Unit - 2

First order partial differential equations

2.1 Non – Linear partial differential Equations of Type I ƒ(p, q) = 0, Type II ƒ(p,q,z)=0, Type III ƒ2 (p, x)= ƒ2 (q,y)

A partial differential equation (PDE)is an equation involving one or more partial derivatives of an (unknown) function, call it u, that depends on two or more variables, often time t and one or several variables in space. The order of the highest derivative is called the order of the PDE. Just as was the case for ODEs, second-order PDEs will be the most important ones in applications.

Just as for ordinary differential equations (ODEs) we say that a PDE is linear if it is of the first degree in the unknown function u and its partial derivatives. Otherwise we call it nonlinear.

The standard methods of solving the differential equations of the following

Types:

(i)                Equations solvable by separation of the variables.

(ii)              Homogeneous equations.

(iii)           Linear equations of the first order.

(iv)            Exact differential equations.

The differential equation of first order and first degree is namely:

1. and
2. 

Example-1. Solve

Solution. We have,

Separating the variables we get

(sin y + y cos y )dy ={ x (2 log x +1} dx

Integrating both the sides we get

Example 2. Solve the differential equation

Solution.

Put,

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-

z/x = a,   z/y = b

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.

Example: Solve

Sol:

We have

Let z = ax + by + c  ... (2)

On substituting the values of p and q in (1), we have

Putting the value of b in (2), we get

Type-2: Equation of the type-

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

Its solution will be-

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

Example: Solve

Sol:

Its solution

Type-3: Equation of the type f(z, p, q) = 0 equations not containing x and y.

Let z be a function of u where

u = x + ay.

u/x = 1                  and                          u/y = a

Then

p = z/x = dz/du . u/x = dz/du

q = z/y = dz/du . u/y = dz/du (a)

On putting the values of p and q in the given equation f (z, p, q) = 0, it becomes

f(z, dz/dy, a dz/du) = 0

Which is an ordinary differential equation of the first order.

Example: Solve

Sol:

Let u = x + by

So that

Substituting these values of p and q in the given equation, we have

Type-4: Equation of the type-

f1 (x, p) = f2 (y, q)

Method-

Let-

f1(x, p) = f2(y, q) = a

f1(x, p) = a, solve it for p.    Let p = F1(x)

f2(y, q) = a,  solve it for q. Let q = F2(y)

Dz = z/x  dx + z/y  dy     dz = p dx + q dy

Dz = F1(x) dx + F2(y) dy           z =

Example: Solve-

x2p2 + y2q2 = z2

Sol.

This equation can be transformed as-

x2/z2  p2 + y2/z2  q2 = 1

(x/z . z/x)2 + (y/Z . z/y)2 = 1

Let

z/z = Z,   x/x = X,     y/y = Y

Log z = Z,         log x = X,            log y = Y

Equation (1) can be written as-

P2 + Q2 = 1        ……….(2)

Let the required solution be-

Z = aX + bY + c

P = Z/X = a,        Q = Z/Y = b

From (2) we have-

a2 + b2 = 1    or  b =

Z = aX + Y + c

Log z = a log x + log y + c

Example: Solve-

Sol.

Let-

That means-

p = x2 + c            and            q = c – y2

Put these values of p and q in

Dz = pdx + qdy = (x2 + c) dx + (c – y2) dy

z = (x3/x  + cx) + (cy – y3/3)  + c1

2.2 Linear partial differential equation Lagrange’s method

Linear Equations of the First Order

A linear partial differential equation of the first order, commonly known as Lagrange’s Linear equation is of the form

Pp + Qq = R              (1)

Where, P, Q and R are functions of x, y, z. This equation is called a quast linear equation. When P, Q and R are independent of z it is known as linear equation.

Such asn equation is obtained by eliminating an arbitrary function from

Where u,v are are some functions of x, y, z.

Differentiating (2) partially with respect to x and y

Eliminating and , we get

Which simplifies to

This is of the same form as (1)

Now suppose u = a and v=b, where a, b are constants, so that

By cross multiplication we have,

The solution of these equations are u = a and v = b

Therefore, is the required solution of (1).

Thus to solve the equation Pp + Qq =R.

(i)form the subsidiary equations

(ii) Solve these simultaneous equations

(iii) write the complete solution as or u=f(v)

Example. Solve    (Kottayam, 2005)

Solution. Rewriting the given equation as

The subsidiary equations are

The first two fractions give

Integrating we get    n    (i)

Again the first and third fraction give xdx = zdz

Integrating, we get

Hence from (i) and (ii), the complete solution is

Example. Solve

Solution. Here the subsidiary equations are

Using multipliers x,y, and z we get each fraction =

which on integration gives

Again using multipliers l, m and n we get each fraction

which on integration gives lx +my +nz = b     (ii)

Hence from (i) and (ii) the required solution is

Example. Solve

Solution. Here the subsidiary equations are

From the last two fractions, we have

Which on integration gives log y = log z + log a or y/z=a     (i)

Using multipliers x, y and z we have

Each fraction

Which on integration gives

Hence from (i) and (ii) the required solution is

Example: Solve (

Sol:

Here we have

(

The auxiliary equations are

Or

Integrating first members of (2), we have

Log (x – y) = log (y –z) + log c1

Similarly from last two members of (2), we have

The required solution is

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

x2/2 = - y2/2 + C1               x2 + y2 = C1                …………….(2)

Now consider last two equations-

On integrating we get-

y2 = yz + C2

y2 –yz = C2                 ……………(3)

From equation (2) and (3)-

x2 + y2 = f(y2 – yz)

Example: Find the general solution of-

x(z2 – y2) z/x + y(x2 – z2) z/y = z(y2 – x2)

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-

x2 + y2 + z2 = C1           ………… (2)

Again equation (1) can be written as-

Or

               dx/x  + dy/y + dz/z  = 0

                   log x + log y + log z = log C2

                   log xyz = log C2

                         xyz = C2              …………(3)                 

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

Xyz = f(x2 + y2 + z2)

2.3 Solution of partial differential equation by method of separation of variables

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

References:

1. E. Kreyszig, “Advanced Engineering Mathematics”, John Wiley & Sons, 2006.
2. P. G. Hoel, S. C. Port And C. J. Stone, “Introduction To Probability Theory”, Universal Book Stall, 2003.
3. S. Ross, “A First Course in Probability”, Pearson Education India, 2002.
4. W. Feller, “An Introduction To Probability Theory and Its Applications”, Vol. 1, Wiley, 1968.
5. N.P. Bali and M. Goyal, “A Text Book of Engineering Mathematics”, Laxmi Publications, 2010.
6. B.S. Grewal, “Higher Engineering Mathematics”, Khanna Publishers, 2000.
7. T. Veerarajan, “Engineering Mathematics”, Tata Mcgraw-Hill, New Delhi, 2010