(b) With the given functional S[y] and constraint C[y], if A is the Lagrange multiplier, then the auxiliary functional is 1 2 S[y] − \C\[v] = [^* dx {ry'² + ½ y² – Axy²}, y(0) = y(1) = 0. The Euler-Lagrange equation is (2xy') - 2½y + A2xy = 0, IC which simplifies to 1 xy" + y′ −=−y + λxy = 0, y' and hence I x²y" + xy' + (λx² – 1)y = 0, Ans with boundary conditions y(0) = y(1) = 0, as required. - (c) Consider the trial function z(x; a) = ax(1 − x), where a is a parameter. Using the Rayleigh-Ritz method in the theory of Sturm-Liouville systems, the smallest eigenvalue λ₁ is given by \₁ = inf S[y], where the infimum is taken over all admissible functions y with C[y] = 1. Now, 1 = dx xz² = a² S dx 2³ (1-2)² 26 dx (x³ – 2x² + x³) hence a² = 60. = = 2 = a ²² ( 1 − } + }) 60 - Then λ1 ≤ S[z] = a² - dx [x(1 − 2x)²+x(1 − x)²] - =-15. 4 -how come of this formula?? 00 (b) (i) the eigenvalue problem 1/3/2 y = 0, y(−1) = y(1) = 0, x²y"+2xy'+(3x² + λ (x² + with eigenvalue λ, can be written as a constrained variational problem with functional 1 S[y] = = [ { -1 and constraint C[{y}] = [" dx (x²y'² — 3x²y²) dx y2 (1+x2)3 with boundary conditions y(−1) = y(1) = 0. = 1, (ii) Assume now that the system in part (b)(i) has eigenvalues > with λk < λk+1, and eigenfunctions yk(x), k = 1, 2, … · Using the trial function z(x; A) = A(1 − x²) show that λι Σ 32 7π' - briefly outlining the ideas of any theory you use: detailed proofs are not required. You may use the integral dx (1 - x²)² Lidz (1 + 23) 3 -1 Π 4
(b) With the given functional S[y] and constraint C[y], if A is the Lagrange multiplier, then the auxiliary functional is 1 2 S[y] − \C\[v] = [^* dx {ry'² + ½ y² – Axy²}, y(0) = y(1) = 0. The Euler-Lagrange equation is (2xy') - 2½y + A2xy = 0, IC which simplifies to 1 xy" + y′ −=−y + λxy = 0, y' and hence I x²y" + xy' + (λx² – 1)y = 0, Ans with boundary conditions y(0) = y(1) = 0, as required. - (c) Consider the trial function z(x; a) = ax(1 − x), where a is a parameter. Using the Rayleigh-Ritz method in the theory of Sturm-Liouville systems, the smallest eigenvalue λ₁ is given by \₁ = inf S[y], where the infimum is taken over all admissible functions y with C[y] = 1. Now, 1 = dx xz² = a² S dx 2³ (1-2)² 26 dx (x³ – 2x² + x³) hence a² = 60. = = 2 = a ²² ( 1 − } + }) 60 - Then λ1 ≤ S[z] = a² - dx [x(1 − 2x)²+x(1 − x)²] - =-15. 4 -how come of this formula?? 00 (b) (i) the eigenvalue problem 1/3/2 y = 0, y(−1) = y(1) = 0, x²y"+2xy'+(3x² + λ (x² + with eigenvalue λ, can be written as a constrained variational problem with functional 1 S[y] = = [ { -1 and constraint C[{y}] = [" dx (x²y'² — 3x²y²) dx y2 (1+x2)3 with boundary conditions y(−1) = y(1) = 0. = 1, (ii) Assume now that the system in part (b)(i) has eigenvalues > with λk < λk+1, and eigenfunctions yk(x), k = 1, 2, … · Using the trial function z(x; A) = A(1 − x²) show that λι Σ 32 7π' - briefly outlining the ideas of any theory you use: detailed proofs are not required. You may use the integral dx (1 - x²)² Lidz (1 + 23) 3 -1 Π 4
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
![(b) With the given functional S[y] and constraint C[y], if A is the Lagrange
multiplier, then the auxiliary functional is
1
2
S[y] − \C\[v] = [^* dx {ry'² + ½ y² – Axy²}, y(0) = y(1) = 0.
