Engineering
Mathematics-III
Dr. P.K. Srivastava
Assistant Professor of Mathematics
Galgotia College of Engineering & Technology
Greater Noida (U.P.)
(An ISO 9001:2008 Certified Company)
Vayu Education of India
2/25, Ansari Road, Darya Ganj, New Delhi-110 002
Engineering Mathematics-III
Copyright ©VAYU EDUCATION OF INDIA
ISBN: 978-93-83137-12-1
First Edition: 2013
Price:
160/-
All rights reserved. No part of this publication may be reproduced, stored in a retrieval
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(An ISO 9001:2008 Certified Company)
Vayu Education of India
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Contents
1. Partial Differentiation and Partial Differential Equation...................... 1-37
2. Partial Differential Equations............................................................. 38-85
3.Fourier Series....................................................................................86-113
4.Laplace Transformation....................................................................114-152
5.Numerical Techniques. ..................................................................... 153-184
6.Numerical Methods for Solution
.
of
Partial Differential Equation. 185-221
Chapter-1
1.1 Functions of Two or More Variables
A symbol z which has a definite value for every pair of values of x and y is called a function
of two independent variables x and y and is written as
z = f (x, y) orI (x, y)
1.2 Limits
“The function f (x, y) is said to tend to limit l as x o a and y o b if and only if the limit l is
independent of the path followed by the point (x, y) as x o a and y o b and we write
lim f ( x , y) = l
x oa
y ob
Or, in circular neighbourhood we define the limit as:
“The function f (x, y) defined in a region R, is said to tend to the limit l as x o a and
y o b if and only if corresponding to a positive number H, another positive number G such
that
f ( x , y) l e for 0 < (x – a)2 + (y – b) < G for every point (x, y) in R”.
1.3 Continuity
“A function f(x, y) is said to be continuous at the point (a, b) if
lim f ( x , y) exists = f(a, b)
xoa
yob
If it is continuous at all points in a region, then it is said to be continuous in that
region. A function which is not continuous at a point in any region is called discontinuous
at that point for same region.
1.4 Partial Derivatives
Let z = f (x, y) be variables x and y. Then partial differentiation of z w.r.t. x keeping y as a
constant is denoted as
wz w f ( x , y )
,
, fx ( x , y ), Dx f ( x , y )
wx
wx
2 2
Engineering
Mathematics–III
A Textbook
of Engineering Mathematics-I
wz
f ( x dx , y ) f ( x , y )
= lim
dx o 0
wx
dx
Similarly, the partial differentiation of z w.r.t. y keeping x as a constant is denoted as
where
wz w f ( x , y )
,
, f y ( x , y), Dy f ( x , y )
wy
wy
In general, fx and fy being functions of x and y, so these can be further differentiated
partially w.r.t. x and y and thus we have
w2 z
w ­ wz ½
® ¾ =
dx ¯ wx ¿
wx 2
or
w2 f
wx 2
or fxx
w
dy
w2z
­ wz ½
® ¾ =
wy wx
¯ wx ¿
or
w2 f
or fxy
wy wx
w
dx
­ wz ½
w2z
® ¾ =
wx w y
¯ wy ¿
or
w2 f
or fyx
wx wy
w ­ wz ½
w2z
® ¾ = 2
dy ¯ wy ¿
wy
It can be verified easily that
or
w2 f
wy2
or fyy
w2z
w2z
=
wx wy
wy wx
Also, we can use the following notations
w2 z
wz
w2 z
w2 z
wz
, q=
, r=
,
s
=
,
t
=
wx w y
wy
x
wx 2
wy 2
If z be a function of number of variables say x1, x2, ... xn; then its partial derivatives
w.r.t. to one of the variables say x1 is denoted as
p =
wz
, keeping others as constant.
wx1
Example 1.1
If z (x + y) = x2 + y2, show that
2
Solution:
§ wz w z ·
§
wz wz ·
¸ = 4 ¨1 ¨
¸
wx wy ¹
© wx wy ¹
©
z (x + y) = x2 + y2
z=
Now
x 2 y2
xy
x 2 2xy y2
wz
( x y ) 2 x ( x 2 y2 ) . 1
=
=
wx
( x y )2
( x y )2
Partial Differentiation and Partial Differential Equation
3
wz
( x y) . 2 y ( x 2 y2 ) . 1
y2 2xy x 2
=
wy =
( x y )2
( x y )2
Also,
2
?
