Dr. P.K. Srivastava Assistant Professor of Mathematics Galgotia College of Engineering & Technology Greater Noida (U.P.)

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1 Engineering Mathematics-III Dr. P.K. Srivastava Assistant Professor of Mathematics Galgotia College of Engineering & Technology Greater Noida (U.P.) (An ISO 9001:008 Certified Company) Vayu Education of India /5, Ansari Road, Darya Ganj, New Delhi

2 Engineering Mathematics-III Copyright VAYU EDUCATION OF INDIA ISBN: First Edition: 013 Price: 160/- All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted, in any form or by any means, electronic, mechanical, photocopying, recording or otherwise, without the prior permission of the Author and Publisher. Printed & bound in India Published by: (An ISO 9001:008 Certified Company) Vayu Education of India /5, Ansari Road, Darya Ganj, New Delhi Ph.: , Fa: vei@veiindia.com Web:

3 Contents 1. Partial Differentiation and Partial Differential Equation Partial Differential Equations Fourier Series Laplace Transformation Numerical Techniques Numerical Methods for Solution of Partial Differential Equation

4 Chapter Functions of Two or More Variables A symbol z which has a definite value for every pair of values of and y is called a function of two independent variables and y and is written as z f (, y) or (, y) 1. Limits The function f (, y) is said to tend to limit l as a and y b if and only if the limit l is independent of the path followed by the point (, y) as a and y b and we write lim a y b f (, y ) l Or, in circular neighbourhood we define the limit as: The function f (, y) defined in a region R, is said to tend to the limit l as a and y b if and only if corresponding to a positive number, another positive number such that fy (, ) l e for 0 < ( a) + (y b) < for every point (, y) in R. 1.3 Continuity A function f(, y) is said to be continuous at the point (a, b) if lim a y b f (, y ) eists f(a, b) 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 (, y) be variables and y. Then partial differentiation of z w.r.t. keeping y as a constant is denoted as z f(, y),, f ( y, ), D fy (, )

5 Engineering A Tetbook Mathematics III of Engineering Mathematics-I z where f( d, y) f(, y) lim d 0 d Similarly, the partial differentiation of z w.r.t. y keeping as a constant is denoted as z f(, y),, fy ( y, ), Dy fy (, ) y y In general, f and fy being functions of and y, so these can be further differentiated partially w.r.t. and y and thus we have z z f or or f d z z f or y y or fy z f z or d y y y or fy z z or f y y y or fyy It can be verified easily that z z y y Also, we can use the following notations p z, q z y z z y z y If z be a function of number of variables say 1,,... n ; then its partial derivatives w.r.t. to one of the variables say 1 is denoted as z, keeping others as constant. 1 Eample 1.1 If z ( + y) + y, show that z z z z 4 1 y y Solution: z ( + y) + y Now z y y z ( y) ( y ).1 ( y) y y ( y)

6 Partial Differentiation and Partial Differential Equation 3 Also, z z y z z y Also, we have z y ( y). y ( y ).1 ( y) y y y y ( y) y y 4( y) ( y) y y ( y) z z y y y y y ( y) ( y) y y 4 ( y) Hence from (1) and (), we have y y y y y y 4 ( y) ( y) 4. ( y )...(1)...() z z y z 41 z y Eample 1. 3 u If u e yz, find the value of yz Solution: Let u e yz...(1) u then ye yz z u y z yz yz e ( ) e ( z)( y) u y z 3 u yz yz e ( yz) yz e (1 yz) ( yz) yz. e e yz (1 + 3yz + y z ) yz