The Euler-Lagrange equation is
(2xy') - 2½y + A2xy = 0,
IC
which simplifies to
1
xy" + y′ −=−y + λxy = 0,
y'
and hence
I
x²y" + xy' + (λx² – 1)y = 0,
Ans
with boundary conditions y(0) = y(1) = 0, as required.
-
(c) Consider the trial function z(x; a) = ax(1 − x), where a is a parameter.
Using the Rayleigh-Ritz method in the theory of Sturm-Liouville
systems, the smallest eigenvalue λ₁ is given by
\₁ = inf S[y],
where the infimum is taken over all admissible functions y with C[y] = 1.
Now,
1
=
dx xz² = a²
S
dx 2³ (1-2)²
26
dx (x³ – 2x² + x³)
hence a² = 60.
=
=
2
= a ²² ( 1 − } + })
60
-
Then
λ1 ≤ S[z] = a²
-
dx [x(1 − 2x)²+x(1 − x)²]
-
=-15.
4
-how
come
of this
formula??
00](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa6c8ed7d-75cc-4e27-869e-3ad6a1efc0b4%2F35a7aa85-5438-41df-8127-37fe866f19d8%2Fwe0b7q1_processed.png&w=3840&q=75)
Transcribed Image Text:(b) With the given functional S[y] and constraint C[y], if A is the Lagrange
multiplier, then the auxiliary functional is
1
2
S[y] − \C\[v] = [^* dx {ry'² + ½ y² – Axy²}, y(0) = y(1) = 0.
The Euler-Lagrange equation is
(2xy') - 2½y + A2xy = 0,
IC
which simplifies to
1
xy" + y′ −=−y + λxy = 0,
y'
and hence
I
x²y" + xy' + (λx² – 1)y = 0,
Ans
with boundary conditions y(0) = y(1) = 0, as required.
-
(c) Consider the trial function z(x; a) = ax(1 − x), where a is a parameter.
Using the Rayleigh-Ritz method in the theory of Sturm-Liouville
systems, the smallest eigenvalue λ₁ is given by
\₁ = inf S[y],
where the infimum is taken over all admissible functions y with C[y] = 1.
Now,
1
=
dx xz² = a²
S
dx 2³ (1-2)²
26
dx (x³ – 2x² + x³)
hence a² = 60.
=
=
2
= a ²² ( 1 − } + })
60
-
Then
λ1 ≤ S[z] = a²
-
dx [x(1 − 2x)²+x(1 − x)²]
-
=-15.
4
-how
come
of this
formula??
00
![(b) (i)
the eigenvalue problem
1/3/2
y = 0, y(−1) = y(1) = 0,
x²y"+2xy'+(3x² +
λ
(x² +
with eigenvalue λ, can be written as a constrained variational
problem with functional
1
S[y] =
=
[ {
-1
and constraint
C[{y}] = ["
dx (x²y'² — 3x²y²)
dx
y2
(1+x2)3
with boundary conditions
y(−1) = y(1) = 0.
=
1,
(ii) Assume now that the system in part (b)(i) has eigenvalues >
with λk < λk+1, and eigenfunctions yk(x), k = 1, 2, … ·
Using the trial function z(x; A) = A(1 − x²) show that
λι Σ
32
7π'
-
briefly outlining the ideas of any theory you use: detailed
proofs are not required.
You may use the integral
dx
(1 - x²)²
Lidz (1 + 23) 3
-1
Π
4](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa6c8ed7d-75cc-4e27-869e-3ad6a1efc0b4%2F35a7aa85-5438-41df-8127-37fe866f19d8%2F8rqo7l_processed.png&w=3840&q=75)
Transcribed Image Text:(b) (i)
the eigenvalue problem
1/3/2
y = 0, y(−1) = y(1) = 0,
x²y"+2xy'+(3x² +
λ
(x² +
with eigenvalue λ, can be written as a constrained variational
problem with functional
1
S[y] =
=
[ {
-1
and constraint
C[{y}] = ["
dx (x²y'² — 3x²y²)
dx
y2
(1+x2)3
with boundary conditions
y(−1) = y(1) = 0.
=
1,
(ii) Assume now that the system in part (b)(i) has eigenvalues >
with λk < λk+1, and eigenfunctions yk(x), k = 1, 2, … ·
Using the trial function z(x; A) = A(1 − x²) show that
λι Σ
32
7π'
-
briefly outlining the ideas of any theory you use: detailed
proofs are not required.
You may use the integral
dx
(1 - x²)²
Lidz (1 + 23) 3
-1
Π
4
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