D-admissible directions and extermal points

  • Thread starter Thread starter braindead101
  • Start date Start date
  • Tags Tags
    Points
braindead101
Messages
158
Reaction score
0
Let X = C[a,b], J(y) = integ(a to b) sin^3(x) + y(x)^2 dx and D={yEX; integ(a to b) y(x)dx = 1}
(a) what are the D-admissible directions for J?
(b) Find all possible (local) external points for J on D?

so far i have:
(let e be epsilon)
lim e->0 J(y+ev) - J(y) / e
= lim e->0 integ (a to b) [ sin^3(x) + (y+ev)^2(x)dx - J(y) ] / e
= lim e->0 integ (a to b) [2ey(x)v(x) + e^2v(x)^2) dx ] / e
= 2 integ(a to b) y(x)v(x) dx
A similar example was done in class, so i just copied it for this question.
G(y) = integ(a to b) y(x) dx = 1
gateaux G(y;v) = integ (a to b) y(x)v(x) dx

is this right so far? how do i go about answering (a) and (b).
 
Physics news on Phys.org
i was wondering if the D-admissible directions for J is just
y+Ev E D
so,
integ (a-b) (y+Ev)(x) dx
= 1+ E integ(a-b) v(x) dx

can i simplify any further?
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top