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6.22.5  Finding conjugate points: conjugate_equation

The conjugate_equation computes conjugate points.

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)2x y(x)−y(x)2
 dx

on D={yC1[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

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