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I'm struggling with the following:

Let a bounded connected domain $U \subset \mathbb{R}^2$ be given. Prove the weak maximum principle for the equation $$u_{xx} + 2u_{yy} + u_{y} =0 \qquad (*)$$on $U$ using the same idea as in the proof of the maximum principle for the heat equation.

[The idea mentioned above about the heat equation $u_t = u_{xx}$ (on $U \subset \mathbb{R}^2$ bounded connected domain) is the one in which one takes the function $v(x,t) := u(x,t) + \epsilon x^2, \ \epsilon >0$ and shows that $v$ can't have a local maximum on $U$].

I tried with the same "support" function (and with similar others), but I can't reach a contraddiction. Maybe I'm just badly handling the first and second order necessary conditions for the maximality - in general if $(x_0,y_0) \in U$ local maximum for a function $f:U\subset \mathbb{R}^2 \to \mathbb{R}$, then $f_x(x_0,y_0)=f_y(x_0,y_0)=0$, $f_{xx}(x_0,y_0)f_{yy}(x_0,y_0) - (f_{xy}(x_0,y_0))^2 \ge 0$ and $f_{xx}(x_0,y_0) \le 0$, right?

In particular if what I've said about the necessary conditions is correct, it is pretty straightforward to show that $(x_0,y_0)$ can't be a maximum for $v(x,y):=u(x,y) + \epsilon x^2$ (here $u$ is clearly the solution of $(*)$) with $v_{xx}(x_0,y_0) < 0$, but I've found no way to get a contraddiction with $v_{xx}(x_0,y_0)=0$.

I'm really confused, and any hint/idea would be appreciated.

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The function $v=u+\epsilon\,x^2$ satisfies the equation $$ v_{xx}+2\,v_{yy}+v_y=2\,\epsilon. $$ If $(x_0,y_0)\in U$ is a local maximum of $v$, then $v_{xx}(x_0,y_0)\le0$, $v_{yy}(x_0,y_0)\le0$, and $v_{y}(x_0,y_0)=0$. Then the left hand side of the equation is $\le0$ while the right hand side is $>0$.

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  • $\begingroup$ I think that there should be $2 \epsilon$ on the RHS, but it doesn't matter. It seems that it works, thanks! $\endgroup$ – gangrene Nov 5 '15 at 16:14

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