Method of Laplace Transform - Mathematics Methods in Physics - Solved Exam, Exams for Mathematical Physics. Aligarh Muslim University
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saidullah20 February 2013

Method of Laplace Transform - Mathematics Methods in Physics - Solved Exam, Exams for Mathematical Physics. Aligarh Muslim University

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This is the Solved Exam of Mathematics Methods in Physics which includes Inverse Laplace Transform, Fourier Cosine Transform, System of Units, Electromagnetic Stress Tensor, First-Order Tensors, Electric and Magnetic Fie...
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AY2008/2009 [PC3274 MATHEMATICAL METHODS IN PHYSICS II SOLUTIONS]

1

a)  + 2 +  =   −  −  + 2  −  +  =   + 2 +   =  + + 1  +   =  +   + + 1 +    + 

 +  

 =  +   + 1 −   +  

+   +  +   =  −        +  1 −   +   + 

b)  +  = sin  −  0 −  0 +  = −  ! 1  + 1"  − + 1 +  = 2  + 1   +   = 2  + 1  + 1 − 1  = 2 − 2  + 1 − 2  + 1  + 1 − 1  = 3 − 1  − 2  + 1 − 2  + 1 

 + 2 +  =  .

 +  = sin ,

 +  −      = 3 − − 2 cos − sin .

Question 1

(a) Use the method of Laplace transform to find a particular integral of the

equation

(b) Show that the non-linear differential equation

given that  0 = 1 and  0 = −1 , can be written as the integral equation

AY2008/2009 [PC3274 MATHEMATICAL METHODS IN PHYSICS II SOLUTIONS]

2

∴  +  −      = 3 − − 2 cos − sin

a) If * has order g, then the order of the subgroups must divide g. Given X & Y belong to * and +~-, then +, -, +.- belong to ℋ. We let - = +ℋ0 for some element ℋ0. Each coset must have distinct elements such that ℋ must have h elements.

Since each member must only be in one set, then g is a multiple of h. [proven]

. 1 = 1,  1 = 11 − 1 , 2 1 = 1 − 11 , 3 1 = 11 , 4 1 = 1 − 1, 5 1 = 11 − 1

676 − 1 ! 676" = 0 6761 + 1 ! 676 − 7" = 0.

Question 2

(a) Show that if * is a finite group of order g, and ℋ is a subgroup of * and of order h, then g is a multiple of h.

(b) Show that the following set of six functions,

with the law of combination as 0 1 • 9 1 = 0:9 1 ; forms a non-Abelian group. Determine the order of each element in the group.

(c) Show that the Euler-Lagrange equation

can be written as

where 7 = 7 , , 1 .

AY2008/2009 [PC3274 MATHEMATICAL METHODS IN PHYSICS II SOLUTIONS]

3

b) The table below shows that the operation is closed. Functions are always associative,

the identity exists, and it is . 1 ; every element has an inverse, and the table is not symmetric. ∴ the set of functions form a non-Abelian group.

. 1  1 2 1 3 1 4 1 5 1 . 1 1 11 − 1 1 − 11 11 1 − 1 11 − 1  1 11 − 1 1 − 11 1 11 − 1 11 1 − 1 2 1 1 − 11 1 11 − 1 1 − 1 11 − 1 11 3 1 11 1 − 1 11 − 1 1 11 − 1 1 − 11 4 1 1 − 1 11 − 1 11 1 − 11 1 11 − 1 5 1 11 − 1 11 1 − 1 11 − 1 1 − 11 1

Elements of

Order 1: . 1 Order 2:  1 , 2 1 Order 3: 3 1 , 4 1 , 5 1

c) We know that 71 = 6761 + 676  + 676  676 = 1 71 − 

  676 − 1 6761

then 676 − 1 ! 676" = 0 1 71 − 

  676 − 1 6761 − 1 ! 676" = 0 71 − < 676 +  1 ! 676"= − 6761 = 0

We also find that 1 ! 676" =  676 +  1 ! 676" substituting back, we have

AY2008/2009 [PC3274 MATHEMATICAL METHODS IN PHYSICS II SOLUTIONS]

4

71 − 1 ! 676" − 6761 = 0 6761 − 1 ! 676 − 7" = 0  ℎ[email protected]

