.F.

Rules for finding P.I.-

Given partial differential equation is-

f(D, D’)z = F(x, y)

1. When F(x, y) =

Put D = a and D’ = b

2. When F(x, y) = sin (ax + by) or cos(ax+ by)

Put

3. When F(x, y) =

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

4. When F(x,y) any function.

Resolve 1/f(D,D’) into partial fractions.

Considering f(D, D’) as a function of D alone

Here c is replaced by y + mx after integration.

Example: Solve-

Sol.

Here we have-

Auxiliary equation will be-

C.F.-

P.I.-

Here

This is the case of failure.

Now-

Example: Solve-

Sol.

The given equations can be written in the form of-

Its auxiliary equation is-

We get-

Put D = 1, D’ = 2, -

Therefore the complete solution is-

Example: Solve-

Sol.

This equation can be written as-

Writing D = m and D’ = 1, then the auxiliary equation will be-

m = -3, 2

C.F.-

Hence the complete solution is-

Example: Solve-

Sol.

The given partial differential equation can be written in the form-

A.E.-

Put y = c + 3x

Hence the complete solution is-