Consider the functional Sy cb S[v] = [[* dx (o² y² – ß³²y² ) $ where a, ẞ are positive constants, b = a/ß, with boundary conditions y(0) = 0, y(b) = 0, and with (non-unique) stationary path y. =sin((B/α)x). Show that S is not continuous at y, with respect to the strong norm on C₁[a, b]. Is S continuous with respect to the weak norm on C₁[a, b]? Let y = y*+h, where h is admissible. Then S[y] = [° ds (a² (y. + h')² = p² (y + h)²) b = S[y*] + S dx (a²h² - 8²h²) how оло Come 88 since the linear term is zero because y✶ is a stationary path. Now let k ≥ 1 be an integer and consider hk (x) = Yk = Y* + hk. The h is admissible since hk (0) sin (kẞx/a) and set = sin (kẞ0/α) = 0 and = sin (kẞb/α) = sinka = 0, and clearly hk is twice hk(b) continuously differentiable. We note that, in the strong norm, ||Yk — Y* || = supo≤x≤b|hk(x)| = 1/√k → 0, as k→ ∞. Now h(x)=(√kß/a) cos(kẞx/α) so S[y] – S[y⭑] = ·b ·b =%" = = 82 B² dx (ah-82h) dx (a²k(8²/a²) cos² (kẞx/α) – (ß²/k) sin² (kẞx/a)) * dx (kcos² (kßx/α) – (1/k) sin² (kßx/a)) dx (k(1+cos(2kßx/a))/2 – (1/k)(sin² (kẞx/a)) = kẞ2b/2+0(1/k) →∞ as k∞. So, although y → y* in the strong norm, S[yk] → S[y*]. It follows that S is not continuous at y* in the strong norm. Now suppose y = y* +h where ||h||1 < 5 and 5 > 0. Then supo≤x≤b|h′(x)|, supo≤x≤b|h(x)| < d, so that |S[y] – Sv.]| = | |º dz (o²h² – ß³²h²) | - Ľ dx (a²|hk|² + ß²|hk|²) ≤ 68² (a² + B²), so that for a given € > 0, we may choose ♪ < - - E (60'+5))" 1/2 so that ||y – y⭑|| < & implies |S[y] – S[y*]|
Consider the functional Sy cb S[v] = [[* dx (o² y² – ß³²y² ) $ where a, ẞ are positive constants, b = a/ß, with boundary conditions y(0) = 0, y(b) = 0, and with (non-unique) stationary path y. =sin((B/α)x). Show that S is not continuous at y, with respect to the strong norm on C₁[a, b]. Is S continuous with respect to the weak norm on C₁[a, b]? Let y = y*+h, where h is admissible. Then S[y] = [° ds (a² (y. + h')² = p² (y + h)²) b = S[y*] + S dx (a²h² - 8²h²) how оло Come 88 since the linear term is zero because y✶ is a stationary path. Now let k ≥ 1 be an integer and consider hk (x) = Yk = Y* + hk. The h is admissible since hk (0) sin (kẞx/a) and set = sin (kẞ0/α) = 0 and = sin (kẞb/α) = sinka = 0, and clearly hk is twice hk(b) continuously differentiable. We note that, in the strong norm, ||Yk — Y* || = supo≤x≤b|hk(x)| = 1/√k → 0, as k→ ∞. Now h(x)=(√kß/a) cos(kẞx/α) so S[y] – S[y⭑] = ·b ·b =%" = = 82 B² dx (ah-82h) dx (a²k(8²/a²) cos² (kẞx/α) – (ß²/k) sin² (kẞx/a)) * dx (kcos² (kßx/α) – (1/k) sin² (kßx/a)) dx (k(1+cos(2kßx/a))/2 – (1/k)(sin² (kẞx/a)) = kẞ2b/2+0(1/k) →∞ as k∞. So, although y → y* in the strong norm, S[yk] → S[y*]. It follows that S is not continuous at y* in the strong norm. Now suppose y = y* +h where ||h||1 < 5 and 5 > 0. Then supo≤x≤b|h′(x)|, supo≤x≤b|h(x)| < d, so that |S[y] – Sv.]| = | |º dz (o²h² – ß³²h²) | - Ľ dx (a²|hk|² + ß²|hk|²) ≤ 68² (a² + B²), so that for a given € > 0, we may choose ♪ < - - E (60'+5))" 1/2 so that ||y – y⭑|| < & implies |S[y] – S[y*]|
Calculus For The Life Sciences
2nd Edition
ISBN:9780321964038
Author:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.
Publisher:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.
Chapter9: Multivariable Calculus
Section9.CR: Chapter 9 Review
Problem 4CR
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How come of hk(x)in part a solution??
![Consider the functional
Sy
cb
S[v] = [[* dx (o² y² – ß³²y² )
$
where a, ẞ are positive constants, b = a/ß, with boundary
conditions y(0) = 0, y(b) = 0, and with (non-unique) stationary
path y. =sin((B/α)x).
