8. Partial Differential Equations – Differential Equations

8

Partial Differential Equations

8.1 INTRODUCTION

The reader is familiar with ordinary differential equations. These equations involve functions of a single variable only and their derivatives. In many problems that arise in geometry, physics, population dynamics, social sciences, medicine and engineering, one has to deal with equations containing functions of more than one variable and partial derivatives with respect to these independent variables. Such equations are called partial differential equations. Thus, a partial differential equation is an equation of the form

 

 

containing independent variables t, x, y, …, an unknown function z = z (x, y, …, t) and partial derivatives with respect to these variables t, x, y, …

8.2 ORDER, LINEARITY AND HOMOGENEITY OF A PARTIAL DIFFERENTIAL EQUATION

8.2.1 Order

The order of a partial differential equation is the order of the highest derivative appearing in it.

8.2.2 Linearity

As in the case of an ordinary differential equation, we say that a partial differential equation is linear if it is of the first degree in the dependent variable (the unknown function) and its partial derivatives and are not multiplied together.

8.2.3 Homogeneity

A linear partial differential equation is called homogeneous if it contains no term free from the unknown function and its derivatives; otherwise, it is called a non-homogeneous equation.

The following examples of partial differential equations with their order and nature noted against each of them are meant to illustrate the points explained above.

 

8.3 ORIGIN OF PARTIAL DIFFERENTIAL EQUATION
  1. Consider the equation

     

     

    where c and a are arbitrary constants. It represents the set of all spheres with their centres on the z-axis.

    Differentiating with respect to x and y we get

     

     

     

    Eliminating the arbitrary constant c from (8.3) and (8.4) we obtain the first-order linear partial differential equation

     

     

    which characterizes the set of all spheres with centres on the z-axis.

  2. Consider again the equation

     

     

    where c and α are arbitrary constants. Equation (8.6) represents the set of all right circular cones whose axes coincide with the z-axis.

    Differentiating (8.6) partially with respect to x and y we get

     

     

    respectively. Eliminating the constants c and α between these equations we again obtain Eq. (8.5).

  3. The spheres and cones are surfaces of revolution which have the line OZ as their axis of symmetry. Now, consider the equation

     

     

    where f is an arbitrary function. Equation (8.9) characterizes surfaces of revolution having z-axis as the axis of symmetry.

    Differentiating Eq. (8.9) partially with respect to x and y we get

     

     

    where u = x2 + y2 and Eliminating f between equations (8.10) and (8.11), we again obtain the first-order linear partial differential equation (8.5).

Formation of PDE Ordinary differential equations are formed by eliminating arbitrary constants only, whereas partial differential equations are formed by eliminating (a) arbitrary constants or (b) arbitrary functions.

We know that the order of an ordinary differential equation is equal to the number of arbitrary constants to be eliminated from a relation.

In the case of partial differential equations, if the number of arbitrary constants to be eliminated is equal to the number of independent variables involved in a relation, one obtains a first-order partial differential equation, and if the number of arbitrary constants to be eliminated is more than the number of independent variables, then one obtains a higher-order partial differential equation.

Further, if one arbitrary function is to be eliminated from a relation, then a first-order partial differential equation is obtained; and if two arbitrary functions are to be eliminated, then a second-order partial differential equation is obtained and so on.

A first-order partial differential equation is of the form

 

 

where x, y are independent variables, z is a dependent variable and and

8.4 FORMATION OF PARTIAL DIFFERENTIAL EQUATION BY ELIMINATION OF TWO ARBITRARY CONSTANTS

Consider a relation of the type

 

 

where a and b are arbitrary constants.

Differentiating (8.13) partially with respect to x and y we have

 

 

We can now eliminate the two arbitrary constants a and b between the Eqs. (8.13)(8.15) and obtain a first order partial differential equation of the form

 

 

We now work out a few examples to show how partial differential equations are formed by eliminating arbitrary constants.

Example 8.4.1   Show by eliminating the arbitrary constants a and b from

 

 

the partial differential equation

 

 

is obtained, which is non-linear.

Solution   Differentiating (8.17) partially with respect to x and y

 

 

Eliminating a and b from (8.17), (8.19) and (8.20) we get

 

z2(p2 + q2 +1) = 1

 

which is non-linear.

 

Examples

Form the partial differential equation by eliminating the arbitrary constants a and b from each of the relations given below:

Example 8.4.2

z = ax + by.

 

Solution   We have

 

 

Differentiating (8.21) partially with respect to x and y we have

 

 

Eliminating a and b from Eqs. (8.21)(8.23) we obtain the first-order partial differential equation

 

 

Example 8.4.3

 

z = (x2 + a2)(y2 + b2).

 

Solution   We have

 

 

Differentiating (8.25) partially with respect to x and y

 

 

 

Eliminating a and b from (8.25)(8.27) we get

 

Example 8.4.4

 

 

Solution   We have

 

 

Differentiating (8.29) partially with respect to x and y

 

 

Eliminating a and b from (8.29)(8.31) we get

 

 

Example 8.4.5

 

z = (x + a)(y + b).

 

Solution   We have

 

 

Differentiating (8.33) partially with respect to x and y

 

 

Eliminating a and b from (8.33)(8.35) we have

 

Example 8.4.6

 

z = (ax + y)2 + b.

 

Solution   We have

 

 

Differentiating (8.37) partially with respect to x and y

 

 

Eliminating a and b from (8.37)(8.39) we get

 

 

which is non-linear.

Example 8.4.7

 

ax2 + by2 + z2 = 1.

 

Solution   We have

 

 

Differentiating (8.41) partially with respect to x and y

 

 

Multiplying (8.42) by x and (8.43) by y and adding

 

EXERCISE 8.1

Form the partial differential equation in each of the following cases, by eliminating the arbitrary constants a and b (Questions 1–10):

  1. z = ax2 + by2.

    Ans: 2z = px + qy

  2. (xa)2 + (yb)2 = z2 cot2 α.

    Ans: p2 + q2 = tan2 α

  3. z = ax + by + a2 + b2.

    Ans: z = px + qy + p2 + q2

  4. z = axy + b.

    Ans: px = qy

  5. [JNTU 2003]

    Ans: 16(p2 + q2)(x + y) = 1

  6. u = a(x + y) + b(xy) + abz + c.

    Ans:

  7. Ans: px + qy = pq

  8. Ans: q = px + p2

  9. [JNTU 2003]

    Ans:

  10. [JNTU 2003]

    Ans: px + qy = p + q

  11. Form the partial differential equation of all spheres of radius a with their centres on the x–y plane.

    Ans: z2(p2 + q2 + 1) = a2

  12. Form the partial differential equation of all planes through the origin.

    Ans: z = px + qy

8.5 FORMATION OF PARTIAL DIFFERENTIAL EQUATIONS BY ELIMINATION OF ARBITRARY FUNCTIONS

(a) Elimination of one arbitrary function of the form z = f (u) where u = u(x, y, z)

 

 

where f (u) is an arbitrary function of u where u = u(x, y, z) a known function of x, y and z.

Differentiating (8.45) partially with respect to x and y

 

 

where ′ on f denotes differentiation with respect to the argument u. Eliminating f from Eqs. (8.46)–(8.47) we obtain a first-order partial differential equation.

(b) Elimination of two arbitrary functions of the form z = f (x) g ( y)

Let

 

 

where f is a function of x and g is a function of y alone.

Differentiating (8.48) partially with respect to x and y, respectively, we have

 

 

where

 

(c) Formation of PDE by elimination of arbitrary function of the form F (u, v) = 0

Consider a relation between x, y and z of the type

 

 

where u and v are known functions of x, y and z; and F is an arbitrary function of u and v. Also, z is a function of x and y.

Differentiating (8.50) by chain rule, with respect to x and y we obtain

 

 

 

Eliminating F between (8.51) and (8.52) we get

 

 

where and so on.

Expanding the determinant in (8.53), collecting the terms and simplifying we have

 

 

where

which are functions of x, y and z and do not contain p and q. This equation is called Lagrange's1 linear equation. In fact, it is a quasi-linear equation since the dependent variable may be present in P, Q and R.

If P and Q are independent of z and R is linear in z the equation (8.54) is called linear.

