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anonymous
 one year ago
Let M be the set of all functions f(x, y) twicedifferentiable 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.
anonymous
 one year ago
Let M be the set of all functions f(x, y) twicedifferentiable 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.

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so we have a functional \(I\) $$I[f]=\iint_D \\nabla f\^2\, dA$$ and we want to maximize this over the set \(M\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oops i mean minimize and note \(\\nabla f\^2=\left(\frac{\partial f}{\partial x}\right)^2+\left(\frac{\partial f}{\partial y}\right)^2\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0recognize 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, $$\nabla^2 f=0$$

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so 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 \(\{(\cos\theta,\sin\theta):0\le \theta< 2\pi\}\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0let'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: $$\frac{d^2 G}{dx^2}k^2G=0\\\frac{d^2 H}{dy^2}+k^2H=0$$

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0hmm, i'm debating whether to backtrack and switch to polar coordinates

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0if 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 \(0\le r<1\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0the 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 $$R''(r)\Theta(\theta)+\frac1rR'(r)\Theta(\theta)+\frac1{r^2}R(r)\Theta''(\theta)=0\\(R''(r)+\frac1r R'(r))\Theta(\theta)=\frac1{r^2}R(r)\Theta''(\theta)\\\frac{r^2 R''(r)+rR'(r)}{R(r)}=\frac{\Theta''(\theta)}{\Theta(\theta)}=k^2$$

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0which decouples into $$r^2\frac{d^2R}{dr^2}+r\frac{dR}{dr}k^2 R=0\\\frac{d^2\Theta}{d\theta^2}+k^2\Theta=0$$

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0thanks alot do u think i could ask u another here

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Give 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) .

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0its supposed to be like this

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1438803742329:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0we're not done decoupling, let me finish that problem first

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0the first equation is an example of a CauchyEuler equation: $$r^2R''+rR'k^2 R=0$$now using the change of variables \(r=e^t\), so \(dr/dt=r \) we have $$\frac{dR}{dt}=\frac{dR}{dr}\cdot\frac{dr}{dt}=r\frac{dR}{dr}\\\frac{d^2R}{dt^2}=\frac{d}{dt}\left(r\frac{dR}{dr}\right)=\frac{dr}{dt}\frac{dR}{dr}+r\frac{d}{dt}\left(\frac{dR}{dr}\right)=r\frac{dR}{dr}+r\frac{d^2R}{dr^2}\cdot\frac{dr}{dt}$$giving us $$\frac{dR}{dt}=r\frac{dR}{dr}\\\frac{d^2R}{dt^2}=r^2\frac{d^2R}{dr}+r\frac{dR}{dr}$$ which gives simply $$\frac{d^2R}{dt^2}k^2R=0$$so \(R(t)=Ae^{kt}+Be^{kt}\) and $$R(r)=Ar^k+Br^{k}$$

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0similarly $$\frac{d^2\Theta}{d\theta^2}+k^2\Theta=0$$gives $$\Theta(\theta)=A\cos(k\theta)+B\sin(k\theta)$$

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so our general solution looks something like: $$g(r,\theta)=\sum_{k=0}^\infty r^k(A\cos(k\theta)+B\sin(k\theta))$$

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so our general solution looks something like: $$g(r,\theta)=\sum_{k=0}^\infty r^k(A_k\cos(k\theta)+B_k\sin(k\theta))$$ and now we impose conditions, \(g(1,\theta)=\cos(2\theta)\) so: $$g(1,\theta)=\sum_{k=0}^\infty (A_k\cos(k\theta)+B_k\sin(k\theta))$$ which tells us that \(A_2=1,A_k=0\) for \(k\ne 2\), and \(B_k=0\) for all \(k\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so $$g(r,\theta)=r^2\cos(2\theta)$$ and in Cartesian coordinates we can find it noting $$\cos(2\theta)=\cos^2(\theta)\sin^2(\theta)=\frac1{r^2}\left(x^2y^2\right)\\r^2\cos(2\theta)=x^2y^2$$ so $$f(x,y)=x^2y^2$$Q.E.D.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0now we focus on the next problem, finding continuous, invertible \(f\) such that $$\int f^{1}(x)\, dx=\frac1{f(x)}$$now consider that we know $$\int_a^b f^{1}(x)\,dx+\int_c^d f(y)\, dy=bdac$$ from a simple geometrical argument where \(f(c)=a,f(d)=b\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so in this case we have $$\int_a^b f^{1}(x)\,dx=bdac\int_c^d f(y)\, dy$$so if we let \(F(z)=\int_0^z f(y)\, dy\) $$\int f^{1}(x)\, dx=xf^{1}(x)F(f^{1}(x))+C$$

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0where are you getting these from?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0theyre really hard challenge questions from my calc 3 class

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so the last euation is the answer? someone told me there is no equation because if we put 0 it becomes non continuous

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0the last one has me stumped right now but i'll take a look at it again

anonymous
 one year ago
Best ResponseYou've already chosen the best response.00 doesn't have to be in the domain of definition

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yeah thats what i thought. it seemed too easy

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0did you mean $$f^{1}(x)=\frac1{f(x)}$$ or did you mean $$\int f^{1}\, dx=\frac1{f(x)}$$

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0y=f(x),x=g(y) , g is f inverse, so x.y = 1, but, (x, y) and (y, x) are reflecting the graph across the line y = x. so the scalar product (x, y) and (y, x) x.y+y.xx2+y2=cos3π4 2x2+y2=−2√2 is impossíble

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0this is wat someone said

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i tagged u in the original

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so for first question would you go with the polar coordinates or first part
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