Finding conjugate points: conjugate_equation

### 6.22.5 Finding conjugate points: conjugate_equation

The conjugate_equation 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

  ∂ y0(t)
  ∂α
  ∂ y0(a)
  ∂β
  | − |
  ∂ y0(a)
  ∂α
  ∂ y0(t)
  ∂β
  | , (4)

  at α=A and β=B, which is zero if and only if t is conjugate to a.

To find any conjugate points, set the returned expression to zero and solve.

Example.
Find a minimum for the functional

F(y)= | ∫ |
π
2
0
| ⎛ ⎝ | 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:
Input:

eq:=euler_lagrange(y’(x)^2-x*y(x)-y(x)^2,y(x))

Output:

d2
dx2
| y | ⎛ ⎝ | x | ⎞ ⎠ | =− |
x
2
| −y | ⎛ ⎝ | x | ⎞ ⎠

The general solution is:
Input:

y0:=dsolve(eq,x,y)

Output:

c0 cosx+c1 sinx − |
x
2

The stationary function depends on two parameters c0 and c1 which are fixed by the boundary conditions:
Input:

c:=solve([subs(y0,x,0)=0,subs(y0,x,pi/2)=0],[c_0,c_1])

Output:

⎡ ⎢ ⎢ ⎣ | ⎡ ⎢ ⎢ ⎣ | 0, |
1
4
| π | ⎤ ⎥ ⎥ ⎦ | ⎤ ⎥ ⎥ ⎦

Input:

conjugate_equation(y0,[c_0,c_1],c[0],x,0)

Output:

sinx

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:
Input:

subs(y0,[c_0,c_1],c[0])

Output:

1
4
| π sinx− |
x
2