Consider the following functional Jx 12 integral02 xt t 2 0

Consider the following functional: J(x) = 1/2 integral_0^2 x^t (t) [2 0 2 2] x(t) dt Find an external x * of J(x) if the boundary conditions are x(0) = [1 -3]^T, x(2) = [0 5]^T. Compute the value of the resulting J (x*).

Solution

Ans-

if I have a functional with a Lagrangian L(t,x(t),y(t),x’(t),y’(t)), meaning two functions x and y of one parameter t. And want to solve the minimization problem

t0Ldtsuperscriptsubscript0tLdt\\int_{0}^{t}Ldt

. Then I get necessary conditions to find extrema by getting the two Euler Lagrange equation

LxddtLx=0LxddtLsuperscriptxnormal-0\\frac{\\partial L}{\\partial x}-\\frac{d}{dt}\\frac{\\partial L}{\\partial x^{{%\\prime}}}=0

and

LyddtLy=0LyddtLsuperscriptynormal-0\\frac{\\partial L}{\\partial y}-\\frac{d}{dt}\\frac{\\partial L}{\\partial y^{{%\\prime}}}=0

now, if i solved these functions. how do i find out, that it is an actual minimum? are there methods to show this in general? i know, that in case of one variable it would be sufficient to show somehow that the lagrangian is convex. but is there a way to do this in this case too? or do i need to calculate a second derivative? if this is necessary, can someone give me a referece,

t0Ldtsuperscriptsubscript0tLdt\\int_{0}^{t}Ldt