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5.19.4  Finding conjugate points : conjugate_equation

conjugate_equation takes four arguments :

The function y0(x) is assumed to be stationary for the problem of minimizing some functional F(y)=∫abf(x,y,y′) dx. The return value is 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.

For example, we 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 :

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

Output :

diff(y(x),x,2)=((-2*y(x)-x)/2)

The general solution is :

y0:=dsolve(eq,x,y)

Output :

c_0*cos(x)+c_1*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])

Output :

[[0,pi/4]]

Input :

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

Output :

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, input :

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

Output :

pi*sin(x)/4-x/2

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