Equation (8.54) is of the type given by Eq. (8.12), which need not be linear, as Example 8.4.1 above will illustrate.

 

Examples

Form the partial differential equation by eliminating the arbitrary function f from each of the following relations [Examples 8.5.1–8.5.8 except (Example 8.5.6) in which functions f and g have to be eliminated.]

Example 8.5.1

 

z = xf (x + y).

 

Solution   We have

 

 

Differentiating (8.55) partially with respect to x and y

 

 

Eliminating f between (8.55)–(8.57) we have

 

 

which is the required partial differential equation.

Example 8.5.2

 

z = xy + f (xy).

 

Solution   We have

 

 

Differentiating (8.59) partially with respect to x and y

 

 

Eliminating f′ between (8.59) and (8.60) we get the partial differential equation

 

Example 8.5.3

 

z = eaxby f (ax + by).

 

Solution   We have

 

 

Differentiating (8.63) partially with respect to x and y

 

 

Eliminating f between (8.63)–(8.65) we have

 

Example 8.5.4

 

 

Solution   We have

 

 

Differentiating (8.67) partially with respect to x and y

 

 

We have from (8.68) and (8.69)

 

Example 8.5.5

 

 

Solution   We have

 

 

Differentiating (8.71) with respect to x and y

 

 

Eliminating f′ between (8.72) and (8.73) we have

 

Example 8.5.6

 

z = f (x + at) + g(xat).       [Osmania 1999]

 

Solution   We have

 

 

Differentiating (8.75) partially with respect to x and t twice

 

 

 

Eliminating the arbitrary functions from (8.77) we have

 

 

which is a partial differential equation of the second order.

Example 8.5.7

 

 

Solution   We have

 

 

Differentiating (8.79) with respect to x and y

 

 

Eliminating f′ between (8.80) and (8.81) we get

 

 

which is the required partial differential equation.

 

Example 8.5.8

 

f(xy + z2, x + y + z) = 0.

[Madras 1995, Kerala 1990 S]

Solution   We have

 

 

where

 

 

We know that the partial differential equation is

 

 

where

 

 

Hence, the required partial differential equation is

 

EXERCISE 8.2

Form the partial differential equations by eliminating the arbitrary functions from the following relations:

  1. Ans: px = qy

  2. xyz = f(x + y + z).       [Osmania 1995, Calicut 1994]

    Ans: x(yz)p + y(zx)q = z(xy)

  3. x + y + z = f(x2 + y2 + z2).           [Madras 2000, Ranchi 1990]

    Ans: (yz)p + (zx)q = xy

  4. f(x2+ y2, zxy) = 0.             [Madurai 1990]

    Ans: pyqx = y2x2

  5. z = f(x2y2).                  [Madras 1991]

    Ans: py + qx = 0

  6. z = yf(x) + xg(y).             [Madras 1993, Karnataka 1993, Madurai 1998]

    Ans:

  7. z = f(y) + φ(x + y).              [JNTU 2005 S (1)]

    Ans:

  8. z = xy + f(x2 + y2).          [JNTU 2005 S (2)]

    Ans: pyqx = y2x2

8.6 CLASSIFICATION OF FIRST-ORDER PARTIAL DIFFERENTIAL EQUATIONS

8.6.1 Linear Equation

Let z = z(x, y) be a function of two variables x and y. A first-order partial differential equation is said to be linear if it is linear in p, q and z, i.e. if it is of the form

 

P(x, y)p + Q(x, y)q = R(x, y)z + S(x, y)

 

where

 

 

8.6.2 Semi-Linear Equation

A first-order partial differential equation is said to be semi-linear if it is linear in p and q and the coefficients of p and q are functions of x and y alone, i.e. if it is of the form

 

 

8.6.3 Quasi-Linear Equation

A first-order partial differential equation is said to be quasi-linear if it is linear in p and q, i.e. if it is of the form

 

 

 

Note that every semi-linear partial differential equation is quasi-linear.

8.6.4 Non-linear Equation

A first-order partial differential equation is said to be non-linear if it does not fall into any one of the above types.

8.7 CLASSIFICATION OF SOLUTIONS OF FIRST-ORDER PARTIAL DIFFERENTIAL EQUATION

Consider a first-order partial differential equation of the form

 

 

Let

 

 

be a continuously differentiable function of x and y in a region D.

Compute and from (8.94) and substitute in (8.93). If this reduces (8.93) into an identity in x and y then (8.94) defines a solution of partial differential equation (8.93).

Equation (8.94) defines a surface in three-dimensional space. It is called an integral surface of partial differential equation (8.93).

There are different types of solutions of Eq. (8.93).

8.7.1 Complete Integral

A two-parameter family of solutions

 

 

is called a complete integral (or complete solution) of (8.93) if in the region D the matrix

 

 

is of rank two.

8.7.2 General Integral

Let a and b be related and suppose

 

 

Substituting this in (8.95) we get

 

 

which is a one-parameter family of solutions of (8.93).

This is a sub-family of the two-parameter family given by (8.95). The envelope of (8.98), if it exists, is obtained by eliminating a between (8.98) and

 

 

If (8.99) can be solved for a then

 

 

Substituting for a in (8.98) we obtain an integral surface as

 

 

If φ is arbitrary then (8.101) is called a general integral (or general solution) of the partial differential equation (8.93).

8.7.3 Particular Integral

When a particular function φ is used in (8.101) we obtain a particular integral or particular solution of the partial differential equation (8.93).

8.7.4 Singular Integral

In some cases we find another integral which satisfies the partial differential equation (8.93) but is not a particular integral of (8.93). If it exists, it is obtained by eliminating a and b from

 

 

and it is the envelope of the two-parameter family of surfaces

 

 

and it is called a singular integral (or a singular solution).

In a sense, a general integral provides a much broader class of solutions of the partial differential equation than does a complete integral.

However, it is possible to derive a general integral when a complete integral is known.

Note that for the partial differential equation

the relation

 

 

 

is a complete integral while the relation

 

 

is a general integral.

8.8 EQUATIONS SOLVABLE BY DIRECT INTEGRATION

We now consider partial differential equations which can be solved by direct integration. While carrying out integration with respect to a variable the other variable is held fixed. So, in place of constant of integration we have to add an arbitrary function of the variable held fixed.

 

Example 8.8.1   Solve

Solution   The given partial differential equation is

 

 

Integrating once partially with respect to x we get

 

 

and integrating (8.110) partially with respect to x we get the solution as

 

 

where f and g are arbitrary functions of y.

Example 8.8.2   Solve

Solution   The given partial differential equation is

 

 

Integrating (8.112) partially with respect to x

 

 

where f(y) is an arbitrary function of y. Integrating (8.113) again partially with respect to y

 

 

where and h(x) are arbitrary functions of y and x, respectively.

Example 8.8.3   Solve

Solution   The given partial differential equation is

 

 

Integrating (8.115) partially with respect to y we get

 

 

where f(x) is an arbitrary function of x. Integrating (8.116) again partially with respect to x we get

 

 

which is the solution of (8.115). We can write (8.117) as

 

 

where h(x) and g(y) are arbitrary functions of x and y, respectively.

Example 8.8.4   Solve

Solution   The given partial differential equation is

 

 

Integrating (8.119) partially with respect to x

 

 

where f(y) is an arbitrary function of y. Integrating (8.120) partially with respect to x again

 

 

where g is an arbitrary function of y. Integrating (8.121) partially with respect to y,

 

 

which can be put in the form

 

 

where f1(x) is an arbitrary function of x and f2(y) and f3(y) are arbitrary functions of y.

Example 8.8.5   Solve

when x = 0.

Solution   The given partial differential equation is

 

 

We have to solve (8.123) under the conditions:

When x = 0,

 

 

If we treat z as a function of x alone, the solution of (8.123) is

 

 

where A and B are constants. Since z is a function of y also we can take z as

 

z = A(y) sinh x + B(y) cosh x

 

 

∴      A( y) = a sin y and B(y) = constant.

 

Hence     z = (a sin y) sinh x + b cosh x,

 

where b is an arbitrary constant.

EXERCISE 8.3
  1. Solve given that when x = 0 and z = 0 when y is an odd multiple of π/2.

