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 nonhomogeneous 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
 Consider the equation
where c and a are arbitrary constants. It represents the set of all spheres with their centres on the zaxis.
Differentiating with respect to x and y we get
Eliminating the arbitrary constant c from (8.3) and (8.4) we obtain the firstorder linear partial differential equation
which characterizes the set of all spheres with centres on the zaxis.
 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 zaxis.
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).
 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 zaxis as the axis of symmetry.
Differentiating Eq. (8.9) partially with respect to x and y we get
where u = x^{2} + y^{2} and Eliminating f between equations (8.10) and (8.11), we again obtain the firstorder 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 firstorder 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 higherorder partial differential equation.
Further, if one arbitrary function is to be eliminated from a relation, then a firstorder partial differential equation is obtained; and if two arbitrary functions are to be eliminated, then a secondorder partial differential equation is obtained and so on.
A firstorder 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 nonlinear.
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
which is nonlinear.
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
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 firstorder partial differential equation
Example 8.4.3
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
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
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 nonlinear.
Example 8.4.7
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):
 z = ax^{2} + by^{2}.
Ans: 2z = px + qy
 (x − a)^{2} + (y − b)^{2} = z^{2} cot^{2} α.
Ans: p^{2} + q^{2} = tan^{2} α
 z = ax + by + a^{2} + b^{2}.
Ans: z = px + qy + p^{2} + q^{2}
 z = axy + b.
Ans: px = qy

[JNTU 2003]
Ans: 16(p^{2} + q^{2})(x + y) = 1
 u = a(x + y) + b(x − y) + abz + c.
Ans:

Ans: px + qy = pq

Ans: q = px + p^{2}

[JNTU 2003]
Ans:

[JNTU 2003]
Ans: px + qy = p + q
 Form the partial differential equation of all spheres of radius a with their centres on the x–y plane.
 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 firstorder 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's^{1} linear equation. In fact, it is a quasilinear 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
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
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
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
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.
[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:

Ans: px = qy
 xyz = f(x + y + z). [Osmania 1995, Calicut 1994]
Ans: x(y − z)p + y(z − x)q = z(x − y)
 x + y + z = f(x^{2} + y^{2} + z^{2}). [Madras 2000, Ranchi 1990]
 f(x^{2}+ y^{2}, z − xy) = 0. [Madurai 1990]
Ans: py − qx = y^{2} − x^{2}
 z = f(x^{2} − y^{2}). [Madras 1991]
Ans: py + qx = 0
 z = yf(x) + xg(y). [Madras 1993, Karnataka 1993, Madurai 1998]
Ans:
 z = f(y) + φ(x + y). [JNTU 2005 S (1)]
Ans:
 z = xy + f(x^{2} + y^{2}). [JNTU 2005 S (2)]
Ans: py − qx = y^{2} − x^{2}
8.6 CLASSIFICATION OF FIRSTORDER PARTIAL DIFFERENTIAL EQUATIONS
8.6.1 Linear Equation
Let z = z(x, y) be a function of two variables x and y. A firstorder 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
where
8.6.2 SemiLinear Equation
A firstorder partial differential equation is said to be semilinear 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 QuasiLinear Equation
A firstorder partial differential equation is said to be quasilinear if it is linear in p and q, i.e. if it is of the form
Note that every semilinear partial differential equation is quasilinear.
8.6.4 Nonlinear Equation
A firstorder partial differential equation is said to be nonlinear if it does not fall into any one of the above types.
8.7 CLASSIFICATION OF SOLUTIONS OF FIRSTORDER PARTIAL DIFFERENTIAL EQUATION
Consider a firstorder 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 threedimensional 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 twoparameter 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 oneparameter family of solutions of (8.93).
This is a subfamily of the twoparameter 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 twoparameter 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.
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 f_{1}(x) is an arbitrary function of x and f_{2}(y) and f_{3}(y) are arbitrary functions of y.
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
∴ 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
 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
 Solve
Ans:
 Solve given that when x = 0, z = e ^{y} and
[Mysore 1987 S, Madras 1993, Karnataka 1994]
 Solve
[Madurai 1988, Mysore 1987]
Ans: u = −e^{−t} sin x + φ (x) + ψ (t)
 Solve
Ans: z = e^{x}· e^{y} + ∫ f(y)dy + φ (x)
 Solve
Ans:
8.9 QUASILINEAR EQUATIONS OF FIRST ORDER
Quasilinear 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 nonlinear.
Theorem 8.9.1 The general solution of the quasilinear partial differential equation
is
where F is an arbitrary function and u(x, y, z) = c_{1} and v(x, y, z) = c_{2} 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, SEMILINEAR AND QUASILINEAR EQUATIONS
If the equation is linear or semilinear 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 zx^{2}p + zy^{2}q = 1 − z^{2}.
Solution The equation can be thrown into the form
which is a semilinear 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 mp − lq = z tan(lx + my).
Solution This is a semilinear 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 yp − xq = 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 x^{2}(y − z)p + y^{2}(z − x)q = z^{2}(x − y).
Solution Lagrange's auxiliary equations are
Using multipliers in turn we get
General solution is
Example 8.10.5 Solve zxp + zyq = 1+ z^{2}.
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 = (x^{2} − y^{2} + z^{2})x.
Solution Lagrange's auxiliary equations are
From the first two ratios we get
x dx − y dy = 0 ⇒ x^{2} − y^{2} = a^{2} (a^{2} 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 xp − yq = 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 (y − z)p + (z − x)q = (x − y).
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
Example 8.10.11 Solve (sec x)p + (sin x − y sec x tan x)q = (a^{2} − z^{2}).
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)p − y(1 + xy)q = z(1 − xy).
Solution Lagrange's auxiliary equations are
From the first and last ratios cancelling (1  xy).
Integrating
On crossmultiplication the first two ratios yield
On integration we get
General solution is
Example 8.10.13 Solve (x^{2} − y^{2} − z^{2})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 (z^{2} − 2yz − y^{2})p + (xy + zx)q = xy − zx.
[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 x^{n}p + y^{n}q = z^{n}.
Solution Lagrange's auxiliary equations
On integration we get two solutions
General solution when
when n ≠ 1 is
Example 8.10.16 Solve x(y − z)p + y(z − x)q = z(x − y).
[JNTU 2002, 2003]
Solution Lagrange's auxiliary equations are
Now, each ratio
Also each ratio
General solution is
Example 8.10.17 Solve (x^{2} − yz)p +(y^{2} − zx)q = z^{2} − xy.
[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 (y − z)p + (x − y)q = z − x.
[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 + (2x − y)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 crossmultiplication,
Integrating
General solution is
Example 8.10.20 Solve (x + 2y^{2})p + yq = − cos z.
Solution Lagrange's auxiliary equations are
From the last two ratios
Integrating we get
From the first two ratios
On crossmultiplication and transposition
Integrating, we get
General solution is
EXERCISE 8.4
 px + qy = z.
Ans:
 px^{2} + qy^{2} = z^{2}.

