13.8.5 Finding conjugate points
The conjugate_equation
command computes conjugate points.
-
conjugate_equation takes four arguments:
-
y0, an expression which depends on an independent
variable and two parameters. The expression y0 is assumed to
represent a stationary function for the problem of minimizing
some functional F(y)=∫abf(x,y,y′) dx.
- [α,β], a list of parameters which y0 depends on.
- [A,B], a list of the values of parameters α and
β, respectively.
- x, the independent variable.
- a, a real number equal to the lower or to the upper bound for x.
- conjugate_equation(y0,[α,β],[A,B],x,a)
returns the expression
at α=A and β=B, which is zero if and only if t is
conjugate to a.
To find any conjugate points, solve for zeros of the returned expression.
Example
Find a minimum for the functional
F(y)= | ∫ | | ⎛
⎝ | y′(x)2−x y(x)−y(x)2 | ⎞
⎠ | dx
|
on D={y∈ C1[0,π/2]:y(0)=y(π/2)=0}. The corresponding Euler-Lagrange equation is:
eq:=euler_lagrange(y'(x)^2-x*y(x)-y(x)^2,y(x)) |
The general solution is:
The stationary function depends on two parameters c0 and c1
which are fixed by the boundary conditions:
c:=solve([subs(y0,x,0)=0,subs(y0,x,pi/2)=0],[c_0,c_1]) |
|
| ⎡
⎢
⎢
⎣ | ⎡
⎢
⎢
⎣ | 0, | | π | ⎤
⎥
⎥
⎦ | ⎤
⎥
⎥
⎦ |
| | | | | | | | | | |
|
conjugate_equation(y0,[c_0,c_1],c[0],x,0) |
The above expression obviously has no zeros in (0,π/2], hence
there are no points conjugate to 0. Since fy′ y′=2>0, where
f(y,y′,x) is the integrand in F(y) (the strong Legendre
condition), y0 minimizes F on D. To obtain y0 explicitly: