The value of the integral for this extremal is x1√ y1 which is larger than the integral of the quadratic we analyzed before for y1 > x1... is the extremal with the smaller integral and i
Trang 2Solution 48.7
C1 Extremals Without loss of generality, we take the vertical line to be the y axis We will consider x1, y1 > 1 With
ds =p1 + (y0)2dx we extremize the integral,
Z x 1 0
dx(y
0
Fy 0 − F ) = 0
y0Fy 0 − F = constFor the given Lagrangian, this is
y0√
0
p1 + (y0)2 −√yp1 + (y0)2 = const,(y0)2√
y −√y(1 + (y0)2) = constp1 + (y0)2,
y − ady
Trang 3±x + b = 2 a(y − a)
y = x
2
4a± bx2a +
b24a + aThe natural boundary condition is
Fy0 x=0 =
√
yy0p1 + (y0)2
x=0
= 0,
y0(0) = 0The extremal that satisfies this boundary condition is
for y1 ≥ x1 The value of the integral is
Z x 1 0
2
dx = x1(x
2
1+ 12a2)12a3/2
By denoting y1 = cx1, c ≥ 1 we have
a = 12
cx1± x1√c2− 1The values of the integral for these two values of a are
√2(x1)3/2−1 + 3c2± 3c√c2− 1
3(c ±√
c2− 1)3/2
Trang 4The values are equal only when c = 1 These values, (divided by √
x1), are plotted in Figure 48.1 as a function of c.The former and latter are fine and coarse dashed lines, respectively The extremal with
a = 12
1 − x2
1)3 The function y = y1 is an admissible extremal for all x1 The value of the integral for this extremal is x1√
y1 which
is larger than the integral of the quadratic we analyzed before for y1 > x1
2.5 3 3.5 4
Trang 5
is the extremal with the smaller integral and is the minimizing curve in C1 for y1 ≥ x1 For y1 < x1 the C1 extremumis,
d
dtfx0− fx = 0 and d
dtfy0 − fy = 0
If one of the equations is satisfied, then the other is automatically satisfied, (or the extremal is straight) With either
of these equations we could derive the quadratic extremal and the y = const extremal that we found previously Wewill find one more extremal by considering the first parametric Euler differential equation
d
dtfx0 − fx = 0d
dt
py(t)x0(t)p(x0(t))2+ (y0(t))2
!
= 0
py(t)x0(t)p(x0(t))2+ (y0(t))2 = constNote that x(t) = const is a solution Thus the extremals are of the three forms,
b2
4a + a.
Trang 6The Erdmann corner conditions require that
Fy0 =
√
yy0p1 + (y0)2,
F − y0Fy0 =√
yp1 + (y0)2−
√y(y0)2
p1 + (y0)2 =
√yp1 + (y0)2
are continuous at corners There can be corners only if y = 0
Now we piece the three forms together to obtain Cp1 extremals that satisfy the Erdmann corner conditions Theonly possibility that is not C1 is the extremal that is a horizontal line from (0, 0) to (x1, 0) and then a vertical line from(x1, y1) The value of the integral for this extremal is
Z y 1 0
The C1
p extremum is the piecewise smooth extremal for y1 ≤ x1p3 + 2√
3/√3and is the quadratic extremal for y1 ≥ x1p3 + 2√
3/√3
Trang 7ds − z = L.
Writing the arc-length differential as ds =p1 + (y0)2dx we minimize
ρg
Z x 2 0
Suppose y(x) and z are the desired solutions and form the comparison families, y(x) + 1η1(x) + 2η2(x), z + 1ζ1+
2ζ2 Then, there exists a constant such that
∂
∂1(I + λJ )
1 , 2 =0 = 0
∂
∂2(I + λJ )
... class="page_container" data-page="31">
0.2 0 .4 0 .6 0.8 0.01
0.02 0.03 0. 04 0.05 0. 06< /small>
Figure 48 .6: Two Term Approximation and Exact Solution.The general solution... 1+αh 2 and c2 = −1+α2αh2 Thus the equation
of the pencil (48 .6) will have the form
All extremals (48 .7) lie above the... solve for c1 and thus obtain z You can derive that there are no solutions unless
L is greater than about 1.9 366 If L is smaller than this, the rope would slip off the pin For