    [Madras 1994 S, Mysore 1999 S]

    Ans: z = (1 + cos x) cos y

  2. Solve

    Ans:

  3. Solve given that when x = 0, z = e y and

    [Mysore 1987 S, Madras 1993, Karnataka 1994]

  4. Solve

    [Madurai 1988, Mysore 1987]

    Ans: u = −et sin x + φ (x) + ψ (t)

  5. Solve

    Ans: z = ex· ey + ∫ f(y)dy + φ (x)

  6. Solve

    Ans:

8.9 QUASI-LINEAR EQUATIONS OF FIRST ORDER

Quasi-linear partial differential equations of first order can be written as

 

 

where P, Q, R are functions of x, y and z. They do not involve p or q. Equation (8.127) is called Lagrange's equation. Here, linear means that p and q appear to the first degree only. This is in contrast to the situation in ordinary differential equations where z must also be of first degree.

Note that partial differential equation

 

 

is linear while the ordinary differential equation

 

 

is non-linear.

Theorem 8.9.1 The general solution of the quasi-linear partial differential equation

is

 

 

 

where F is an arbitrary function and u(x, y, z) = c1 and v(x, y, z) = c2 form a solution of the equations

 

 

Proof If

 

 

satisfy Eq. (8.132) then

 

 

 

must be compatible with (8.132) so that we must have

 

 

From (8.136) and (8.137) we have

 

 

Now, differentiating (8.131) with respect to x and y

 

 

 

Eliminating and from (8.139) and (8.140) we get

 

where are functions of x, y, z.

Substituting from (8.138) into (8.141) we see that (8.131) is a solution of (8.127) if u and v are given by (8.133).

8.10 SOLUTION OF LINEAR, SEMI-LINEAR AND QUASI-LINEAR EQUATIONS

If the equation is linear or semi-linear or can be written in these forms then we can solve them as follows.

8.10.1 All the Variables are Separable

Example 8.10.1   Solve zx2p + zy2q = 1 − z2.

Solution   The equation can be thrown into the form

 

 

which is a semi-linear equation.

Lagrange's auxiliary equations are

 

 

Integrating the first and last equations we have

 

 

 

General solution is

 

8.10.2 Two Variables are Separable

Example 8.10.2   Solve mplq = z tan(lx + my).

Solution   This is a semi-linear equation Lagrange's auxiliary equations are

 

 

From the first equation we have

 

 

Using this in the last equation

 

Integrating we have

 

 

General solution is

 

Example 8.10.3   Solve ypxq = xyz + xy.

Solution   The given equation is linear. Lagrange's auxiliary equations are

 

 

From the first equation we have

 

 

Integrating we get

 

 

From the last equation

 

 

Integrating,

 

 

General solution is

 

8.10.3 Method of Multipliers

In addition to the above methods, we can apply the following method called method of multipliers.

Example 8.10.4   Solve x2(yz)p + y2(zx)q = z2(xy).

Solution   Lagrange's auxiliary equations are

 

Using multipliers in turn we get

 

 

 

 

General solution is

 

Example 8.10.5   Solve zxp + zyq = 1+ z2.

Solution   Lagrange's auxiliary equations are

 

 

Taking the first two ratios,

 

 

 

Taking the last two ratios,

 

 

General solution is

 

Example 8.10.6   Solve yp + xq = (x2y2 + z2)x.

Solution   Lagrange's auxiliary equations are

 

 

From the first two ratios we get

x dxy dy = 0 ⇒ x2y2 = a2 (a2 is an arbitrary constant)

Considering the last two ratios we have

 

 

Integrating,

 

 

where b is an arbitrary constant.

General solution is

 

Example 8.10.7   Solve yp + xq =(x + y)z.

Solution   Lagrange's auxiliary equations are

 

 

From the first two ratios

 

 

Again, each ratio

 

 

On integration we have

 

 

General solution is

 

Example 8.10.8   Solve xpyq = xyz.

Solution   Lagrange's auxiliary equations are

 

 

From the first two ratios

 

 

From the last two ratios

 

 

Integrating

 

 

General solution is

 

Example 8.10.9   Solve zp + yq = x.

Solution   Lagrange's auxiliary equations are

 

 

From the first and last ratios

 

 

Integrating we get

 

Each of the ratios (8.180)

Integrating,

 

 

General solution is

 

Example 8.10.10   Solve (yz)p + (zx)q = (xy).

Solution   Lagrange's auxiliary equations are

 

 

Taking (1, 1, 1) as Lagrange's multipliers, each ratio

 

 

Taking (x, y, z) as Lagrange's multipliers, each ratio

 

 

General solution is

 

Example 8.10.11   Solve (sec x)p + (sin xy sec x tan x)q = (a2z2).

Solution   Lagrange's auxiliary equations are

 

 

From the first and last ratios

 

 

On integration, we get

 

 

From the first two ratios

 

 

 

Integrating we have

 

 

General solution is

 

Example 8.10.12   Solve x(1 − xy)py(1 + xy)q = z(1 − xy).

Solution   Lagrange's auxiliary equations are

 

 

From the first and last ratios cancelling (1 - xy).

Integrating

 

 

On cross-multiplication the first two ratios yield

 

 

On integration we get

 

 

General solution is

 

Example 8.10.13   Solve (x2y2z2)p + 2xyq = 2xz.

[Bangalore 1990, Gorakhpur 1991, Andhra 1989]

Solution   Lagrange's auxiliary equations are

 

Considering the last two ratios, we have On integration, we get

 

 

Using x, y, z as multipliers, we get that each ratio

 

 

 

This gives, on integration,

 

 

From (8.197) and (8.198), the general solution is

 

Example 8.10.14   Solve (z2 − 2yzy2)p + (xy + zx)q = xyzx.

[Bangalore 1990, Madras 1997 S]

Solution   Lagrange's auxiliary equations are

 

 

Using x, y, z as multipliers, each ratio

 

 

Considering the last two ratios, we have

 

 

On integrating and simplifying

 

 

The general solution

 

 

Example 8.10.15   Solve xnp + ynq = zn.

Solution   Lagrange's auxiliary equations

 

 

On integration we get two solutions

 

 

 

General solution when

 

 

when n ≠ 1 is

 

Example 8.10.16   Solve x(yz)p + y(zx)q = z(xy).

[JNTU 2002, 2003]

Solution   Lagrange's auxiliary equations are

 

 

Now, each ratio

 

 

Also each ratio

 

 

General solution is

 

Example 8.10.17   Solve (x2yz)p +(y2zx)q = z2xy.

[AMIE 1997, Madras 1994 S, 1998; Bhopal 1991, Karnataka 1990, Ranchi 1996]

Solution   Langrange's auxiliary equations are

 

 

Each ratio

 

 

 

Each ratio

 

 

 

From (8.216) and (8.217) we can write the general solution as

 

Example 8.10.18   Solve (yz)p + (xy)q = zx.

[Punjab 1987 S]

Solution   Lagrange's auxiliary equations are

 

 

 

 

From (8.220) and (8.221) we write the general solution as

 

Example 8.10.19   Solve xp + (2xy)q = (1 − z).

Solution   Lagrange's auxiliary equations are

 

From the first and last ratios

Integrating we get

 

From the first two ratios

We have, on cross-multiplication,

 

 

Integrating

 

 

General solution is

 

 

Example 8.10.20   Solve (x + 2y2)p + yq = − cos z.

Solution   Lagrange's auxiliary equations are

 

From the last two ratios

Integrating we get

 

 

From the first two ratios

On cross-multiplication and transposition

 

 

Integrating, we get

 

 

General solution is

 

EXERCISE 8.4
  1. px + qy = z.

    Ans:

  2. px2 + qy2 = z2.

    Ans:

  3. Ans: F(x2y2, x2z2) = 0

  4. Ans: F(x3y3, x2z2) = 0

  5. z(xpyq) = y2x2.

    Ans: F(x2 + y2 + z2, xy) = 0

  6. px(z − 2y2) = (zqy)(zy2 − 2x3).

    Ans:

  7. x(x + y)py(x + y)q = −(xy)(2x +2y + z).