2
2
2
2
§ wz w z ·
°­ x 2xy y y x 2xy ½°
¾
¸ = ®
¨
( x y )2
© wx wy ¹
°¿
¯°
?
­ 2 x 2 y2
°
§ wz w z ·
2
¸ = ®
¨
°¯ x y
© wx wy ¹
2
=
2
½
°
¾
°¿
4 ( x y )2
( x y )2
...(1)
Also, we have
­° ª x 2 2xy y2 º ª y2 2xy x 2 º ½°
­
wz wz ½
4 ®1 ¾ = 4 ®1 «
»«
»¾
2
2
wx wy ¿
¯
¼ ¬ (x y)
¼ °¿
¯° ¬ ( x y)
2
2
2
2
2
2
°­ x y 2xy x 2xy y y 2xy x °½
= 4®
¾
( x y )2
¯°
¿°
2
2
( x y )2
°­ x 2xy y ½°
4
.
= 4®
=
¾
2
( x y )2
°¯ ( x y)
°¿
Hence from (1) and (2), we have
...(2)
2
§ wz w z ·
§
wz wz ·
¸ = 4 ¨1 ¨
¸
wx wy ¹
© wx wy ¹
©
Example 1.2
If u = exyz, find the value of
Solution: Let
then
w3 u
wx wy wz
u = exyz
wu
= xyexyz
wz
w2 u
= e xyz ( x ) e xyz ( xz ) ( xy )
wy wz
w2 u
= e xyz ( x x 2 yz )
wy wz
w3 u
= e xyz (1 2xyz ) ( x x 2 yz ) yz . e xyz
wx wy wz
= exyz (1 + 3xyz + x2 y2 z2)
...(1)
4 4
Engineering
Mathematics–III
A Textbook
of Engineering Mathematics-I
Example 1.3
If u = f (y / x), show that
wu
wu
x
y
dx
dy = 0
Solution: Let
u = f(y / x)
wu
= fc (y / x) . – y / x2
Now
dx
wu
y
= f c( y / x )
dx
x
Also, from (1), we have
1
wu
= f c( y / x ) .
x
dy
y
wu
= f c( y / x )
x
dy
Now, adding (2) and (3), we get
wu
wu
x
y
=0
dx
dy
y
...(1)
...(2)
...(3)
Example 1.4
If u = log (x3 + y3 + z3 – 3xyz), show that
2
§ w
w
w ·
9
¸ u = ¨
x
y
z
w
w
w
( x y z )2
¹
©
Solution: Let
then from (1), we have
u = log (x3 + y3 + z3 – 3xyz)
...(1)
1
wu
. 3x 2 3 yz
= 3
3
3
dx
x y z 3xyz
=
3 x2 y z
x 3 y3 z3 3xyz
...(2)
Similarly, we have
3 y2 xz
wu
dy = x 3 y3 z 3 3xyz
...(3)
3 z 2 xy
wu
= 3
and
wz
x y3 z 3 3xyz
By adding (2), (3) and (4), we get
...(4)
3 x 2 y2 z 2 xy yz zx
wu wu wu
wx wy wz =
x 3 y3 z 3 3xyz
Partial Differentiation and Partial Differential Equation
=
3 x 2 y2 z 2 xy yz zx
x yz
x 2 y2 z 2 xy yz zx
§ w
w
w ·
3
¸u =
¨
x
y
z
w
w
w
x yz
©
¹
2
§ w
w
w ·
·
§ w
w
w ·§
3
¸ u = ¨
¨
¸
¸¨
© wx wy wz ¹
w
w
w
x
y
z
x
y
z
©
¹©
¹
· w §
· w §
·
w §
3
3
3
=
¸
¨
¸
¨
¸
¨
wx © x y z ¹ wy © x y z ¹ wz © x y z ¹
3
3
3
9
=
2
2
2 =
( x y z )2
(x y z )
(x y z)
(x y z )
EXERCISES
1. I f
xx
yy
zz
= c, show that of x = y = z,
w2z
= – (x log ex)–1
wx w y
2. If V = (x2 + y2 + z2)–1/2, we show that
w2 V
wx 2
w2 V
wy2
w2 V
wz 2
=0
­
½°
xy
1 °
3. If u = tan ®
¾ , show that
°¯ (1 x 2 y2 ) °¿
1
w2 u
=
wx w y
1 x 2 y2
3/2
4. If z = f (x + ay) + f (x – ay), prove that
w2 z
wy2
2
= a
w2 z
wx 2
5. If u = sin–1 {x / y} + tan–1 {y / x} then find the value of
x
wu
wu
y
=0
wx
wy
6. If z = eax + by . f (ax – by), prove that
5
6 6
Engineering
Mathematics–III
A Textbook
of Engineering Mathematics-I
b
wz
wz
a
= 2abz
wx
wy
7. If z = x2 tan–1 (y / x) – y2 tan–1 (x / y), prove that
x 2 y2
w2z
= 2
wx w y
x y2
8. If u = log (x2 + y2) + tan–1 {y / x}, show that
w 2u
wx 2
w2u
wy 2
=0
§ xz ·
9. If u = exyz f ¨ ¸ , prove that
© y ¹
x
wu
wu
y
wx
wy = 2xy zu,
wu
wu
z
= 2xy zu,
wy
wz
Also, reduce that
y
x
10. If u =
x
w2 u
w2 u
= y
wz wx
wz wy
y z x
, show that
z x y
wu
wu
wu
y
z
wx
wy
wz = 0
1.5 Total Differentiation
If u = f(x, y), where x = I(t) and y = \(t), then we find the value of u interms of t. Hence we
can regard u as a function of t alone. Then ordinary differential coefficient of u w.r.t. t, i.e.,
du
is called total differential coefficient of u.