7 =  + B − 41B − 4B + D  − 1 − B 1 676 = 676 1 2 + 2D − D1 = 0 2 + 2D − D1 = E

 = 12 + 2D E + D1  = 12 1 + D FE1 + D1

 2 + GH

using the boundary conditions,

 0 = 0,  1 = 1, 0 = G2 1 + D ⇒ G = 0 1 = 12 1 + D !E + D2" 4 + 4D = 2E + D E = 4 + 3D2

1 676B = 676B 1 2B − 41 − 2DB = −4 2 − 2D B − 41 = −41 +  B = 2 − 2D B = 2 − 2D 1 +  using the boundary conditions,

B 0 = 0, B 1 = 1,  = 0,  = 2 − 2D ∴ B = 1

J =   + B − 41B − 4B . 1

  − 1 − B . 1 = 2,

Question 3

Find the extremal of the following functional

subjected to the constraint

given that  0 = 0, B 0 = 0,  1 = 1 and B 1 = 1 . Calculate the corresponding value of the integral I.

AY2008/2009 [PC3274 MATHEMATICAL METHODS IN PHYSICS II SOLUTIONS]

5

∴  = 12 1 + D !4 + 3D2 1 + D2 1" = 14 D1 + D 1 + 14 !4 + 3D1 + D " 1 = 14 1 − 1 + D1 + D 1 + 14 !1 + 3 + 3D1 + D " 1 = 14 !1 − 11 + D" 1 + 14 !3 + 11 + D" 1

substituting K = ..LM, we have  = 14 1 − K 1 + 14 3 + K 1, B = 1  = 12 1 − K 1 + 14 3 + K , B = 1   − 1 − B . 1 = 2  F<12 1 − K 1 + 14 3 + K =

 − 1 <12 1 − K 1 + 14 3 + K = − 1H .

 1 = 2  !14 1 − K 1 + 14 1 − K 3 + K 1 + 116 3 + K  − 12 1 − K 1 − 14 3 + K 1 − 1"

.  1 = 2

 4 K − 1 1 − 4K 3 + K 1 + 3 + K  − 16 . 1 = 32 <43 K − 1 12 − 2K 3 + K 1 +  3 + K  − 161=

. = 32 4 K − 1 − 6K 3 + K + 3 3 + K  − 48 = 96 4K − 4 − 18K − 6K + 27 + 18K + 3K − 48 − 96 = 0 K − 121 = 0, K = ±11, D = 1±11 − 1 = − 1011 , − 1211 and we have

 = − 52 1 + 72 1 ?T  = 31 − 21  = −51 + 72 ?T  = 61 − 2 For  = − 4 1 + U 1, J =   + B − 41B − 4B . 1

=  !−51 + 72"  + 1 − 81. 1

AY2008/2009 [PC3274 MATHEMATICAL METHODS IN PHYSICS II SOLUTIONS]

6

= V− 115 !−51 + 72" 2 + 1 − 41W

.

= − 11140 + 343120 = 112 For  = 31 − 21, J =   + B − 41B − 4B . 1

=  61 − 2  + 1 − 81. 1 = < 118 61 − 2 2 + 1 − 41=

.

= − 599 − 49 = −7

a)

i.

 E = X 12Y  sin 1 0Z[ \

 1

 1 = ]sin 1 , 0 < 1 < Y0, ? ℎ[email protected]_  ̀

 1 = 1Y  cos E1 + cos E 1 − Y 1 − E a

 E

bT 12 cd ∙ cd = cd × GTg cd + cd ∙ bT cd.

Question 4

(a)

i. Represent the following function as an exponential Fourier

transform

ii. Show that your results can be written as

(b) Use the method of tensor to establish the following vector identity,

AY2008/2009 [PC3274 MATHEMATICAL METHODS IN PHYSICS II SOLUTIONS]

7

= X 12Y V  0Z[

E − 1 _E sin 1 + cos 1 W \

= X 12Y F 0Z\ + 11 − E H

ii.

 1 = X 12Y   E 0Z[E a

a

= X 12Y  X 12Y F 0Z\ + 11 − E H 0Z[E

a a

= 12Y   0Z [\ + 0Z[1 − E

a a E

= 1Y  cos E1 + cos E 1 − Y 1 − E a

 E

b)

cd × ∇ × cd + cd ∙ ∇ cd = i09jijkl9 6l61k + 9 60619 = mn0kn9l − n0ln9ko9 6l61k + 9 60619 = 9 69610 − 9 60619 + 9 60619 = 9 69610 = 12 F9 69610 + 69610 9H = 12 6610 m99o = 12 ∇ cd ∙ cd

Solutions provided by: A/Prof Paul Lim (Q1, Q4) and John Soo (Q2, Q3)

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