    Ans: F(xy, (x + y + z)(x + y)) = 0

  8. x2(yz)p + y2(zx)q = z2(xy).          [Madurai 1990]

    Ans: F(xyz, (x−1 + y−1 + z−1) = 0

  9. x(y2z2)p + y(z2x2)q = z(xy2).

    Ans: F(xyz, (x2 + y2 + z2)) = 0

  10. (mzny)p +(nxlz)q = lymx.          [AMIE 1990, Madras 1994 S]

    Ans: F(lx + my + nx, x2 + y2 + z2) = 0

  11. (yzx)p +(x + yz)q = x2 + y2.

    Ans: F(x2y2 + z2, xyz) = 0

  12. (bc)a−1yzp + (ca)b−1zxq = (ab)c−1xy.

    Ans: F(ax2 + by2 + cz2, a2x2 + b2y2 + c2z2) = 0

  13. (y2 + z2x2)p − 2xyq = −2zx.

    Ans:

  14. p tan x + q tan y = tan z.              [Andhra 1990, Kerala 1987 S]

    Ans:

  15. xeyp + (1 − ey)q = zey.

    Ans:

8.11 NON-LINEAR EQUATIONS OF FIRST ORDER

The most general partial differential equation of the first order in two independent variables is of the form

 

 

which may not be linear.

We do not propose to study any general method of integration of Eq. (8.231) but consider only four standard forms which admit integration by very short processes. Also, many equations can be reduced to one or the other of these four forms.

The general integral, the singular integral and the complete integral must be indicated in each case. Otherwise, the equation is not considered fully solved.

Standard Form I: pq-equation: Equation of the form f (p, q) = 0 where x, y, z do not occur explicitly.

To solve such an equation we put p = a and find the value of q in terms of a, i.e. q = f (a). Then we substitute these values in

 

 

so that we have dz = a dx + f (a) dy.

Integrating, we get the complete integral as

 

 

where a and c are arbitrary constants.

Note 8.11.1 We can put q = a instead of p = a and proceed to obtain the complete integral.

Example 8.11.2   Solve pq = k.

Solution   The complete integral is

 

 

where a and c are arbitrary constants

The general integral is obtained by eliminating a between the equations

 

 

where c has been replaced by φ (a) and

 

 

obtained by differentiating with respect to a.

The singular integral, if it exists, is determined from the Eq. (8.234)

 

 

 

obtained from (8.234) on differentiating it with respect to a and c, respectively.

The inconsistency of the last equation shows that singular integral does not exist in this case.

Example 8.11.3   Solve pq = xα yβ zγ.

Solution   The given equation is

 

 

We can write (8.239) as

 

 

Case (i) α ≠ 1, β ≠ −1, γ ≠ −2

 

 

 

Equation (8.240) now becomes,

 

 

whose complete integral is

 

Z = aX + bY + c where ab = 1

 

 

Case (ii) α = −1, β = −1, γ = −2, we have

 

 

 

The CI is

 

 

 

To find the General Integral

Case (i) Writing c = φ (a) in the complete integral at (8.242) we have

 

 

Differentiating (8.244) with respect to a we get

 

 

The general integral is the eliminant of a between (8.244) and (8.245).

Case (ii) Writing c = φ (a) in the complete Integral at (8.243)

 

 

Differentiating (8.246) with respect to a

 

 

Eliminant of a between (8.246) and (8.247) is the general integral.

To Find the Singular Integral

Case (i) The singular integral, if it exists, is obtained from the following equations:

 

 

Differentiating partially with respect to a and c, respectively, we get

 

 

The inconsistency in the last equation shows that the singular integral does not exist in this case.

Case (ii) We can show that in this case also singular integral does not exist.

Example 8.11.4   Find a complete integral of

 

 

Solution   Put p = a, then and the equation dz = p dx + q dy becomes

 

 

whose solution is

 

 

which is the required complete integral.

Standard Form II: zpq-equation: Equation of the form f (z, p, q) = 0 where the independent variables do not appear explicitly.

In this case put

 

 

The given equation is f (z, p, q) = 0 or solving for p we have

 

 

Substituting in

 

 

Example 8.11.5   Solve

 

 

Solution   Put q = ap in the given partial differential equation

 

 

 

Now, dz = p dx + q dy = p dx + ap dy = p d(x + ay)

 

 

Example 8.11.6   Solve

 

 

Solution   Putting q = ap in (8.262) we get

 

 

 

Standard Form III: Separable equation f (x, p) = g ( y, q).

We can take

 

 

Solving for p and q we get

 

 

Substituting in

 

 

Integrating we get the CI as

 

Example 8.11.7   Solve p + q + x + y.

Solution   We can write the equation as

 

 

so that

 

 

Substituting in

 

 

we have

 

 

Integrating after multiplying throughout by 2, the complete integral is

 

 

where c is an arbitrary constant.

Example 8.11.8   Solve zpq = xy.

Solution   The equation can be written as

 

 

 

 

 

Substituting in the equation

 

We have which on integrating yields,

 

where c is an arbitrary constant.

The general integral and singular integral (if it exists) is obtained as explained in Section 8.7.

Standard Form IV: Clairaut's2 Equation z = px + qy + f (p, q).

A partial differential equation of the type

 

 

is called a Clairaut's equation.

Complete Integral A complete integral of an equation of type (8.282) is obtained by replacing p and q by arbitrary constants a and b, respectively. Thus, a complete integral of Eq. (8.282) is

 

 

One can readily verify that (8.283) is a solution of Eq. (8.282)

Also, writing (8.283) as

 

 

we observe that the matrix

 

 

is of rank two. Hence, (8.284) is indeed a complete integral of equation.

General integral and singular integral (if it exists) are obtained as explained in Section 8.7.

Singular Integral The complete integral is

 

 

Differentiating (8.286) partially with respect to a and b we get

 

 

 

respectively. Eliminating a and b between (8.287), (8.288) we have singular solution of (8.282) as

 

z + xy = 0

 

Examples

Find complete integral in each of the following cases:

Example 8.11.9

 

Solution   The given partial differential equation

 

 

is an equation of Clairaut's type.

The complete integral of (8.289) is

 

 

where a and b are arbitrary constants.

Example 8.11.10   

Solution   The given partial differential equation

 

 

is an equation of Clairaut's type.

The complete integral of this equation is

 

Example 8.11.11   2q(zpxqy) = 1+ q2.  [JNTU 2005 S]

The given partial differential equation can be written as

 

 

which is an equation of Clairaut's type.

The complete integral of this equation is

 

Example 8.11.12   Find the singular integral in Example 8.11.9.

Solution   The complete integral is

 

 

Differentiating this with respect to a and b, respectively, we get

 

 

Eliminating a and b from these equations we get the singular integral as

 

EXERCISE 8.5
  1. q2 = ep/α.

    Ans: z = ax + ea/2a y + c

  2. p + q = pq.

    Ans:

  3. p2 + q2 = 1.                              [Osmania 2000]

    Ans:

  4. x2p2 + y2q2 = z.            [Osmania 2000 s, Madras 1998]

    Ans:

  5. z = pq.

    Ans: (x + ay + c)2 = 4az

  6. zq = p2.

    Ans: z = ceax + a2y

  7. p(1 + q)qz.                                [Kerala 1990 S, Punjab 1986]

    Ans: log (az − 1) = (x + ay + c)

  8. z = p2 + q2.           [Calicut 1994, Andhra 1990, Madurai 1990]

    Ans: 4z (1 + a2) = (x + ay + c)2

  9. z(p2q2) = xy.     [Madras 1991, Calicut 1991, Karnataka 1990]

    Ans: z3/2 = (x + a)3/2 + (y + a)3/2 + c

     

  10. q = xyp2.

    Ans:

     

  11. yp + xq + pq = 0.

    [Marathwada 1993]

    Ans:

     

  12. p + q = sin x + sin y.

    [Madras 1993, Karnataka 1993]

    Ans: z = a(xy) − (cos x + cos y) + c

     

  13. z = px + qy + p2 + q2.

    Ans: z = ax + by + a2 + b2

     

  14. Ans:

     

  15. z = px + qy + sin(p + q).