dt
du
Now, to find
without substituting the values of x and y in f(x, y), we establish the
dt
following formula:
wu dx wu dy
du
= wx ˜ dt wy dt
dt
Proof: We have u = f(x, y)
...(1)
Now, giving the increment Gt to t, we suppose that the corresponding increments in x,
y, and u be Gx, Gy and Gu respectively. Then
Partial Differentiation and Partial Differential Equation
7
u + Gu = f(x + Gx, y + Gy)
?
Gu = f(x + Gx, y + Gy) – f(x, y)
= f(x + Gx, y + Gy) – f(x, y + Gy)+ f(x, y + Gy) – f(x, y)
?
du
f ( x dx , y dy ) f ( x , y dy) dx f ( x , y dy) f ( x , y) ˜ dy
˜
=
dy
dt
dt
dx
dt
Now taking limits as Gt o 0, Gx and Gy also o 0, we have
lim f ( x dx , y dy ) f ( x , y dy) dx
du
= dx o 0
d
yo0
dt
dt
dx
f ( x , y dy) f ( x , y ) dy
+ dlim
˜
y o0
dt
dy
=
wf ( x , y ) dx wf ( x , y ) dy
˜
wx
dt
wy
dt
which is the desired formula.
Note: 1. If t = x, then
wu wu dy
du
˜
=
wx wy dx
dx
2. If u = f(x, y, z) and x, y, z all being functions in t, then, we have
du
wu dx wu dy wu dz
˜
=
dt
wx dt wy dt wz dt
3. If f(x, y) = c be an implicit relation between x and y then we have
0=
df
dx
wf wf dy
˜
wx wy dx
dy
wf / wx
=
dx
wf / wy
Ÿ
4. If f(x, y) = 0 then
d2 y
dx 2
= q 2r 2 pqs p2t
q3
Example 1.5
dy
dx
Solution: Let f(x, y) = x3 + 3x2y + 6xy2 + y3 = 1
Then from (1), we have
If x3 + 3x2y + 6xy2 + y3 = 1, find
wf ( x , y)
= 3x2 + 6xy + 6y2
wx
...(1)
8 8
Engineering
Mathematics–III
A Textbook
of Engineering Mathematics-I
and
wf ( x , y)
= 3x2 + 12xy + 3y2
wy
Hence
3( x 2 2xy) 2 y2
dy
x 2 2xy 2 y2
= =
dx
3( x 2 4 xy y2 )
x 2 4 xy y2
Example 1.6
§x·
sin ¨ ¸ , x
© y¹
direct substitution.
Solution: We have
Given u
et and y = t2, find
du
as a function of t. Verify your result by
dt
du
wu dx wu dy
˜
˜
=
dt
wx dt wy dt
§x· t 1
§x·§x·
= cos ¨ y ¸ ˜ e ˜ y cos ¨ y ¸ ¨ 2 ¸ ˜ 2t
© ¹
© ¹© y ¹
du
et
t 2
t 2
t 2
= cos( e / t ) ˜ e / t 2 cos ( e / t ) ˜ 3
dt
t
­ (t 2) ½ t
t 2
= ® 3 ¾ e cos ( e / t )
¯ t
¿
Also
?
