## anonymous one year ago 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.

1. anonymous

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

2. anonymous

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

3. anonymous

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, $$\nabla^2 f=0$$

4. anonymous

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 $$\{(\cos\theta,\sin\theta):0\le \theta< 2\pi\}$$

5. anonymous

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: $$\frac{d^2 G}{dx^2}-k^2G=0\\\frac{d^2 H}{dy^2}+k^2H=0$$

6. anonymous

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

7. anonymous

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 $$0\le r<1$$

8. anonymous

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 9. anonymous 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 10. anonymous thanks alot do u think i could ask u another here 11. anonymous its much easier 12. anonymous 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) . 13. anonymous its supposed to be like this 14. anonymous |dw:1438803742329:dw| 15. anonymous we're not done decoupling, let me finish that problem first 16. anonymous ohh ook 17. anonymous the first equation is an example of a Cauchy-Euler equation: r^2R''+rR'-k^2 R=0now 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}$$18. anonymous similarly$$\frac{d^2\Theta}{d\theta^2}+k^2\Theta=0$$gives$$\Theta(\theta)=A\cos(k\theta)+B\sin(k\theta)$$19. anonymous so our general solution looks something like:$$g(r,\theta)=\sum_{k=0}^\infty r^k(A\cos(k\theta)+B\sin(k\theta))$$20. anonymous so our general solution looks something like:$$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$$ 21. anonymous so$$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^2-y^2\right)\\r^2\cos(2\theta)=x^2-y^2$$so$$f(x,y)=x^2-y^2$$Q.E.D. 22. IrishBoy123 . 23. anonymous now we focus on the next problem, finding continuous, invertible $$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=bd-ac$$from a simple geometrical argument where $$f(c)=a,f(d)=b$$ 24. anonymous i see 25. anonymous so in this case we have$$\int_a^b f^{-1}(x)\,dx=bd-ac-\int_c^d f(y)\, dy$$so if we let $$F(z)=\int_0^z f(y)\, dy$$$$\int f^{-1}(x)\, dx=xf^{-1}(x)-F(f^{-1}(x))+C$$26. anonymous where are you getting these from? 27. anonymous theyre really hard challenge questions from my calc 3 class 28. anonymous so the last euation is the answer? someone told me there is no equation because if we put 0 it becomes non continuous 29. anonymous the last one has me stumped right now but i'll take a look at it again 30. anonymous later maybe hmm 31. anonymous oh ok thanks 32. anonymous 0 doesn't have to be in the domain of definition 33. anonymous yeah thats what i thought. it seemed too easy 34. anonymous did you mean$$f^{-1}(x)=\frac1{f(x)}$$or did you mean$$\int f^{-1}\, dx=\frac1{f(x)}

35. anonymous

first one

36. anonymous

y=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

37. anonymous

this is wat someone said

38. anonymous

i tagged u in the original

39. anonymous

idk do u agree

40. anonymous

so for first question would you go with the polar coordinates or first part