So that S.121] =/dx ((21)²-2²). S₂21²2] = √((22²). S[21, 22] = S1[21] + S2[22]. Hence the resulting 'uncoupled' EL equations are given by z₁+10z₁=0,₂ = 0. how come ?
So that S.121] =/dx ((21)²-2²). S₂21²2] = √((22²). S[21, 22] = S1[21] + S2[22]. Hence the resulting 'uncoupled' EL equations are given by z₁+10z₁=0,₂ = 0. how come ?
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
Related questions
Question
![Problem
Find two coupled Euler-Lagrange equations for the functional
S[v₁, v2] = * dx ((vj)² + (v₂)² – (v₁ + 3y2)²).
Using the linear transformations
Changing Dependent Variables
Z1 = y1+3y2, Z2 = a1 + y2
find the value of the constant a such that the functional can be written as
the sum of two functionals: one depending only on 2₁, the other only on
Z2.
(M820 Tutorial 2)
Hence
Changing Dependent Variables
S[y₁, y₂] =
-S
From the first part we have y + (y1 + 3y2) = 0, y₂ +3(y1 + 3y2) = 0.
Using the linear transformation
Z₁ = y₁ + 3y2₁ Z₂ = 2y₂ + y₂
we have can write y₁ and y2 In terms of 2₁ and 22, provided the matrix
Rearranging leads to
(M820 Tutorial 2)
S[Z₁, Z₂] =
S[Z1, Z2] =
dx ((v₂)² + (v₂)²-(y+3y2)²).
Y₁ =
Z1 - 322
1-3a
y₂ =
17th January 2024
32₁-22
32-1
- / * (( 4 = ³ ) ² + ( * = ³ ) ² - 4).
dx
1-3a
− / dx ( ( (2, ³² + 9(2₂)³² + 9(z;)³² + (2₂)³²)
100
13
a 1
The only terms involving both dependant variables and their derivatives are the ziz₂ terms:
-6z/z₂-2aziz.
Taking a = -3 (note that then 1-320) kills these 'cross terms', and we have
Is Invertible.
-
- zí
17th January 2024
3/25
5/25
Changing Dependent Variables
S[y₁, y2] = dx ((y₁)² + (⁄₂)² − (y₁ + 3y₂)²).
We use HB 4.2 pg 17: The EL equations are
d OF
dx ay
Hence
(M820 Tutorial 2)
So that
(294)-(
The solutions are
Hence the two required functionals are
Changing Dependent Variables
) − (−2(y₁ + 3y2)) = 0
Y₁ + (y₁ + 3y₂) = 0, and
=
OF
Byk
(22) + 6(y₁ + 3y2) = 0
y₂ + 3(y₁ + 3y₂) = 0.
k = 1, 2.
5₁1²₁] = √((2)²2-2²),
$₂122] = /dx (16 (²3)²).
dx
S[Z1, Z2] S1 [21] + S2[22].
Hence the resulting 'uncoupled' EL equations are given by
z₁+10z₁ = 0, Z₂ = 0.
COS V
Z₁ = A. cos √ 10x + B. sin✓/10x, Z₂ = Cx + D.
Translating these back to give solutions in terms of the original dependent variables gives
1
y₁ = (21-322)
10
1
=((A. cos ✓/10x + B. sin ✓/10x) - 3(Cx+D))
A
10
B
10x +
sin ✓10x
10
3C
17th January 2024
how come
?
3D
10* 10
1/25
Here I'd absorb the constant factors and write this as y₁ = A cos√10x + B sin√10x + Cx+D.
(M820 Tutorial 2)
17th January 2024
6/25](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F59d44c96-efb1-4f3c-83b3-5a6a84cf94cb%2Fa1edefab-bcfd-4f10-bfa8-b962a7d4e962%2Fd20xmqb_processed.png&w=3840&q=75)
Transcribed Image Text:Problem
Find two coupled Euler-Lagrange equations for the functional
S[v₁, v2] = * dx ((vj)² + (v₂)² – (v₁ + 3y2)²).
Using the linear transformations
Changing Dependent Variables
Z1 = y1+3y2, Z2 = a1 + y2
find the value of the constant a such that the functional can be written as
the sum of two functionals: one depending only on 2₁, the other only on
Z2.