§x·
u = sin ¨ ¸
© y¹
§ et ·
sin ¨ 2 ¸
¨t ¸
© ¹
§ et · t 2 . et 2et . t
§ et ·
(t 2) t
du
cos
.
e
cos
¨
¸
=
=
¨¨ t 2 ¸¸
¨ t2 ¸
t4
t3
dt
© ¹
© ¹
EXERCISES
1. If u = x2 – y2 + sin yz, where y = ex, and z = log x, find du/dx.
2. Find du/dx, if u = sin(x2 + y2), where (a2x2 + b2y2) = c2
dy
if (i) ax2 + 2hxy + by2 = 1, (ii) yx + xy = c.
dx
4. If u = x log xy, where x3 + y3 + 3xy = 1, find du/dx.
5. Find the partial differential coefficients of x2y w.r.t. x and y, and its total differential
coefficient w.r.t. x when x and y are connected by the relation x2 + xy + y2 = 1.
3. Find
6. If f(x, y) = 0, I(y, z),= 0, show that
wf wf dz
˜
˜
wy wz dx
wx wf
˜ .
wy wy
Partial Differentiation and Partial Differential Equation
7. If xy = yx, show that
dy
dx
9
y( y x log y)
.
x ( x y log x )
ANSWERS
1. 2x ( 2 y z cos yz ) e x ( y cos yz ) / x
2. 2x {cos ( x 2 y2 )} (1 a 2 / b2 )
3. (i) ( ax hy)/(hx by); (ii) ( y x log y yx y 1 ) / ( xyx 1 xy log x )
4. 1 log xy x ( x 2 y) / y ( x y2 )
5. If f = x2y, then
df
dx
2xy, w 2 f / wx 2
differential coefficients are zero.
wf
wy
2 y,
df
dx
0,
2xy w2 f
wx wy
2x ,
w3 f
wx 2 wy
2 , and all the higher
x 2 (2x y)
.
x 2y
1.6 Homogeneous Functions; Euler’s Theorem
Definition: An expression of the form a0xn + a1xn–1y + a2 xn–2y2 + ... + an yn where each
term is of degree n is called a homogeneous function in degree n. The above expression can
also be written as
­°
x n ®a0 a1
°¯
§ y·
¨ ¸ a2
©x¹
2
§ y·
¨ ¸ ... an
©x¹
n½
§ y·
¨ ¸
©x¹
°
¾ = xn f(y/x)
°¿
If the function f(x1, x2, ..., xm) can be expressed in the form
§x x
x ·
x rn F ¨ 1 , 2 , ..., m ¸ ,
xr ¹
© xr xr
then f(x1, x2, ..., xm ) is called a homogeneous function of m variables in degree n.
1.7 Euler’s Theorem on Homogeneous Functions
If f(x, y) be a homogeneous function of x and y of degree n, then
x
wf
wf
y
wx
wy = n f
Proof: Let f(x, y) = xn F(y/x)
be a homogeneous function in degree n. Then from (1), we have
wf
§ y·
§ y· § y·
n 1
˜ F ¨ ¸ xn F c¨ ¸ ˜ ¨ 2 ¸
= nx
wx
©x¹
©x¹ © x ¹
...(1)
10 10
Engineering
A TextbookMathematics–III
of Engineering Mathematics-I
§ y·
§ y·
n 1
F ¨ ¸ x n 2 ˜ yF c ¨ ¸
= nx
©x¹
©x¹
Also from (1), we have
wf
§ y· 1
x n F c¨ ¸ ˜
=
wy
©x¹ x
...(2)
§y·
x n 1 F c ¨ ¸
©x ¹
...(3)
Now, multiplying (2) by x and (3) by y and then adding, we get
x
wf
wf
§ y·
§ y·
§ y·
y
n x n F ¨ ¸ x n 1 y ˜ F c ¨ ¸ x n 1 ˜ y ˜ F c ¨ ¸
=
wx
wy
©x¹
©x¹
©x¹
§ y·
n
= n ˜ x F ¨ ¸ = n ˜ f ( x , y)
©x¹
Hence the result proved.