7 4 4 Engineering A Tetbook Mathematics III of Engineering Mathematics-I Eample 1.3 If u f (y / ), show that u u y d 0 Solution: Let u f(y / )...(1) u Now f (y / ). y / d u y f ( y / ) d...() Also, from (1), we have u 1 f( y / ). u y y f ( y / )...(3) Now, adding () and (3), we get u u y 0 d Eample 1.4 If u log ( 3 + y 3 + z 3 3yz), show that 9 u y z ( y z) Solution: Let u log ( 3 + y 3 + z 3 3yz)...(1) then from (1), we have u yz d y z 3yz 3 y z y z 3yz...() Similarly, we have u 3 y z y z 3yz...(3) and u 3 z y z y z 3yz...(4) By adding (), (3) and (4), we get u u u 3 y z y yz z y z y z 3yz

8 Partial Differentiation and Partial Differential Equation 5 u y z y z u 3 y z y yz z y z y z y yz z 3 y z 3 y z y z y z y y z z y z ( y z) ( y z) ( y z) 9 ( y z) EXERCISES 1. If y y z z c, show that of y z, z ( log e) 1 y. If V ( + y + z ) 1/, we show that V V V y z 0 3. If u tan 1 y, show that (1 y ) u y 1 1 y 4. If z f ( + ay) + f ( ay), prove that y z a 5. If u sin 1 { / y} + tan 1 {y / } then find the value of u u y 0 y 6. If z e a + by. f (a by), prove that z 3/

9 6 6 Engineering A Tetbook Mathematics III of Engineering Mathematics-I z z b a y abz 7. If z tan 1 (y / ) y tan 1 ( / y), prove that z y y y 8. If u log ( + y ) + tan 1 {y / }, show that 9. If u e yz f u u y 0 z, prove that y u u y y u u y z y z Also, reduce that u y zu, y zu, u z y z y y z 10. If u, show that z y u u u y z y z Total Differentiation If u f(, y), where (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. Now, to find du without substituting the values of and y in f(, y), we establish the following formula: du u d u y Proof: We have u f(, y)...(1) Now, giving the increment t to t, we suppose that the corresponding increments in, y, and u be, y and u respectively. Then

10 Partial Differentiation and Partial Differential Equation 7 u + u f( +, y + y) u f( +, y + y) f(, y) f( +, y + y) f(, y + y)+ f(, y + y) f(, y) du f( d, y ) f(, y ) d d Now taking limits as t 0, and y also 0, we have du lim f ( d, y ) f(, y ) d d 0 0 d which is the desired formula. Note: 1. If t, then + lim d y 0 f(, y ) f(, y) f(, y) d f(, y) y du d u u y d. If u f(, y, z) and, y, z all being functions in t, then, we have du ud u u dz y z 3. If f(, y) c be an implicit relation between and y then we have df f f 0 d y d d f / f / y 4. If f(, y) 0 then Eample 1.5 d y d qrpqs pt 3 q f (, y ) f(, y) If y + 6y + y 3 1, find d Solution: Let f(, y) y + 6y + y (1) Then from (1), we have f (, y) 3 + 6y + 6y

11 8 8 Engineering A Tetbook Mathematics III of Engineering Mathematics-I and f (, y) y 3 + 1y + 3y Hence d 3( y) y 3( 4 y y ) y y 4y y Eample 1.6 Given u sin, e y direct substitution. Solution: We have du t and y t, find du u d u y as a function of t. Verify your result by du 1 cos cos t y y y y e t t t t t e 3 cos( e / t ) e / t cos ( e / t ) t ( t ) t t e cos ( e / t ) 3 t t e Also u sin sin y t du t t t e t. e e. t cos. 4 t t t ( t ) t e e cos 3 t t EXERCISES 1. If u y + sin yz, where y e, and z log, find du/d.. Find du/d, if u sin( + y ), where (a + b y ) c 3. Find d if (i) a + hy + by 1, (ii) y + y c. 4. If u log y, where 3 + y 3 + 3y 1, find du/d. 5. Find the partial differential coefficients of y w.r.t. and y, and its total differential coefficient w.r.t. when and y are connected by the relation + y + y 1. f f dz f 6. If f(, y) 0, (y, z), 0, show that. y z d y y