Show that S is not continuous at y, with respect to the strong
norm on C₁[a, b].
Is S continuous with respect to the weak norm on C₁[a, b]?](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa6c8ed7d-75cc-4e27-869e-3ad6a1efc0b4%2F3259949a-6b07-4ebe-8594-ee2c8a5e4e1f%2Fwm8doo8_processed.png&w=3840&q=75)
Transcribed Image Text:Consider the functional
Sy
cb
S[v] = [[* dx (o² y² – ß³²y² )
$
where a, ẞ are positive constants, b = a/ß, with boundary
conditions y(0) = 0, y(b) = 0, and with (non-unique) stationary
path y. =sin((B/α)x).
Show that S is not continuous at y, with respect to the strong
norm on C₁[a, b].
Is S continuous with respect to the weak norm on C₁[a, b]?
![Let y = y*+h, where h is admissible. Then
S[y] = [° ds (a² (y. + h')² = p² (y + h)²)
b
= S[y*] +
S
dx (a²h² - 8²h²)
how
оло
Come 88
since the linear term is zero because y✶ is a stationary path. Now let
k ≥ 1 be an integer and consider hk (x)
=
Yk = Y* + hk. The h is admissible since hk (0)
sin (kẞx/a) and set
=
sin (kẞ0/α) = 0 and
=
sin (kẞb/α)
= sinka =
0, and clearly hk is twice
hk(b)
continuously differentiable.
We note that, in the strong norm,
||Yk — Y* || = supo≤x≤b|hk(x)| = 1/√k → 0, as k→ ∞. Now
h(x)=(√kß/a) cos(kẞx/α) so
S[y] – S[y⭑]
=
·b
·b
=%"
=
=
82
B²
dx (ah-82h)
dx (a²k(8²/a²) cos² (kẞx/α) – (ß²/k) sin² (kẞx/a))
*
dx (kcos² (kßx/α) – (1/k) sin² (kßx/a))
dx (k(1+cos(2kßx/a))/2 – (1/k)(sin² (kẞx/a))
= kẞ2b/2+0(1/k) →∞
as k∞. So, although y → y* in the strong norm, S[yk] → S[y*]. It
follows that S is not continuous at y* in the strong norm.
Now suppose y = y* +h where ||h||1 < 5 and 5 > 0. Then
supo≤x≤b|h′(x)|, supo≤x≤b|h(x)| < d, so that
|S[y] – Sv.]| = | |º dz (o²h² – ß³²h²) |
-
Ľ
dx (a²|hk|² + ß²|hk|²)
≤ 68² (a² + B²),
so that for a given € > 0, we may choose ♪ <
-
-
E
(60'+5))"
1/2
so that
||y – y⭑|| < & implies |S[y] – S[y*]| <e. It follows that S is continuous at
y* in the weak norm.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa6c8ed7d-75cc-4e27-869e-3ad6a1efc0b4%2F3259949a-6b07-4ebe-8594-ee2c8a5e4e1f%2Fx3b3yo_processed.png&w=3840&q=75)
Transcribed Image Text:Let y = y*+h, where h is admissible. Then
S[y] = [° ds (a² (y. + h')² = p² (y + h)²)
b
= S[y*] +
S
dx (a²h² - 8²h²)
how
оло
Come 88
since the linear term is zero because y✶ is a stationary path. Now let
k ≥ 1 be an integer and consider hk (x)
=
Yk = Y* + hk. The h is admissible since hk (0)
sin (kẞx/a) and set
=
sin (kẞ0/α) = 0 and
=
sin (kẞb/α)
= sinka =
0, and clearly hk is twice
hk(b)
continuously differentiable.
We note that, in the strong norm,
||Yk — Y* || = supo≤x≤b|hk(x)| = 1/√k → 0, as k→ ∞. Now
h(x)=(√kß/a) cos(kẞx/α) so
S[y] – S[y⭑]
=
·b
·b
=%"
=
=
82
B²
dx (ah-82h)
dx (a²k(8²/a²) cos² (kẞx/α) – (ß²/k) sin² (kẞx/a))
*
dx (kcos² (kßx/α) – (1/k) sin² (kßx/a))
dx (k(1+cos(2kßx/a))/2 – (1/k)(sin² (kẞx/a))
= kẞ2b/2+0(1/k) →∞
as k∞. So, although y → y* in the strong norm, S[yk] → S[y*]. It
follows that S is not continuous at y* in the strong norm.
Now suppose y = y* +h where ||h||1 < 5 and 5 > 0. Then
supo≤x≤b|h′(x)|, supo≤x≤b|h(x)| < d, so that
|S[y] – Sv.]| = | |º dz (o²h² – ß³²h²) |
-
Ľ
dx (a²|hk|² + ß²|hk|²)
≤ 68² (a² + B²),
so that for a given € > 0, we may choose ♪ <
-
-
E
(60'+5))"
1/2
so that
||y – y⭑|| < & implies |S[y] – S[y*]| <e. It follows that S is continuous at
y* in the weak norm.
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