Exercise 10.11 (0) dx the stationary paths of the functional S[y] = ["dz F (x,y,3′,3′″), y(a) = A‚_y(b) =B, satisfy the Euler-Lagrange equation OF d OF OF + = dx მყ :0, y(a)=A‚_y(b)=B, Fy"|,=Fy"|, = 0. (b) Apply the result found in part (a) to the functional defined in equation (10.1) (page 219), with p = constant and the boundary conditions y(0) = y(L) = 0, to derive the associated Euler-Lagrange equation and show that its solution is y(x) = pg 24K - x (L − x) (L²+xL — x²). E[y] La de dx = (()) Sol. (10.1) (a) The Gâteaux differential is given in equation (10.14) and since h(a) = h(b) = 0 this reduces to AS \y,h] = [h b OF ·b OF + dx dr² ǝy" d ´ƏF dx Oy = h'(b) = Using the subset of varied paths for which h'(a) satisfies the Euler-Lagrange equation (OFF) - (34) d OF + = dx² ay" dx dy ay OF + მყ :) h. 0, we see that y(x) 0, y(a) = A, y(b) = B. The other boundary conditions are obtained by considering those paths for which h'(a) = 0 and those for which h′(b) = 0, which gives Fy"|a = Fy" |b = 0. 1 112 2 (b) In this problem F = KY pgy and the appropriate Euler-Lagrange equation is ky (4) = pg, with the boundary conditions are y(0) = y(L) = 0. The general solution of this equation that satisfies the conditions y(0) = 0 is y(x) = pg −xª + Ax³ + Bx² + Cx. 24K But y(L) y"(x) = = pg 2K 0 and since Fy" Ky"(x) we also have y″(0) = y"(L) = x²+6Ax+2B the condition at x = 0 gives B = 0. Since = 0. Then the conditions at x= L give A = pgL 12k and 0= pgLª pgLª 24K 12K +CL giving C = pgL³ 24K -how ?? Come so that y(x) = pg 24K (x² - 2Lx³ + L³x) = pg 24K -x(L−x) (L²+xL — x²). -
Exercise 10.11 (0) dx the stationary paths of the functional S[y] = ["dz F (x,y,3′,3′″), y(a) = A‚_y(b) =B, satisfy the Euler-Lagrange equation OF d OF OF + = dx მყ :0, y(a)=A‚_y(b)=B, Fy"|,=Fy"|, = 0. (b) Apply the result found in part (a) to the functional defined in equation (10.1) (page 219), with p = constant and the boundary conditions y(0) = y(L) = 0, to derive the associated Euler-Lagrange equation and show that its solution is y(x) = pg 24K - x (L − x) (L²+xL — x²). E[y] La de dx = (()) Sol. (10.1) (a) The Gâteaux differential is given in equation (10.14) and since h(a) = h(b) = 0 this reduces to AS \y,h] = [h b OF ·b OF + dx dr² ǝy" d ´ƏF dx Oy = h'(b) = Using the subset of varied paths for which h'(a) satisfies the Euler-Lagrange equation (OFF) - (34) d OF + = dx² ay" dx dy ay OF + მყ :) h. 0, we see that y(x) 0, y(a) = A, y(b) = B. The other boundary conditions are obtained by considering those paths for which h'(a) = 0 and those for which h′(b) = 0, which gives Fy"|a = Fy" |b = 0. 1 112 2 (b) In this problem F = KY pgy and the appropriate Euler-Lagrange equation is ky (4) = pg, with the boundary conditions are y(0) = y(L) = 0. The general solution of this equation that satisfies the conditions y(0) = 0 is y(x) = pg −xª + Ax³ + Bx² + Cx. 24K But y(L) y"(x) = = pg 2K 0 and since Fy" Ky"(x) we also have y″(0) = y"(L) = x²+6Ax+2B the condition at x = 0 gives B = 0. Since = 0. Then the conditions at x= L give A = pgL 12k and 0= pgLª pgLª 24K 12K +CL giving C = pgL³ 24K -how ?? Come so that y(x) = pg 24K (x² - 2Lx³ + L³x) = pg 24K -x(L−x) (L²+xL — x²). -
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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![Exercise 10.11 (0)
dx
the stationary paths of the functional
S[y] = ["dz F (x,y,3′,3′″), y(a) = A‚_y(b) =B,
satisfy the Euler-Lagrange equation
OF
d OF OF
+
=
dx
მყ
:0, y(a)=A‚_y(b)=B, Fy"|,=Fy"|, = 0.