Ans: F(x^{2} − y^{2}, x^{2} − z^{2}) = 0

Ans: F(x^{3} − y^{3}, x^{2} − z^{2}) = 0
 z(xp − yq) = y^{2} − x^{2}.
Ans: F(x^{2} + y^{2} + z^{2}, xy) = 0
 px(z − 2y^{2}) = (z − qy)(z − y^{2} − 2x^{3}).
Ans:
 x(x + y)p − y(x + y)q = −(x − y)(2x +2y + z).
Ans: F(xy, (x + y + z)(x + y)) = 0
 x^{2}(y −z)p + y^{2}(z −x)q = z^{2}(x −y). [Madurai 1990]
Ans: F(xyz, (x^{−1} + y^{−1} + z^{−1}) = 0
 x(y^{2} − z^{2})p + y(z^{2} − x^{2})q = z(x − y^{2}).
Ans: F(xyz, (x^{2} + y^{2} + z^{2})) = 0
 (mz − ny)p +(nx − lz)q = ly − mx. [AMIE 1990, Madras 1994 S]
Ans: F(lx + my + nx, x^{2} + y^{2} + z^{2}) = 0
 (y − zx)p +(x + yz)q = x^{2} + y^{2}.
Ans: F(x^{2} − y^{2} + z^{2}, xy − z) = 0
 (b − c)a^{−1}yzp + (c − a)b^{−1}zxq = (a − b)c^{−1}xy.
Ans: F(ax^{2} + by^{2} + cz^{2}, a^{2}x^{2} + b^{2}y^{2} + c^{2}z^{2}) = 0
 (y^{2} + z^{2} − x^{2})p − 2xyq = −2zx.
Ans:
 p tan x + q tan y = tan z. [Andhra 1990, Kerala 1987 S]
Ans:
 xe^{y}p + (1 − e^{y})q = ze^{y}.
Ans:
8.11 NONLINEAR 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: pqequation: 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 = x^{α} y^{β} z^{γ}.
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
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: zpqequation: 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's^{2} 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
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(z − px − qy) = 1+ q^{2}. [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
 q^{2} = e^{−p/α}.
Ans: z = ax + e^{−a/2a} y + c
 p + q = pq.
Ans:
 p^{2} + q^{2} = 1. [Osmania 2000]
Ans:
 x^{2}p^{2} + y^{2}q^{2} = z. [Osmania 2000 s, Madras 1998]
Ans:
 z = pq.
Ans: (x + ay + c)^{2} = 4az
 zq = p^{2}.
Ans: z = ce^{ax} + a^{2}y
 p(1 + q)qz. [Kerala 1990 S, Punjab 1986]
Ans: log (az − 1) = (x + ay + c)
 z = p^{2} + q^{2}. [Calicut 1994, Andhra 1990, Madurai 1990]
Ans: 4z (1 + a^{2}) = (x + ay + c)^{2}
 z(p^{2} − q^{2}) = x − y. [Madras 1991, Calicut 1991, Karnataka 1990]
Ans: z^{3/2} = (x + a)^{3/2} + (y + a)^{3/2} + c
 q = xyp^{2}.
Ans:
 yp + xq + pq = 0.
[Marathwada 1993]
Ans:
 p + q = sin x + sin y.
[Madras 1993, Karnataka 1993]
Ans: z = a(x − y) − (cos x + cos y) + c
 z = px + qy + p^{2} + q^{2}.
Ans: z = ax + by + a^{2} + b^{2}