In general, if f(x1, x2, .... xm) be a homogeneous function of degree n, then
x1
wf
wf
wf
x2
... xm
wx1
wx2
wxm
n˜ f
Example 1.7
Verify Euler’s Theorem when
f(x, y, z) = axy + byz + czx
Solution: Let f(x, y, z) = axy + byz + czx
then from (1), we have
wf
wf
= ay cz Ÿ x
wx
wx
Again from (1), we have
wf
wf
= ax bz Ÿ y
wy
wy
...(1)
axy czx
...(2)
axy byz
...(3)
wf
wf
bzy czx
= by cx Ÿ z
wy
wz
Then adding (2), (3) and (4), we have
wf
wf
wf
x
2 ( axy byz czx )
y
z
wx
wy
wz
= 2f(x, y, z)
which verifies Euler’s Theorem in this case.
and
...(4)
Example 1.8
If u
­° x 4 y4 ½°
wu
wu
log e ®
y
¾ , show that x
x
y
wx
wy
°¿
¯°
3
(U.P.T.U., 2000)
Partial Differentiation and Partial Differential Equation 11
Solution: We have
­° x 4 y4 ½°
log
¾
e ®
u=
°¯ x y °¿
eu =
Ÿ
x 4 y4
xy
4
­°
§ y · ½°
®1 ¨ ¸ ¾
© x ¹ ¿°
°
3 ¯
eu = x ˜
­
§ y ·½
®1 ¨ ¸ ¾
© x ¹¿
¯
Ÿ
§ y·
x 3 f ¨ ¸ (say) = z
©x¹
Ÿ z is a homogeneous function of degree 3.
? By Euler’s formula, we have
wz
wz
x
y
3˜ z
wx
wy
But
z = eu
wz
u wu
Ÿ
= e
wx
wx
wz
w
u u
and
= e
wy
wy
Then from (1), using (2), we have
Ÿ
...(1)
...(2)
­ wu
wu ½
y
eu ®x
¾ = 3eu
w
wy ¿
x
¯
wu
wu
x
y
wx
wy = 3. Proved.
Example 1.9
If z be a homogeneous function of degree n, show that
(i) x
(ii) x
w2 z
w2 z
wx wy
(n 1)
wz
,
wx
w2 z
w2 z
y 2
wx wy
wy
( n 1)
wz
, and
wy
wx
2
y
2
w2 z
2 w z
y
n( n 1) z.
wx w y
wx 2
wy2
Proof: By Euler’s Theorem, we know that
2
(iii) x
w2 z
2xy
x
wz
wz
y
wx
wy = n . z
(i) Differentiating (1) partially w.r.t. x, we get
...(1)
12 12
Engineering
A TextbookMathematics–III
of Engineering Mathematics-I
wz
w2 z
w2 z
wz
x 2 y˜
= n˜
w
x
wx
w
x
w
y
wx
Ÿ
w2z
wz
= (n 1)
2
wx wy
wx
wx
(ii) Again differentiating (1) w.r.t. y, we get
x
x
Ÿ
w2z
y
...(2)
w2 z
w 2 z wz
wz
y 2 = n˜
wx wy
wy
wy
wy
wz
w2 z
w2 z
y 2 = (n 1)
Proved.
wy
wx wy
wy
(iii) Multiplying (2) by x and (3) by y and then adding, we get
x
x2
w2 z
wx 2
2xy
...(3)
­ wz
wz ½
w2 z
w2z
y ¾
y2 2 = ( n 1) ® x
wy ¿
wx w y
wy
¯ wx
= (n – 1) n . z
= n(n – 1)z Proved.
EXERCISES
1. Verify Euler’s Theorem in the following cases:
(i) 3x2yz + 5xy2z + 4z4 = f(x, y, z)
(ii) f(x, y) = ax2 + 2hxy + by2
(iii) u = (x1/4 + y1/4)/(x1/5 + y1/5)
(iv) u = x2(x2 – y2)3/(x2 +y2)3
2. If u
log
wu
wu
x 2 y2
y
, prove that x
wx
wy
xy
3. If u = sin–1{(x2 + y2)/(x + y)}, show that x
4. If cos u = ( x y ) /( x 5. If u
1.
wu
wu
y
wx
wy
y ) , prove that x
tan u.
wu
wu 1
y
cot u
wx
wy 2
3
3
°­ x y °½
tan1 ®
¾ , prove that
¯° x y °¿
wu
wu
sin 2u
y
wx
wy
(b) x2 uxx + 2xy uxy + y2 yyy = 2cos 3u sin u
(a) x
6. If u = x3 + y3 + z3 + 3xyz, show that x
wu
wu
wu
y
z
wx
wy
wz
3u
0.