(M820 Tutorial 2)
Hence
Changing Dependent Variables
S[y₁, y₂] =
-S
From the first part we have y + (y1 + 3y2) = 0, y₂ +3(y1 + 3y2) = 0.
Using the linear transformation
Z₁ = y₁ + 3y2₁ Z₂ = 2y₂ + y₂
we have can write y₁ and y2 In terms of 2₁ and 22, provided the matrix
Rearranging leads to
(M820 Tutorial 2)
S[Z₁, Z₂] =
S[Z1, Z2] =
dx ((v₂)² + (v₂)²-(y+3y2)²).
Y₁ =
Z1 - 322
1-3a
y₂ =
17th January 2024
32₁-22
32-1
- / * (( 4 = ³ ) ² + ( * = ³ ) ² - 4).
dx
1-3a
− / dx ( ( (2, ³² + 9(2₂)³² + 9(z;)³² + (2₂)³²)
100
13
a 1
The only terms involving both dependant variables and their derivatives are the ziz₂ terms:
-6z/z₂-2aziz.
Taking a = -3 (note that then 1-320) kills these 'cross terms', and we have
Is Invertible.
-
- zí
17th January 2024
3/25
5/25
Changing Dependent Variables
S[y₁, y2] = dx ((y₁)² + (⁄₂)² − (y₁ + 3y₂)²).
We use HB 4.2 pg 17: The EL equations are
d OF
dx ay
Hence
(M820 Tutorial 2)
So that
(294)-(
The solutions are
Hence the two required functionals are
Changing Dependent Variables
) − (−2(y₁ + 3y2)) = 0
Y₁ + (y₁ + 3y₂) = 0, and
=
OF
Byk
(22) + 6(y₁ + 3y2) = 0
y₂ + 3(y₁ + 3y₂) = 0.
k = 1, 2.
5₁1²₁] = √((2)²2-2²),
$₂122] = /dx (16 (²3)²).
dx
S[Z1, Z2] S1 [21] + S2[22].
Hence the resulting 'uncoupled' EL equations are given by
z₁+10z₁ = 0, Z₂ = 0.
COS V
Z₁ = A. cos √ 10x + B. sin✓/10x, Z₂ = Cx + D.
Translating these back to give solutions in terms of the original dependent variables gives
1
y₁ = (21-322)
10
1
=((A. cos ✓/10x + B. sin ✓/10x) - 3(Cx+D))
A
10
B
10x +
sin ✓10x
10
3C
17th January 2024
how come
?
3D
10* 10
1/25
Here I'd absorb the constant factors and write this as y₁ = A cos√10x + B sin√10x + Cx+D.
(M820 Tutorial 2)
17th January 2024
6/25
Expert Solution
This question has been solved!
Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.
Step by step
Solved in 4 steps with 14 images
Recommended textbooks for you
Advanced Engineering Mathematics
Advanced Math
ISBN:
9780470458365
Author:
Erwin Kreyszig
Publisher:
Wiley, John & Sons, Incorporated
Numerical Methods for Engineers
Advanced Math
ISBN:
9780073397924
Author:
Steven C. Chapra Dr., Raymond P. Canale
Publisher:
McGraw-Hill Education
Introductory Mathematics for Engineering Applicat…
Advanced Math
ISBN:
9781118141809
Author:
Nathan Klingbeil
Publisher:
WILEY
Advanced Engineering Mathematics
Advanced Math
ISBN:
9780470458365
Author:
Erwin Kreyszig
Publisher:
Wiley, John & Sons, Incorporated
Numerical Methods for Engineers
Advanced Math
ISBN:
9780073397924
Author:
Steven C. Chapra Dr., Raymond P. Canale
Publisher:
McGraw-Hill Education
Introductory Mathematics for Engineering Applicat…
Advanced Math
ISBN:
9781118141809
Author:
Nathan Klingbeil
Publisher:
WILEY
Mathematics For Machine Technology
Advanced Math
ISBN:
9781337798310
Author:
Peterson, John.
Publisher:
Cengage Learning,