Ans:
 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 Euler^{3} 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
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 e^{mx}, e^{my} 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 c_{3} = c_{4} = … = 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 c_{1} + c_{2} = 0,
Since e^{αx}−e^{−αx} = 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 = e^{mx}, Y = e^{my}, the auxiliary equations are
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:

Ans: u(x,t) = sin x − e^{−t} sin x

Ans: u = e^{2x−5y}

Ans:
 .
Ans:
 .
Ans:
 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 SECONDORDER PARTIAL DIFFERENTIAL EQUATIONS
8.13.1 Introduction
Many physical and engineering applications such as fluid flow, heat transfer and wave motion involve secondorder partial differential equations and hence we take up a study of these equations and their solution by separation of variables method.
The general secondorder 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 quasilinear. 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 secondorder linear partial differential equation in two independent variables x and y may be written as
If G ≠ 0 then Eq. (8.334) is caled nonhomogeneous and if G = 0 then it is called homogeneous.
8.13.2 Classification of Equations
We call the quantity ∆ = B^{2} – 4AC the discriminant and classify Eq. (8.334) as hyperbolic, parabolic or elliptic according as ∆ > 0, = 0 or < 0. The following are wellknown examples of these three types.
Hyperbolic Type
1. Onedimensional wave equation
Here A = a^{2}, B = 0, C = –1 and ∆ = B^{2} – 4AC = 4a^{2} > 0.
Parabolic Type
2. Onedimensional heatflow equation
Here A = a^{2}, B = C = 0 and ∆ = B^{2} – 4AC = 0.
Elliptic Type
3. (a) Twodimensional Laplace equation
Here A = 1, B = 0, C = 1 and ∆ = B^{2} – 4AC = – 4 < 0.
(b) Poisson's equation
Equations (8.335)–(8.337) are homogeneous while Eq. (8.338) is nonhomogeneous.
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 u_{n} 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 elliptictype equation is always a boundary value problem. The boundary conditions may be one of the following three types:
 Dirichlet problem (First boundary value problem)
The solution is prescribed along the boundary.
 Neumann problem (Second boundary value problem)
The derivative of the solution is prescribed along the boundary.
 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 zerocondition and nonhomogeneous if it is a nonzero condition.
8.13.4 Solution of Onedimensional Heat Equation (or diffusion equation)
Consider a long and thin wire, rod or bar OA of length l and of constant crosssection and homogeneous heatconducting material. Let the bar be placed along the xaxis 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 xdirection only. Therefore, the temperature u of the bar depends on x and t only. The initial boundary value problem consists of onedimensional heat equation:
where a^{2} 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 = −p^{2} 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
which holds if
The constant p_{n} is an eigenvalue and the function sin p_{n}x 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 halfrange Fourier series in (0, l). The b_{n} 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 = −p^{2} 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 halfrange Fourier sine series in [0, 2] we get the solution. The Fourier coefficients b_{n} 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 steadystate 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 ‘a^{2}’ 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 halfrange Fourier series expansion in (0, l) for the function (100/l)x. Therefore, B_{n} 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 halfrange Fourier sine series in (0, 100) and determine B_{n}.
Now, B_{n} 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(l−x)/l^{2}.
Solution The initial boundary value problem consists of the following:
 Partial differential equation for conduction of heat:
 Boundary conditions:
 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 halfrange Fourier sine series in (0, l) for f(x) = α x(l − x)/l^{2}. The constants B_{n} 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
 Partial differential equation for conduction of heat
 Boundary conditions: (Insulation at both ends)
 Initial condition:
The general solution of Eq. (8.402) is of the form
(Note: we have added the constant A_{0}/2 since in this case we have halfrange 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 steadystate equation: d^{2}u/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 halfrange Fourier sine series expansion in (0, l) for the function u (x, 0) and so the constants B_{n} are given by
Finally, the solution of the problem is
EXERCISE 8.7
 Solve such that
 θ is finite as t → ∞.
 when x = 0 and θ = 0 when x = l for all t.
 θ = 0_{0} when t = 0 for all x in 0 < x < l.
Ans:
 A bar 10 cm long with insulated sides has its ends A and B maintained at temperatures 50°C and 100°C, respectively, until steadystate 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]
 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:
 The ends A and B of a rod 20m long have temperatures at 30°C and 80°C, respectively, until steadystate 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:
 Find the solution of onedimensional 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:
 Solve subject to the conditions:
 u (0, t) = 0
Ans:
 The ends A and B of a rod 30cm long have their temperatures kept at 20°C and 80°C until steadystate 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:
 Find the solution of under the conditions
 u(0, t) = 0 = u(l, t) for all t.
Ans:
8.13.5 Onedimensional Wave Equation
Consider the vibrations of an elastic string placed along the xaxis, 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 xaxis at x = 0 and x = l. The string is displaced, then released from rest to vibrate in the xyplane. We want to find the displacement function y (x, t), whose graph is a curve in the xyplane 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 lefthand member is a function of x only and the righthand 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
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 C_{n}’s to satisfy
This series is the halfrange 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 onedimensional wave equation
with boundary conditions
and initial conditions
The solution is given by
where C_{n} 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 = λ_{n} = 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 termbyterm differentiation. So, we get
Now, the initial velocity condition yields
This is the halfrange 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 halfrange 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 halfrange 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
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 y_{1} (x, t) and y_{2} (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 nonzero initial displacement and velocity functions.
Example 8.13.17 An elastic string of length l, fastened at its ends on the xaxis 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 y_{tt} = a^{2}y_{xx} 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
 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:
 A string of length l is stretched and fastened to two fixed points. Find y(x, t) satisfying the wave equation y_{tt} = a^{2}y_{xx} when it is given as:
8.13.9 Laplace's Equation or Potential Equation or Twodimensional Steadystate Heat Flow Equation
The twodimensional heat conduction equation is given by
In the case of steadystate 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 twodimensional. 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 yplane, with the boundary conditions
Solution Let
Substituting in (8.508) we get
where the separation constant is taken as negative to get nontrivial solutions. The boundary value problem reduces to solution of the ordinary differential equations
under the conditons
The general solution of Eq. (8.515) is
By (8.517) we get A = 0 and by (8.518) we get sin λb = 0
λ_{n} = 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 C_{n}.
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 halfrange Fourier sine series expansion of f(y) in (0, b) and the constants C_{n} 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 C_{n} 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 steadystate temperature at any point of the plate.
[JNTU 2002, 2003s]
Solution Let u(x, y) be the steadystate 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 steadystate 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 u_{0} 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
 u(0, y) = 0;
 u(π, y) = 0 for all y
 u(x, 0) = u_{0};
 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) = (c_{n}e^{ny} + D_{n}e^{–ny}) sin nx where BC and BD have been replaced by c_{n} and d_{n}, 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 halfrange Fourier sine series expansion for u_{0} in (0, π) and hence the Fourier constants d_{n} 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 u_{0} until steadystate 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:
 u(0, y) = 0
 u(a, y) = 0 for 0 ≤ y ≤ a
 u(x, 0) = 0
 u(x, a) = u_{0} for 0 ≤ x ≤ a.
The general solution of Eq. (8.545) is
By condition (8.545) we get C_{1} = 0 so that Eq. (8.546) becomes
By condition (8.546) we get u(a, y) = 0 = C_{2}sin pa(C_{3}e^{py} + C_{4}e^{−py}) ⇒ p = (nπ/a)(n = 1,2,3), so that we have
Now, condition (8.547) implies that C_{4} = − C_{3} (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 halfrange Fourier sine series expansion of u_{0} in (0, a), and the Fourier coefficients are obtained by
Substituting for C_{n} in (8.551) we have
Temperature at the centre = (50, 50) is
Example 8.13.22 Find the steadystate temperature in a rectangular plate 0 ≤ x ≤ a, 0 ≤ y ≤ b 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 twodimensional Laplace's equation. (∂^{2}u/∂x^{2}) + (∂^{2}u/∂y^{2}) = 0 in the region 0 < x < a, 0 < y < b bounded by a metal plate with the following boundary conditions:
 u(0, y) = u(x, 0) = x(x, b) = 0 and u(a, y) = g(y) for 0 < y < b.
where
 Solve problem 1 completely by taking g (y) = 100.
Ans:

Ans:
 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 steadystate temperature at any point of the plate.
[JNTU 2003S]
Ans:
 Find the solution of problem 4 if f(x) = 100.
Ans:
 Solve (∂^{2}u/∂x^{2}) + (∂^{2}u/∂y^{2}) = 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) = sin^{2} x.