Question

Let M be the set of all functions f(x, y) twice-differentiable in both
x and y, defined on the unit disk D = {(x, y)|x^2 + y^2 < 1}, and such that on the
boundary circle θ ∈ [0, 2π):
f(x = cos θ, y = sin θ) = cos(2θ), 0 ≤ θ < 2π.
Find the function h ∈ M which minimizes the integral
I[f] = double integral absolute value gradient f squared

double integral||∇*(f)||^2
dA,
which means that h ∈ M and
I[h] ≤ I[f], ∀f ∈ M.

Answer 1

so we have a functional \(I\) $$I[f]=\iint_D \|\nabla f\|^2\, dA$$ and we want to maximize this over the set \(M\)

Answer 2

oops i mean minimize and note \(\|\nabla f\|^2=\left(\frac{\partial f}{\partial x}\right)^2+\left(\frac{\partial f}{\partial y}\right)^2\)

Answer 3

recognize that this functional is equivalent to the Dirichlet energy:

https://en.wikipedia.org/wiki/Dirichlet%27s_energy

and minimizing the Dirichlet energy is equivalent to solving the Laplace equation, $$\nabla^2 f=0$$

Answer 4

so we have that $$\frac{\partial^2 f}{\partial x^2}+\frac{\partial^2 f}{\partial y^2}=0$$ given the boundary condition \(f(\cos\theta,\sin\theta)=\cos2\theta\) on the unit circle \(\{(\cos\theta,\sin\theta):0\le \theta< 2\pi\}\)

Answer 5

let's attack using separation of variables: \(f(x,y)=G(x)H(y)\) so that we can decouple our equation: $$\frac{\partial^2 f}{\partial x^2}=G''(x)H(y)\\\frac{\partial^2 f}{\partial y^2}=G(x)H''(y)$$so we have $$G''(x)H(y)+G(x)H''(y)=0\\G''(x)H(y)=-G(x)H''(y)\\\frac{G''}{G}=-\frac{H''}H=k^2$$which gives us two decoupled linear ODEs: $$\frac{d^2 G}{dx^2}-k^2G=0\\\frac{d^2 H}{dy^2}+k^2H=0$$

Answer 6

hmm, i'm debating whether to backtrack and switch to polar coordinates

Answer 7

if we do switch to polar coordinates, consider we'd have \(g(r,\theta)=f(r\cos\theta,r\sin\theta)\) in which case our Dirichlet boundary condition becomes \(g(1,\theta)=\cos(2\theta)\) and we're interested in solving \(\nabla^2 g=0\) over \(0\le r<1\)

Answer 8

the Laplacian in polar coordinates looks like $$\nabla^2 g=\frac{\partial^2 g}{\partial^2 r}+\frac1r\frac{\partial g}{\partial r}+\frac1{r^2}\frac{\partial^2 g}{\partial^2\theta}=0$$
and then we can separate variables like \(g(r,\theta)=R(r)\Theta(\theta)\) so that: $$\frac{\partial g}{\partial r}=R'(r)\Theta(\theta)\\\frac{\partial^2 g}{\partial^2 r}=R''(r)\Theta(\theta)\\\frac{\partial^2 g}{\partial\theta^2}=R(r)\Theta''(\theta)$$ giving us $$R''(r)\Theta(\theta)+\frac1rR'(r)\Theta(\theta)+\frac1{r^2}R(r)\Theta''(\theta)=0\\(R''(r)+\frac1r R'(r))\Theta(\theta)=-\frac1{r^2}R(r)\Theta''(\theta)\\\frac{r^2 R''(r)+rR'(r)}{R(r)}=-\frac{\Theta''(\theta)}{\Theta(\theta)}=k^2$$

Answer 9

which decouples into $$r^2\frac{d^2R}{dr^2}+r\frac{dR}{dr}-k^2 R=0\\\frac{d^2\Theta}{d\theta^2}+k^2\Theta=0$$

Answer 10

thanks alot
do u think i could ask u another here

Answer 11

its much easier

Answer 12

Give an example (or prove that none exists) of a real function f(x)
which is continuous, invertible, and satisfies the following identity everywhere on
its domain of definition:
f
−1
(x) = 1
f(x)
.

Answer 13

its supposed to be like this