(b) Apply the result found in part (a) to the functional defined in equation (10.1)
(page 219), with p = constant and the boundary conditions y(0) = y(L) = 0, to
derive the associated Euler-Lagrange equation and show that its solution is
y(x) =
pg
24K
-
x (L − x) (L²+xL — x²).
E[y]
La de
dx
= (())
Sol.
(10.1)
(a) The Gâteaux differential is given in equation (10.14) and since h(a) = h(b) = 0
this reduces to
AS \y,h] = [h
b
OF
·b
OF
+ dx
dr² ǝy"
d ´ƏF
dx Oy
=
h'(b)
=
Using the subset of varied paths for which h'(a)
satisfies the Euler-Lagrange equation
(OFF) - (34)
d
OF
+
=
dx² ay" dx dy ay
OF
+
მყ
:)
h.
0, we see that y(x)
0, y(a) = A, y(b) = B.
The other boundary conditions are obtained by considering those paths for which
h'(a)
= 0 and those for which h′(b) = 0, which gives Fy"|a = Fy" |b = 0.
1 112
2
(b) In this problem F = KY pgy and the appropriate Euler-Lagrange equation
is ky (4)
=
pg, with the boundary conditions are y(0) = y(L) = 0. The general
solution of this equation that satisfies the conditions y(0) = 0 is
y(x) =
pg
−xª + Ax³ + Bx² + Cx.
24K
But y(L)
y"(x) =
=
pg
2K
0 and since Fy" Ky"(x) we also have y″(0) = y"(L) =
x²+6Ax+2B the condition at x = 0 gives B
= 0. Since
= 0. Then the conditions
at x=
L give
A =
pgL
12k
and 0=
pgLª pgLª
24K 12K
+CL giving C =
pgL³
24K
-how ??
Come
so that
y(x) =
pg
24K
(x² - 2Lx³ + L³x) =
pg
24K
-x(L−x) (L²+xL — x²).
-](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa6c8ed7d-75cc-4e27-869e-3ad6a1efc0b4%2Ff882e955-3aba-4573-94c8-1eee17e50a4b%2Ft6lp4gh_processed.png&w=3840&q=75)
Transcribed Image Text:Exercise 10.11 (0)
dx
the stationary paths of the functional
S[y] = ["dz F (x,y,3′,3′″), y(a) = A‚_y(b) =B,
satisfy the Euler-Lagrange equation
OF
d OF OF
+
=
dx
მყ
:0, y(a)=A‚_y(b)=B, Fy"|,=Fy"|, = 0.
(b) Apply the result found in part (a) to the functional defined in equation (10.1)
(page 219), with p = constant and the boundary conditions y(0) = y(L) = 0, to
derive the associated Euler-Lagrange equation and show that its solution is
y(x) =
pg
24K
-
x (L − x) (L²+xL — x²).
E[y]
La de
dx
= (())
Sol.
(10.1)
(a) The Gâteaux differential is given in equation (10.14) and since h(a) = h(b) = 0
this reduces to
AS \y,h] = [h
b
OF
·b
OF
+ dx
dr² ǝy"
d ´ƏF
dx Oy
=
h'(b)
=
Using the subset of varied paths for which h'(a)
satisfies the Euler-Lagrange equation
(OFF) - (34)
d
OF
+
=
dx² ay" dx dy ay
OF
+
მყ
:)
h.
0, we see that y(x)
0, y(a) = A, y(b) = B.
The other boundary conditions are obtained by considering those paths for which
h'(a)
= 0 and those for which h′(b) = 0, which gives Fy"|a = Fy" |b = 0.
1 112
2
(b) In this problem F = KY pgy and the appropriate Euler-Lagrange equation
is ky (4)
=
pg, with the boundary conditions are y(0) = y(L) = 0. The general
solution of this equation that satisfies the conditions y(0) = 0 is
y(x) =
pg
−xª + Ax³ + Bx² + Cx.
24K
But y(L)
y"(x) =
=
pg
2K
0 and since Fy" Ky"(x) we also have y″(0) = y"(L) =
x²+6Ax+2B the condition at x = 0 gives B
= 0. Since
= 0. Then the conditions
at x=
L give
A =
pgL
12k
and 0=
pgLª pgLª
24K 12K
+CL giving C =
pgL³
24K
-how ??
Come
so that
y(x) =
pg
24K
(x² - 2Lx³ + L³x) =
pg
24K
-x(L−x) (L²+xL — x²).
-
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