    Ans: z = ax + by + sin(a + b)

     

8.12 EULER'S METHOD OF SEPARATION OF VARIABLES

When we model scientific, engineering, biotechnological and other processes there arise initial and boundary value problems involving partial differential equations. The general solutions of these partial differential equations are in the form of arbitrary functions which are not suitable for determining the exact solutions satisfying the given initial and boundary conditions.

The method of separation of variables, due to Euler3 is a simple, yet powerful technique in breaking up a partial differential equation into an ordinary differential equation, which can be easily solved using the known methods.

For a partial differential equation in the function u = u(x, y) where x and y are independent variables, we assume that the solution is separable, that is,

 

 

where X(x) is a function of x alone and Y(y) is a function of y alone. Substituting in the given partial differential equation, separating the variables and assuming each side must be equal to the same constant λ (say), we obtain two ordinary differential equations, whose solution gives the solution of the problem. The method is best illustrated through examples.

Example 8.12.1   Solve by the method of separation of variables given that u(x, 0) = 6e−3x.

Solution   We assume the solution to be

 

 

where X(x) is a function of x alone and T(t) is a function of t alone. We have

 

 

where denotes differentiation with respect to the argument.

Substituting in the given partial differential equation, we get

 

 

Dividing both sides of (8.303) by 2XT we get

 

 

 

Solving (8.304) we get

 

 

and solving (8.305) we get

 

 

Combining (8.306) and (8.307) we may write the solution as

 

 

The latter equation gives λ = −2

Substituting in the general solution we get

 

u(x,y) = 6e−3xe−2t = 6e−(3x + 2t)

 

which is the required solution.

Example 8.12.2   Solve by the method of separation of variables given that u = 3e−y − 3e−5y when x = 0.

Solution   Let

 

 

where X(x) is a function of x alone and Y(y) is a function of y alone. Calculating the derivatives a nd substituting in (8.309) we have

 

 

Thus, we have two ordinary differential equations

 

 

Taking trial solution of the form emx, emy for the above equations we obtain auxiliary equations

 

 

respectively.

The solution for X is of the form

 

 

and the solution for Y is of the form

 

 

The solution for u can be taken as

 

 

where we have used the principle of superposition of solutions since the equation is linear.

Putting x = 0 in (8.316) we get

 

 

and c3 = c4 = … = 0

Equating like terms on both sides

 

 

The required solution is

 

Example 8.12.3   Solve subject to conditions. when x = 0.

Solution   Let u = X(x)Y(y) be a solution of

 

 

Substituting in the Eq. (8.317)

 

 

We obtain the ordinary differential equations X″ − (2 + k)X = 0, Y″kY = 0, where k is a constant.

Taking the solutions as

 

 

Auxillary equations

 

 

∴ The solution is of the form

If we take the solution as

 

 

The condition x = 0 ⇒ u = 0 which gives c1 + c2 = 0,

 

 

Since eαxeαx = 2 sinh αx, and we have to satisfy another condition by the principle of superposition of solutions, we may take

 

 

 

 

Since The second solution is

 

 

Example 8.12.4   Solve

 

Solution   Let

where X is a function of x alone and Y is a function of y alone, be a solution of Eq. (8.329). Calculating the derivatives and substituting in (8.329) we have

 

 

Thus, we obtain two ordinary differential equations

 

 

where k is a constant. Taking trial solutions, X = emx, Y = emy, the auxiliary equations are

 

m2 − 2mk = 0, m + k = 0

 

 

where A and B are two arbitrary constants, consistent with the order of the differential equation.

EXERCISE 8.6

Solve the following equations by separation of variables:

  1. Ans: u(x,t) = sin xet sin x

  2. Ans: u = e2x−5y

  3. Ans:

  4. .

    Ans:

  5. .

    Ans:

  6. Find a solution of in the form u = X(x)Y (y). Solve the equation subject to the conditions u = 0 and when x = 0 for all values of y.    [Andhra 2000, Nagpur 1997]

    Ans:

8.13 CLASSIFICATION OF SECOND-ORDER PARTIAL DIFFERENTIAL EQUATIONS

8.13.1 Introduction

Many physical and engineering applications such as fluid flow, heat transfer and wave motion involve second-order partial differential equations and hence we take up a study of these equations and their solution by separation of variables method.

The general second-order partial differential equation may be written in the form

 

 

Equation (8.333) is called semilinear if A, B and C are functions of the independent variables x and y only. On the other hand, if A, B and C are functions of x, y, u, then (8.333) is called quasi-linear. If A, B and C are functions of x and y and H is a linear function of u, then Eq. (8.333) is called linear. The general second-order linear partial differential equation in two independent variables x and y may be written as

 

 

If G ≠ 0 then Eq. (8.334) is caled non-homogeneous and if G = 0 then it is called homogeneous.

8.13.2 Classification of Equations

We call the quantity ∆ = B2 – 4AC the discriminant and classify Eq. (8.334) as hyperbolic, parabolic or elliptic according as ∆ > 0, = 0 or < 0. The following are well-known examples of these three types.

Hyperbolic Type

1. One-dimensional wave equation

 

 

Here A = a2, B = 0, C = –1 and ∆ = B2 – 4AC = 4a2 > 0.

Parabolic Type

2. One-dimensional heat-flow equation

 

 

Here A = a2, B = C = 0 and ∆ = B2 – 4AC = 0.

Elliptic Type

3. (a) Two-dimensional Laplace equation

 

 

Here A = 1, B = 0, C = 1 and ∆ = B2 – 4AC = – 4 < 0.

(b) Poisson's equation

 

 

Equations (8.335)(8.337) are homogeneous while Eq. (8.338) is non-homogeneous.

8.13.3 Initial and Boundary Value Problems and their Solution

The unique solution corresponding to a particular physical problem is obtained by use of additional information arising from the physical situation. If this information is given on the boundary as boundary conditions, the resulting problem is called a boundary value problem (BVP). If this information is given at one instant as initial conditions, the resulting problem is called an initial value problem (IVP).

The principle of superposition of solutions is applicable as long as the equation is linear and homogeneous. That is, if un is a solution for each n then , which is a linear combination of the solutions {un}is also a solution of the equation.

The hyperbolic and parabolic types of equations are either initial value problems or initial and boundary value problems, whereas the elliptic-type equation is always a boundary value problem. The boundary conditions may be one of the following three types:

  1. Dirichlet problem (First boundary value problem)

    The solution is prescribed along the boundary.

  2. Neumann problem (Second boundary value problem)

    The derivative of the solution is prescribed along the boundary.

  3. Mixed problem (Third boundary value problem)

    The solution and its derivative are prescribed along the boundary.

Any of the above conditions is called homogeneous if it is a zero-condition and non-homogeneous if it is a non-zero condition.

8.13.4 Solution of One-dimensional Heat Equation (or diffusion equation)

Consider a long and thin wire, rod or bar OA of length l and of constant cross-section and homogeneous heat-conducting material. Let the bar be placed along the x-axis with one end O coinciding with the origin and the other end A at a distance l from O (Figure 8.1).

 

 

Figure 8.1 Heat conduction along a bar

 

Suppose that the lateral surface of the bar is perfectly insulated, so that heat flow is along the x-direction only. Therefore, the temperature u of the bar depends on x and t only. The initial boundary value problem consists of one-dimensional heat equation:

 

 

where a2 is the thermal diffusivity. The boundary conditions at the ends O and A are

 

 

respectively. The initial temperature distribution in the bar is

 

 

where f(x) is a given function of x. The solution by the method of separation of variables reduces the initial boundary value problem (IBVP) to that of two ordinary differential equations.

Assume that the solution u (x, t) is separable

 

 

where X(x) is a function of x only and T(t) is a function of t only. Differentiating (8.343) with respect to t and x we get

 

 

where′ denotes differentiation with respect to the corresponding independent variable. Substituting into (8.343) we have

 

 

Since the L.H. member is a function of x only and the R.H. member is a function of t only, both sides must be equal to the same constant λ, say. So, we obtain two ordinary differential equations

 

 

Three cases arise: Case 1. λ > 0;    Case 2. λ = 0; Case 3. λ < 0.

Case 1 λ > 0. The general solution is

 

 

Equation (8.348) shows that the solution has unbounded temperature for large t due to exponential growth, which is not physically possible.

Case 2 λ = 0. The general solution is

 

 

which is independent of time. This is also not possible.