12 Partial Differentiation and Partial Differential Equation 9 7. If y y, show that y( y log y). d ( y log ) 1. ( y zcos yz) e ( ycos yz)/. {cos( y )}(1 a / b ) ANSWERS y (i) ( a hy)/( h by); (ii) ( y log y y )/ ( y ylog ) 4. 1 log y ( y) / y ( y ) 5. If f df f y, then y, f / y, 0, d y f, y 3 f, and all the higher y differential coefficients are zero. df ( ) y y. d y 1.6 Homogeneous Functions; Euler s Theorem Definition: An epression of the form a 0 n + a 1 n 1 y + a n y a n y n where each term is of degree n is called a homogeneous function in degree n. The above epression can also be written as n n y y y a0 a1 a... an n f(y/) If the function f( 1,,..., m ) can be epressed in the form n 1 m r F,,...,, r r r then f( 1,,..., m ) is called a homogeneous function of m variables in degree n. 1.7 Euler s Theorem on Homogeneous Functions If f(, y) be a homogeneous function of and y of degree n, then f f y y n f Proof: Let f(, y) n F(y/)...(1) be a homogeneous function in degree n. Then from (1), we have f n1 y n y y n F F

13 10 10 Engineering A Tetbook Mathematics III of Engineering Mathematics-I Also from (1), we have f y n1 y n y n F yf n y 1 n1 y F F...()...(3) Now, multiplying () by and (3) by y and then adding, we get f f n y n1 y n1 y y y n F yf yf n y n F n f(, y) Hence the result proved. In general, if f( 1,,... m ) be a homogeneous function of degree n, then Eample 1.7 f f f... m nf 1 1 Verify Euler s Theorem when f(, y, z) ay + byz + cz Solution: Let f(, y, z) ay + byz + cz...(1) then from (1), we have f f ay cz ay cz...() Again from (1), we have f f a bz y ay byz...(3) y y m f f and by c z bzy cz z y Then adding (), (3) and (4), we have f f f y z ( ay byz cz) y z f(, y, z) which verifies Euler s Theorem in this case. Eample (4) If 4 4 y u log e, y u u show that y 3 y (U.P.T.U., 000)

14 Partial Differentiation and Partial Differential Equation 11 Solution: We have e u y u log e y 4 4 y y y 1 e u 3 3 y f (say) y z 1 z is a homogeneous function of degree 3. By Euler s formula, we have z z y 3 z y But z e u z u u e and z u u e y y Then from (1), using (), we have u u u e y y u u y y 3. Proved. Eample 1.9 If z be a homogeneous function of degree n, show that (i) z z z y ( n 1), y (ii) z z z y ( n 1), and y y y z z z (iii) y y n( n 1) z. y y Proof: By Euler s Theorem, we know that...(1)...() z z y y n.z...(1) (i) Differentiating (1) partially w.r.t., we get

15 1 1 Engineering A Tetbook Mathematics III of Engineering Mathematics-I z z y z y z z y y z n z ( n 1)...() (ii) Again differentiating (1) w.r.t. y, we get z y z z z n y y y y z z z y ( n 1) y y y Proved....(3) (iii) Multiplying () by and (3) by y and then adding, we get z z z y y y y z z ( n 1) y y (n 1) n. z n(n 1)z Proved. EXERCISES 1. Verify Euler s Theorem in the following cases: (i) 3 yz + 5y z + 4z 4 f(, y, z) (ii) f(, y) a + hy + by (iii) u ( 1/4 + y 1/4 )/( 1/5 + y 1/5 ) (iv) u ( y ) 3 /( +y ) 3 y u u. If u log, prove that y 1. y y 3. If u sin 1 {( + y )/( + y)}, show that u u y tan u. y 4. If cos u ( y)/( y), prove that u u 1 y cot u 0. y y 5. If u tan, prove that y u u (a) y sinu y (b) u + y u y + y y yy cos 3u sin u 6. If u 3 + y 3 + z 3 u u u + 3yz, show that y z 3u y z

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