Author Topic: Web Bonus Problem to Week 8 (#4)  (Read 316 times)

Victor Ivrii

  • Administrator
  • Elder Member
  • *****
  • Posts: 1332
  • Karma: 0
    • View Profile
    • Personal website of Victor Ivrii

Xi Yue Wang

  • APM346
  • Full Member
  • *
  • Posts: 30
  • Karma: 0
    • View Profile
Re: Web Bonus Problem to Week 8 (#4)
« Reply #1 on: November 30, 2015, 02:52:22 AM »
For part a):
     Given $\Delta^2 u = 0$, we have $u_{xxxx} + 2u_{xxyy} + u_{yyyy} = 0$, we make Fourier transform of $x \mapsto k$, $u(x,y)\mapsto \hat{u}(k,y)$.
     Then we get,$$k^4\hat{u} -2k^2\hat{u}_{yy}+\hat{u}_{yyyy} = 0$$
Solve this ODE, we get$$\hat{u}(k,y) = (A(k)+B(k)y)e^{-|k|y} + (C(k)+D(k)y)e^{|k|y}$$
Because the second term is unbounded, we discard the term $(C(k)+D(k)y)e^{|k|y}$.
Then, $$\hat{u}(k,y) = (A(k)+B(k)y)e^{-|k|y}$$
Given condition that, $$u|_{y=0} = f(x), u_y|_{y=0} = g(x)$$
Then we have $$\hat{u}|_{y=0} = \hat{f}(k),\hat{u}_y|_{y=0} = \hat{g}(k)\\\hat{u}(k,0) = A(k) = \hat{f}(k)\\\hat{u}_y(k,0) = -|k|A(k) + B(k) = \hat{g}(k)$$
Hence, we get $$B(k) = \hat{g}(k)+|k|\hat{f}(k)$$
Then,$$\hat{u}(k,y) = (A(k)+B(k)y)e^{-|k|y}\\= (\hat{f}(k)+ \hat{g}(k)+|k|\hat{f}(k))e^{-|k|y}\\u(x,y)=\int_{-\infty}^{\infty}  (\hat{f}(k)+ \hat{g}(k)+|k|\hat{f}(k))e^{-|k|y}e^{ikx} dk$$

For part b):

     Similarly, we get Fourier transform of x,$$\hat{u}(k,y) = (A(k)+B(k)y)e^{-|k|y} + (C(k)+D(k)y)e^{|k|y}$$
Because the second term is unbounded, we discard the term $(C(k)+D(k)y)e^{|k|y}$.
Then, $$\hat{u}(k,y) = (A(k)+B(k)y)e^{-|k|y}$$
Given condition that, $u_{yy}|_{y=0} = f(x), \Delta u_y|_{y=0} = g(x)$

          we get, $\hat{u}_{yy}|_{y=0} = \hat{f}(k), (\hat{u}_{xxy}+\hat{u}_{yyy})|_{y=0} = \hat{g}(k)$(not sure about this condition)

Then plug in the conditions, we have $$\hat{u}_{yy}(k,0) = |k|^2A(k) -2|k|B(k) = \hat{f}(k) \\(\hat{u}_{xxy}+\hat{u}_{yyy})(k,0)= ?$$
I may figure it out later.