Case 3 λ < 0. We write λ = −p2 where p is real. The general solution in this case is

 

 

The boundary condition u (0, t) = 0 implies that A = 0 so that we have

 

 

The boundary condition u (l, t) = 0 requires that

 

sin pl = 0

 

which holds if

 

 

The constant pn is an eigenvalue and the function sin pnx is an eigenfunction. Now, we can write the solution, using the principle of superposition, as

 

 

The initial condition (8.342) will be satisfied at t = 0 if

 

 

that is, if f(x) can be expanded in a convergent half-range Fourier series in (0, l). The bn are given by

 

Note 8.13.5 Solution of partial differential equations by separation of variables method cannot be applied in all cases. It is only a certain special set of boundary conditions that allows us to separate the variables.

Example 8.13.6   A long copper rod with insulated lateral surface has its left end maintained at a temperature of 0°C and its right end at x = 2m maintained at 100°C. Find the temperature u (x,t) if the initial condition

 

Solution   We have to solve the partial differential equation for heat conduction

 

 

under the boundary conditions

 

 

and the initial condition

 

 

Assuming seperation of variables in the form

 

 

we get

 

 

where λ is the separation constant. In this problem the eigenvalue λ = 0 is important. The solution for λ = 0 is

 

 

 

The boundary conditions u (0, t) = 0 and u (2, t) = 100 imply that b = 0 and ac =50. Then

 

 

Now taking up the case of exponential decay of temperature namely λ = −p2 where p is real we have the general solution as

 

 

Superimposing the above two solutions we obtain a more general solution

 

 

The condition u(0, t) = 0 requires that A= 0 and the condition u (l, t) = u (2, t) = 100 demands that

 

 

This will be satisfied if

 

 

The solution, by the principle of superposition of solutions, is

 

 

       This must satisfy the initial condition (8.359) and hence

 

 

Expanding [f(x) − 50x] in a half-range Fourier sine series in [0, 2] we get the solution. The Fourier coefficients bn are given by

 

 

 

 

 

Finally, the solution is

 

Example 8.13.7   An insulated rod of length l has its ends A and B maintained at 0°C and 100°C, respectively, until steady-state conditions prevail. If B is suddenly reduced to 0°C and maintained at 0°C, find the temperature at a distance x from A at time t.

[JNTU 2003 (Set 4)]

Solution   Let u(x, t) be the temperature at time t at a distance x from A. The equation for the conduction of heat is

 

 

where ‘a2’ is the diffusivity of the material of the rod.

In the steady state when u depends only on x we get from (8.372):

 

 

whose general solution is

 

 

Boundary conditions are u(0) = 0 and u(l) = 100. So we obtain b = 0 and a = 100/l. This gives u(x) = (100/l)x at time t = 0. Thus, we have the initial condition

 

 

Boundary conditions for unsteady flow are:

 

 

Now, we have to solve Eq. (8.372) under the conditions(8.375)–(8.377) A solution of (8.372) is of the form

 

 

 

Equation (8.378) becomes

 

 

 

By the principle of superposition of solutions we may write the solution as

 

 

Imposing the initial condition (8.375) on the solution (8.382) we have

 

 

which is the half-range Fourier series expansion in (0, l) for the function (100/l)x. Therefore, Bn are given by

 

 

The solution for the problem is

 

Example 8.13.8   A homogeneous rod of conducting material of length 100 cm has its ends kept at zero temperature, and the temperature is initially

 

 

Find the temperature u (x, t) at any time.  [JNTU 2004s (Set 3)]

Solution   We have to solve the differential equation for the conduction of heat

 

 

under the boundary and initial conditions

Boundary conditions:

 

 

 

Initial condition:

 

 

A solution of Eq. (8.386) may be taken as

 

 

   u (0, t) = 0 ⇒ A = 0 equation (8.390) becomes

 

 

 

By the principle of superposition of solutions we may take the solution as

 

 

Imposition of the initial condition (8.389) on (8.393) yield,

 

 

We now expand u(x, 0) in a half-range Fourier sine series in (0, 100) and determine Bn.

Now, Bn are given by

 

 

 

Finally, the required solution of the problem is

 

 

Example 8.13.9   Find the temperature u(x, t) in a homogeneous bar of heat conducting material of length l cm with its ends kept at zero temperature and initial temperature given by α x(lx)/l2.

Solution   The initial boundary value problem consists of the following:

  1. Partial differential equation for conduction of heat:

     

     

  2. Boundary conditions:

     

     

  3. Initial condition:

     

     

  A general solution of (8.395) is

 

 

Boundary condition (8.396) is satisfied if we set A = 0 and boundary condition (8.397) is satisfied if

 

 

By the principle of superposition of solutions we may write

 

 

By the imposition of the initial condition (8.398) on (8.401) we get

 

 

which is the half-range Fourier sine series in (0, l) for f(x) = α x(lx)/l2. The constants Bn are given by

 

 

 

Hence the temperature distribution in the bar is given by

 

Example 8.13.10   Find the temperature in a thin metal rod of length l with both ends insulated and with initial temperature in the rod sin (αx/l).

Solution   The initial boundary value problem consists of the following

  1. Partial differential equation for conduction of heat

     

     

  2. Boundary conditions: (Insulation at both ends)

     

     

  3. Initial condition:

     

     

The general solution of Eq. (8.402) is of the form

 

 

(Note: we have added the constant A0/2 since in this case we have half-range Fourier cosine series expansion)

Differentiating (8.406) we have

 

 

Boundary condition (8.403) is satisfied if we set B = 0. Also, boundary condition (8.404) is satisfied if

 

 

Therefore, by the principle of superposition, the solution may be taken as

 

 

Here

 

 

 

 

The temperature distribution in the rod is

 

 

Example 8.13.11   A homogeneous rod of conduction material of length l has its ends kept at zero temperature. The temperature at the centre is T and it falls uniformly to zero at the two ends. Find the temperature at any time t.

Solution   The initial boundary value problem consists of solving the partial differential equation for heat conduction:

 

 

under the boundary conditions

 

 

and the initial condition that u (x,0) falls uniformly to zero at the ends.

To find this condition we have to solve the steady-state equation: d2u/dx = 0 whose general solution is

 

 

If C = l/2 is the centre of the rod then for the portion AC of the rod we have

 

 

For the portion CB of the rod we have

 

 

Consequently the initial condition is

 

 

By the method of separation of variables the solution of (8.414) may be put in the form

 

 

Boundary condition (8.415) is satisfied if we set A = 0, and boundary condition (8.416) is satisfied if we set sin pl = 0

 

 

Thus, by the principle of superposition of solutions, we may write the general solution as

 

 

Imposing the initial condition (8.420) we must have

 

 

Equation (8.424) is a half-range Fourier sine series expansion in (0, l) for the function u (x, 0) and so the constants Bn are given by

 

 

 

 

Finally, the solution of the problem is

 

EXERCISE 8.7
  1. Solve such that
    1. θ is finite as t → ∞.
    2. when x = 0 and θ = 0 when x = l for all t.
    3. θ = 00 when t = 0 for all x in 0 < x < l.

    Ans:

  2. A bar 10 cm long with insulated sides has its ends A and B maintained at temperatures 50°C and 100°C, respectively, until steady-state conditions prevail. The temperature at A is suddenly raised to 90°C and at the same time that at B is lowered to 60°C. Find the temperature distribution in the bar at time t.

     

    [Mysore 1997, Warangal 1996]

    Ans:

  3. Solve with boundary conditions u(x,0) = 3 sin n π x u(0,t) = 0, u(l, t) = 0 when 0 < x < 1, t > 0.

     

    [Osmania 1995, kerala 1990]

    Ans:

  4. The ends A and B of a rod 20m long have temperatures at 30°C and 80°C, respectively, until steady-state conditions prevail. The temperatures of the ends are changed to 40°C and 60°C, respectively. Find the temperature distribution in the rod at time t.

     

    [Kerala 1995, Madras 1991]

    Ans:

  5. Find the solution of one-dimensional heat equation under the boundary conditions u (0, t) = u (l, t) = 0 and the initial conditions u (x, 0) = x, 0 < x < l, l being the length of the rod.

