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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.

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  1. anonymous
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    hmm, i'm debating whether to backtrack and switch to polar coordinates

  7. anonymous
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    thanks alot do u think i could ask u another here

  11. anonymous
    • one year ago
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    its much easier

  12. anonymous
    • one year ago
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    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
    • one year ago
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    its supposed to be like this

  14. anonymous
    • one year ago
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    |dw:1438803742329:dw|

  15. anonymous
    • one year ago
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    we're not done decoupling, let me finish that problem first

  16. anonymous
    • one year ago
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    ohh ook

  17. anonymous
    • one year ago
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    the first equation is an example of a Cauchy-Euler 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 $$R(r)=Ar^k+Br^{-k}$$

  18. anonymous
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    .

  23. anonymous
    • one year ago
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    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
    • one year ago
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    i see

  25. anonymous
    • one year ago
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    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
    • one year ago
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    where are you getting these from?

  27. anonymous
    • one year ago
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    theyre really hard challenge questions from my calc 3 class

  28. anonymous
    • one year ago
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    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
    • one year ago
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    the last one has me stumped right now but i'll take a look at it again

  30. anonymous
    • one year ago
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    later maybe hmm

  31. anonymous
    • one year ago
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    oh ok thanks

  32. anonymous
    • one year ago
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    0 doesn't have to be in the domain of definition

  33. anonymous
    • one year ago
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    yeah thats what i thought. it seemed too easy

  34. anonymous
    • one year ago
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    did you mean $$f^{-1}(x)=\frac1{f(x)}$$ or did you mean $$\int f^{-1}\, dx=\frac1{f(x)}$$

  35. anonymous
    • one year ago
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    first one

  36. anonymous
    • one year ago
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    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
    • one year ago
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    this is wat someone said

  38. anonymous
    • one year ago
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    i tagged u in the original

  39. anonymous
    • one year ago
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    idk do u agree

  40. anonymous
    • one year ago
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    so for first question would you go with the polar coordinates or first part

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