Please help me with finding the function u, which gives the minimal value of the integral from 0 to 1:
Int(u'(x)^2+u(x))dx--> min
with conditions:
Int(u^2(x))dx=1 (integral is from 0 to 1)
u'(0)=u'(1)=0
Where u has a continuous derivative.
Please help me with finding the function u, which gives the minimal value of the integral from 0 to 1:
Int(u'(x)^2+u(x))dx--> min
with conditions:
Int(u^2(x))dx=1 (integral is from 0 to 1)
u'(0)=u'(1)=0
Where u has a continuous derivative.
That is a "calculus of variations" problem. Calculus of variations - Wikipedia, the free encyclopedia
Do you know the corresponding "Euler-Lagrange" equation?