    Ans:

  6. Solve subject to the conditions:
    1. u (0, t) = 0

    Ans:

  7. The ends A and B of a rod 30cm long have their temperatures kept at 20°C and 80°C until steady-state conditions prevail. The temperature of the end B is suddenly reduced to 60°C and

    kept so while the end A is raised to 40°C. Find the temperature distribution of the rod at time t.

    Ans:

  8. Find the solution of under the conditions
    1. u(0, t) = 0 = u(l, t)   for all t.

    Ans:

8.13.5 One-dimensional Wave Equation

Consider the vibrations of an elastic string placed along the x-axis, stretched to length l between two fixed points x = 0 and x = l. First we consider the problem when there is an initial displacement but no initial velocity (string released from rest). Next we consider motion of a string with an initial velocity but no initial displacement (string given an initial blow, but from its horizontal stretched position). Finally, we consider the case of both initial velocity and initial displacement.

8.13.6 Vibrating String with Zero Initial Velocity

Consider an elastic string of length l, fastened at its ends on the x-axis at x = 0 and x = l. The string is displaced, then released from rest to vibrate in the xy-plane. We want to find the displacement function y (x, t), whose graph is a curve in the xy-plane showing the shape of the string at time t. The boundary value problem for the displacement function y (x, t) consists in the solution of the partial differential equation

 

 

under the boundary conditions:

 

 

and the initial conditions:

 

 

The graph of f(x) is the position of the string before release.

The separation of variables method consists of attempting a solution of the form y(x,t) = X(t) T(t) where X(x) is a function of x only and T(t) is a function of t only. Substituting into the wave equation we obtain

 

 

where ′ denotes differentiation with respect to the respective independent variable. Then

 

 

The left-hand member is a function of x only and the right-hand member is a function of t only. The equality is possible only if both the quantities are equal to the same constant, which we take as −λ2 < 0 for convenience. This is called the separation constant. (Taking the separation constant as 0 or positive real number leads to trivial solutions.)

We now have

 

 

we obtain two ordinary differential equations

 

 

The boundary conditions y (0, t) = 0, and y (l, t) = 0 yield X(0) = 0 and X(l) = 0, respectively. The general solutions of equations are

 

 

 

The condition X(0) = 0 implies that A = 0 and the condition X(l) = 0 imples that sin λl = 0 ( B ≠ 0). Therefore, , which are the eigenvalues of the problem. The corresponding eigenfunctions are

 

 

The initial condition (the string is released from rest)

 

 

Hence

 

 

Therefore, we obtain

 

 

Now we take the solutions for the problem as

 

yn(x,t) = Cn sin λnx cos λn at

 

 

Each of these functions satisfies the wave equation, both boundary conditions and the initial condition . We need to satisfy the condition y (x, 0) = f(x). This is achieved by an infinite superposition of solutions in the form

 

 

 

we must choose the Cn’s to satisfy

 

 

This series is the half-range Fourier sine series of f(x) in [0, l]. The Fourier constants are given by

 

Example 8.13.12   A string AB of length l is fastened at both ends A and B. At a distance ‘b’ from the end A, the string is transeversely displaced to a distance ‘d’ and released from rest when it is in this position. Find the solution for the initial displacement function and zero initial velocity.

Solution   Let y (x, t) be the displacement of the string. The initial displacement is given by APB.

 

 

String with transverse displacement d at a point M (b, 0)

 

 

 

The problem is to solve one-dimensional wave equation

 

 

with boundary conditions

 

 

and initial conditions

 

 

The solution is given by

 

 

where Cn are given by

 

 

Hence, the displacement of the string at any point x and time t is given by

 

Example 8.13.13   A string of length l fastened at both ends A = (0,0) and B = (l, 0) undergoes initially a transversal displacement given by

 

 

and is released at rest when it is in this position. Find the displacement function y(x, t) for the subsequent motion.

Solution   The problem consists of solving the wave equation

 

 

under the boundary conditions

 

 

and the initial conditions:

Initial displacement:

 

 

Initial velocity:

 

 

The solution of Eq. (8.454) under the boundary conditions (8.455) and (8.456) and the zero initial velocity is

 

 

where

 

 

 

The solution for the initial displacement given by (8.457) and zero initial velocity is

 

8.13.7 Vibrating String with Given Initial Velocity and Zero Initial Displacement

Next, we consider the case when the string is released from its horizontal position with zero initial displacement but with an initial velocity given at x by g (x). The boundary value problem for the displacement function is

 

 

 

 

By the method of separation of variables we set y(x, t) = x (t) T(t) and obtain ordinary differential equations

 

 

The boundary conditions are same as before and hence we obtain eigenvalues

 

 

and the corresponding eigenfunctions are constant multiples of

 

 

with the values of λ as λ = λn = nπ/l the differential equation for T becomes

 

 

whose general solution is T(t) = A cos(nπat/l) + B sin(nπat/l)

The initial condition of zero initial displacement gives

 

 

Since T(0) = A = 0 solutions for T(t) are constant multiples of sin nπat/l. Thus, for n = 1, 2, 3, ... we have functions

 

 

Each of these functions satisfies the wave equation, the boundary conditions and the zero initial displacement condition. In order to satisfy the initial velocity condition we invoke the superposition principle and write

 

 

We assume that the series admits of term-by-term differentiation. So, we get

 

 

Now, the initial velocity condition yields

 

 

This is the half-range Fourier series expansion of g(x) on [0, l]. Here, the entire coefficient of sin nπx/l is the Fourier sine coefficient of g(x) on [0, l] so that we have

 

 

Example 8.13.14   Find the displacement y(x, t) of a string stretched between two fixed points at a distance 2l apart when the string is initially at rest in equilibrium position and the points of the string are given initial velocity g(x), given by

 

Solution   We have to solve the wave equation

 

 

under the boundary conditions:

 

 

and the initial conditions:

initial displacement:

 

 

and initial velocity:

 

 

The solution is

 

 

where

 

 

Hence, the displacement function is given by

 

Example 8.13.15   A tightly stretched string with fixed end points is initially at rest in its equilibrium position, and each of its points is given a velocity v, which is given by

 

 

Find the displacement y (x, t).

 

[JNTU 1994S, 2001, 2002]

Solution   We have to solve the wave equation

 

 

under the boundary conditions:

 

 

and the initial conditions:

zero initial displacement:

 

 

prescribed initial velocity:

 

 

Solving Eq. (8.486) by the method of separation of variables under the conditions is given by

 

 

Differentiating (8.491) partially with respect to ‘t’ we obtain

 

 

 

This is the half-range Fourier sine series expansion for v(x) in [0, l]. Therefore, we have

 

 

Hence, the solution of the problem is

 

 

Example 8.13.16   Find the solution for the above problem when the string is released from its horizontal position with an initial velocity given by g(x) = x(l + cos πx/l).

Deduce the result for a = 1 and l = π.

Solution   Following the procedure of the above problem, the solution in the present case is obtained as

 

 

The prescribed initial velocity in this case is

 

 

Differentiating (8.496) partially with respect to ‘t’ we have

 

 

Imposing the initial veloctiy condition (8.497) we obtain

 

 

This is the half-range Fourier series expansion for g(x) in [0, l]. Therefore, we have

 

 

For evaluation of the second intergral, we have to separate the cases.

 

Case (1) n = 1 and Case (2) n ≠ 1

 

Case 1

n = 1

 

Case 2

n ≠ 1

 

 

 

Therefore, the solution for g (x) = x (1 + cos πx/l) as the initial velocity function is

 

 

By taking a = 1 and l = π in the above result, we obtain the solution for this particular case as

 

8.13.8 Vibrating String with Initial Displacement and Initial Velocity

Consider the motion of the string with both initial displacement given by f(x) and initial velocity given by g(x). We have to now solve two separate problems, one with initial displacement f(x) and zero initial velocity and the other with zero displacement and initial velocity g(x). Let y1 (x, t) and y2 (x, t) be the respective solutions of the two problems and let

 

 

Then y satisfies the wave equation and the boundary conditions. Further,

 

 

and

 

 

Thus, y(x, t) is the solution in the case of non-zero initial displacement and velocity functions.

