Web Bonus Problem –– Week 1

[Jingxuan Zhang]

Necessity:$$u_{xxyy}=f_{yy}=u_{yyxx}=g_{xx}.$$

[Tristan Fraser]

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

$u_{xxyy}=f_{yy}=u_{yyxx}=g_{xx}$.

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

[Victor Ivrii]

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]

$u(0,y)=\phi(y)$, $u_x(0,y)=\psi(y)$ 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]

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]

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