You said, conventional calculus is used to extremize function and variational calculus is used to extermize Functional. The function returned in calculus of variation is minimum at every point in the domain OR does the minimum is in average sense due to fact we integrate the functional in domain?
At 18:45 how do you get to \tilde{u'} in eq (12) while in the sentence above there is no prime? And also at 15:03 you say that the necessary condition for \phi(0) \leq \phi(\epsilon) means that \phi'(0) = 0 but what happens when 0 is the boundary for phi such that phi is still continuous and differentiable but does not have a zero derivative (for isntance a linear function y = a*x + b : x = [0,1], a>0). Very nice video btw
To your first question: The 2nd integral above Eq. 12 should have \tilde{u'} (it's a typo). To your 2nd question: the domain of \phi(\epsilon) is the space of \epsilon (not x), so there is no boundary in the sense that you describe.
You said, conventional calculus is used to extremize function and variational calculus is used to extermize Functional. The function returned in calculus of variation is minimum at every point in the domain OR does the minimum is in average sense due to fact we integrate the functional in domain?
Hi, Professor. Any recommended textbook for learning variational of calculus?
At 18:45 how do you get to \tilde{u'} in eq (12) while in the sentence above there is no prime? And also at 15:03 you say that the necessary condition for \phi(0) \leq \phi(\epsilon) means that \phi'(0) = 0 but what happens when 0 is the boundary for phi such that phi is still continuous and differentiable but does not have a zero derivative (for isntance a linear function y = a*x + b : x = [0,1], a>0).
Very nice video btw
To your first question: The 2nd integral above Eq. 12 should have \tilde{u'} (it's a typo).
To your 2nd question: the domain of \phi(\epsilon) is the space of \epsilon (not x), so there is no boundary in the sense that you describe.