### 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′)d x.
  - [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

  , (5)

  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)=

∫

π
2

0

⎛
⎝

y′(x)2−x y(x)−y(x)2

⎞
⎠

d x

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))

d2
d x2

y

⎛
⎝

x

⎞
⎠

=−

x
2

−y

⎛
⎝

x

⎞
⎠

The general solution is:

y0:=dsolve(eq,x,y)

c0 cos x+c1 sin x−

x
2

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,

1
4

π

⎤
⎥
⎥
⎦

⎤
⎥
⎥
⎦

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

sin x |  |  |  |  |  |  |  |  |  |

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:

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

1
4

π  sinx−

x
2