1000 Solved Problems in Modern Physics

(Grace) #1

48 1 Mathematical Physics


on settingk=1 andλ 1 = 5

− 2 C 11 + 2 C 21 =0(2)
4 C 11 − 4 C 21 =0(3)

ThusC 21 =C 11 =a= 1
The substitution ofk=2 andλ 2 =−1 yields
(3+1)C 12 + 2 C 22 = 0
4 C 12 + 2 C 22 = 0
orC 22 =− 2 C 12
We may setC 12 =1 so thatC 22 =− 2
ThusC=

(

11

1 − 2

)

(modal matrix)
The inverse ofCis easily found to be

C−^1 =

( 2

3

1
3
1
3 −

1
3

)

Eigen vectors:

(

a 11 −λ a 12
a 21 a 22 −λ

)(

x 1
x 2

)

= 0

Putλ=λ 1 =5;

(

3 − 52

41 − 5

)(

x 1
x 2

)

= 0 →− 2 x 1 + 2 x 2 = 0 →x 1 =x 2

The normalized invariant vector is√^12

(

1

1

)

Putλ=λ 2 =−1;

(

3 −(−1) 2

41 −(−1)

)(

x 1
x 2

)

= 0 → 4 x 1 + 2 x 2 = 0 →
x 2 =− 2 x 1
The second invariant eigen normalized eigen vector is√^15

(

1

− 2

)

(d)C−^1 AC=

(

2 / 31 / 3

1 / 3 − 1 / 3

)(

32

41

)(

11

1 − 2

)

=

(

50

0 − 1

)

1.3.5 MaximaandMinima


1.33 Letf=y=x^3 − 3 x+ 3 = 0
Let the root bea
Ifx=a=− 2 ,y=+ 1
Ifx=a=− 3 ,y=− 15
Thusx=alies somewhere between−2 and−3.
Forx=a− 2 .1,y= 0 .039, which is close to zero.
Assume as a first approximation, the root to bea=v+h
Putv=− 2. 1

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