Example 8.13.17   An elastic string of length l, fastened at its ends on the x-axis at x = 0 and x = λ, is given initial displacement f(x) and initial velocity g(x). Find the displacement function y(x, t) by solving the wave equation ytt = a2yxx under the conditions: y(0, t) = y(l, t) = 0, y(x, 0) = f(x) = x for

 

 

and

 

 

Solution   From Eqs. (8.505) and (8.506), we obtain the solution for the present problem as

 

EXERCISE 8.8
  1. Find the displacement of a string stretched between two fixed points at a distance 2l apart when the string is initially at rest in equilibrium position and the points of the string are given an initial velocity g(x), which are given by .

    Ans:

     

  2. A string of length l is stretched and fastened to two fixed points. Find y(x, t) satisfying the wave equation ytt = a2yxx when it is given as:
    1. Initial displacement y(x, 0) = f(x) = b sin πx/l.

      Ans:

    2. Initial triangular deflection:

       

      Ans:

    3. Initial velocity:

      Ans:

    4. Initial velocity:

      Ans:

8.13.9 Laplace's Equation or Potential Equation or Two-dimensional Steady-state Heat Flow Equation

The two-dimensional heat conduction equation is given by

 

 

In the case of steady-state heat flow ∂u/∂t = 0 and equation reduces to

 

 

The solution u(x, y) of the above equation can be obtained by the method of separation of variables in a rectangular region both in the Dirichlet problem as well as in Neumann's problem. A rectangular thin plate with its two faces insulated is considered so that the heat flow is two-dimensional. The boundary conditions are prescribed on the four edges of the plate.

Example 8.13.18   Solve Laplace's equation

 

 

in the rectangle; 0 < x < a, 0 < y < b in the x y-plane, with the boundary conditions

 

 

 

 

Solution   Let

Substituting in (8.508) we get

 

 

where the separation constant is taken as negative to get non-trivial solutions. The boundary value problem reduces to solution of the ordinary differential equations

 

 

under the conditons

 

 

 

 

The general solution of Eq. (8.515) is

 

Y(y) = A cos λy + B sin λy

 

By (8.517) we get A = 0 and by (8.518) we get sin λb = 0

 

 

λn = nπ/b are the eigenvalues and the corresponding eigenfunctions are

 

 

Now the general solution of (8.516) is

 

 

using (8.519) we get C = 0 so that

 

 

Therefore, the solution of Eq. (8.508) satisfying the boundary conditions (8.509), (8.510) and (8.511) is

 

 

where we have replaced D by Cn.

By the principle of superposition we write the solution as

 

 

Lastly we have condition (8.512) namely u(a, y) = f(y) to be satisfied. This gives

 

 

This is a half-range Fourier sine series expansion of f(y) in (0, b) and the constants Cn are given by

 

 

Thus, this harmonic function u(x, y) satisfying Laplace's equation (8.508) and the boundary conditions (8.509)(8.512) is given by (8.523) where the constants Cn are determined by (8.525) for any specific function f(y).

Example 8.13.19   A retangular plate is bounded by the lines x = 0, y = 0, x = a and y = b and the edge temperatures are u(0, y) = u(x, b) = u(a, y) and u(x, 0) = 5 sin(5πx/a) + 3 sin(3πx/a). Find the steady-state temperature at any point of the plate.

 

[JNTU 2002, 2003s]

Solution   Let u(x, y) be the steady-state temperature at any point p(x, y) of the rectangular plate. We have to solve Laplace's equation

 

 

under the boundary conditions

A suitable solution of Eq. (8.526), by the method of separation of variables, satisfying boundary conditions (8.527)–(8.529) for each n = 1, 2, 3, ... is

 

 

By the principle of superposition of solution we may write the general solution as

 

 

Imposing condition (8.530) we get

 

 

Equating the coefficients of like terms on either side we get

 

 

Hence, the required solution of steady-state temperature is

 

Example 8.13.20   An infinitely long plane uniform plate is bounded by two parallel edges and an end at right angles to them. The breadth in π. This end is maintained at a temperature u0 at all points and the other edges are at zero temperature. Determine the temperature at any point of the plate in the steady state.

 

[JNTU 2002, 2005 (Set 4)]

 

 

Temperature in an infinitely long plate

Solution   Let u(x, y) be the temperature at any point p(x, y) of the plate. Then the steady state temperature distribution is given by Laplace's equation

 

 

with the boundary conditions

  1. u(0, y) = 0;
  2. u(π, y) = 0   for all y
  3. u(x, 0) = u0;
  4. u(x, ∞) = 0   for 0 < x < π

The general solution of Eq. (8.536) is

 

 

By condition (8.536): u(0, y) we get A = 0. Now Eq. (8.537) becomes

 

 

By condition (8.537): u(π, y) = 0 we get

 

 

so that the solution becomes u(x, y) = (cneny + Dneny) sin nx where BC and BD have been replaced by cn and dn, respectively.

By condition (8.539):

 

 

Finally, the general solution satisfying the conditions (8.536), (8.537) and (8.539) is (by the principle of superposition)

 

 

Now applying boundary condition (8.538):

 

 

Equation (8.542) is a half-range Fourier sine series expansion for u0 in (0, π) and hence the Fourier constants dn are given by

 

 

Now (8.541) reduces to

 

 

which is the required solution.

Example 8.13.21   The temperature u (x, y) is maintained at 0°C along three edges of a square plate of side 100 cm and the fourth edge is maintained at a constant temperature u0 until steady-state conditons prevail. Find the temperature at any point (x, y) of the plate and also at the centre of the plate.

 

[JNTU 2003s (Set 4)]

 

Solution   Let the side of the plate be a = 100 cm. The temperature function u (x, y) satisfies Laplace's equation

 

 

with the boundary conditons:

  1. u(0, y) = 0
  2. u(a, y) = 0 for 0 ≤ ya
  3. u(x, 0) = 0
  4. u(x, a) = u0 for 0 ≤ xa.

The general solution of Eq. (8.545) is

 

 

By condition (8.545) we get C1 = 0 so that Eq. (8.546) becomes

 

 

By condition (8.546) we get u(a, y) = 0 = C2sin pa(C3epy + C4e−py) ⇒ p = (/a)(n = 1,2,3), so that we have

 

 

Now, condition (8.547) implies that C4 = − C3     (8.549)

 

The solution may therefore be written as

 

 

for each n. By the principle of superposition we may write

 

 

Imposing the condition (8.548) we have

 

 

Equation (8.552) is a half-range Fourier sine series expansion of u0 in (0, a), and the Fourier coefficients are obtained by

 

 

Substituting for Cn in (8.551) we have

 

 

Temperature at the centre = (50, 50) is

 

Example 8.13.22   Find the steady-state temperature in a rectangular plate 0 ≤ xa, 0 ≤ yb when the sides x = 0, x = a and y = b are insulated while the edge y = 0 is kept at temperature (c/a)cosπx/a.

 

Solution   We have to solve Laplace's equation

 

 

under the boundary conditions

 

 

and

 

 

By separating the variables we get the following two ordinary differential equations:

 

 

these solutions are

 

 

The above boundary conditions imply that

 

 

Thus

 

Hence, the required solution is

 

 

where

 

 

EXERCISE 8.9

Solve the two-dimensional Laplace's equation. (∂2u/∂x2) + (∂2u/∂y2) = 0 in the region 0 < x < a, 0 < y < b bounded by a metal plate with the following boundary conditions:

  1. u(0, y) = u(x, 0) = x(x, b) = 0 and u(a, y) = g(y) for 0 < y < b.

    Ans:

    where

  2. Solve problem 1 completely by taking g (y) = 100.

    Ans:

  3. Ans:

  4. A square plate has its faces and the edge y = 0 insulated. Its edges x = 0 and x = π are kept at zero temperature and its fourth edge y = π is kept at temperature f (x). Find the steady-state temperature at any point of the plate.

     

    [JNTU 2003S]

    Ans:

  5. Find the solution of problem 4 if f(x) = 100.

    Ans:

  6. Solve (∂2u/∂x2) + (∂2u/∂y2) = 0 for 0 < x < π, 0 < y < π given that

     

    u(0, y) = u(π, y) = u(π, y) = u(x, π) = 0 and u(x, 0) = sin2 x.

    Ans: