Author Topic: Web Bonus Problem –– Week 1  (Read 2853 times)

Jingxuan Zhang

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Re: Web Bonus Problem –– Week 1
« Reply #30 on: January 07, 2018, 07:18:57 PM »
Necessity:$$u_{xxyy}=f_{yy}=u_{yyxx}=g_{xx}.$$

Tristan Fraser

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Re: Web Bonus Problem –– Week 1
« Reply #31 on: January 07, 2018, 07:56:11 PM »
Would one approach to prove necessity would be to simply plug the results and show that the remaining terms demand that $f_{yy} = g_{xx}$


$ u_{xx} = f $ (1) and to keep things general, let's then write $u_{x} = \int f dx + \phi(y) $ and $u = \int\int f dxdx + x\phi(y) + \psi(y) $ (I). Then $  u_{y} = \int\int f_y dxdx + x\phi_{y}(y) + \psi_y(y) $ and $u_{yy} = \int\int f_{yy}dxdx + x\phi_{yy}(y) + \psi_{yy}(y) = g $
Then $u_{yy} = g $ , $u_{y} = \int g dy + \alpha(x) $ , $ u = \int \int g dydy  + y\alpha(x) + \beta(x) $ , $ u_x = \int\int g_x dydy + y\alpha_x (x) + \beta_x (x) $ , $ u_{xx} = \int\int g_{xx}dydy + y\alpha_{xx} (x) + \beta_{xx}(x) = f $

From there: $ f_{yy} =  u_{xxyy} = \partial{y}\partial{y} \int\int g_{xx}dydy +\partial{y}\partial{y} (y\alpha_{xx} (x)) + \partial{y}\partial{y}\beta_{xx}(x)  = g_{xx} + 0 $
and $ g_{xx} = u_{yyxx} = \partial{x}\partial{x} \int\int f_{yy}dxdx + \partial{x}\partial{x} (x\phi(y)_{yy} + \psi(y)_{yy}) = f_{yy} + 0  $

Or as Jingxuan said
Quote
$u_{xxyy}=f_{yy}=u_{yyxx}=g_{xx}$.

And we can conclude that if $f_{yy} \neq g_{xx} $ then we wouldn't be able to get this result?




Victor Ivrii

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Re: Web Bonus Problem –– Week 1
« Reply #32 on: January 07, 2018, 09:19:50 PM »
Jingxuan
Correct and minimalist proof of necessity. Obviously $u_{xx}=y^2$, $u_{yy}=-x^2$ does not satisfy and solution does not exist.

To prove that this compatibility condition $f_{yy}=g_{xx}$ is sufficient (under additional assumption that domain is convex, you can figure out generalization) solve $u_{xx}=f$, $u(0,y)=\phi(y)$, $u_x(0,y)=\psi(y)$ where $\phi$, $\psi$ so far are unknown and will be chosen later. Such $u$ obviously exist for any $\phi$, $\psi$.

What about $v:=u_{yy}-g=0$? Obviously $v_{xx}=0$ and we need to satisfy $v(0,y)=0$, $v_x(0,y)=0$, which are, in fact, $\phi '' -g(0,y)=0$, $\psi ''- g_x(0,y)=0$, and we can satisfy these two equations.  Four arbitrary constants $C_1=\phi(0)$, $C_2=\phi'(0)$, $C_3=\psi(0)$ and $C_4=\psi'(0)$ appear.

Sheng Gao

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Re: Web Bonus Problem –– Week 1
« Reply #33 on: January 08, 2018, 11:24:26 PM »
$u(0,y)=\phi(y)$, $u_x(0,y)=\psi(y)$ where $\phi$, $\psi$ so far are unknown and will be chosen later. Such $u$ obviously exist for any $\phi$, $\psi$.

What about $v:=u_{yy}-g=0$? Obviously $v_{xx}=0$ and we need to satisfy $v(0,y)=0$, $v_x(0,y)=0$, which are, in fact, $\phi '' -g(0,y)=0$, $\psi ''- g_x(0,y)=0$, and we can satisfy these two equations.  Four arbitrary constants $C_1=\phi(0)$, $C_2=\phi'(0)$, $C_3=\psi(0)$ and $C_4=\psi'(0)$ appear.
May I ask ask about how we get $u(0,y)=\phi(y)$, $u_x(0,y)=\psi(y)$? Since I cannot follow up starting here.....

Ioana Nedelcu

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Re: Web Bonus Problem –– Week 1
« Reply #34 on: January 09, 2018, 12:36:43 AM »
For the necessity $u_{xxyy}=f_{yy}=u_{yyxx}=g_{xx}$, can we say that this is because mixed partial derivatives of the same type, ie same number of differentiations with respect to the same variables, are equal? (By Green's Theorem)

Victor Ivrii

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Re: Web Bonus Problem –– Week 1
« Reply #35 on: January 09, 2018, 03:32:30 AM »
Explanation: 1) $u_{xxyy}=u_{yyxx}$ due to the property of partial derivatives; in other words operators $\partial_x$ and $\partial_y$ (of partial differentiations) commute.

2) Proving that $f_{yy}=g_{xx}$ (*) is sufficient (at least in convex domains): we denote $v:= (u_{yy}-g)$ and if $u_{xx}=f$ we conclude that $v_{xx}=0$ due to (*).

Therefore if $v=v_x=0$ for $y=0$, then $v=0$ identically. But $u=\phi(y) +\psi(y)x$ and $v=\phi '' (y)+\psi'' (y)x-g(x,y)$ which indeed implies $v(0,y)=\phi''(y)-g(0,y)$ $v_x(0,y)=\psi''(y)-g_x(0,y)$.

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This topic is closed now; I will give points over coming weekends
« Last Edit: January 09, 2018, 03:35:09 AM by Victor Ivrii »