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Let $\Omega\subset C^0$ a bounded domian in $\mathbb{R}^2$. Let $u\in C^2(\Omega)\cap C(\overline{\Omega})$ be a non negative classical solution of $$ (1+x^2)u_{xx}-2xu_{xy}+(1+u)u_{yy}-(1+u^2)u_x+(1+u_x)u_y-u=1\text{ in }\Omega,\\u(x,y)=\frac{\sin^2(x)}{1+y^2}\text{ on }\partial\Omega. $$ Show that $(0\leq) y\leq 1$ for $(x,y)\in\Omega$.

I already asked this here: Do I have to use a maximum principle?

But this time my question concerns a special method for solving this, I think it is called something like linearization by freezing the coefficients.

I do not know more about that, but I heard that it works by inserting the assumed solution $u$ in the coefficients of the main part of the PDE and the aim is to get a semi-linear (or linear?) PDE on which one can apply maximum principle.

Do you know something about that method?

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The idea behind the "freezing" technique is to introduce a new "source term" $v$ and replace enough appearances of $u$ by $v$ such that the resulting equation is linear - for example in this case we could try

$$F_v(u)=(1+x^2)u_{xx} - 2xu_{xy}+(1+v)u_{yy} - (1+v^2)u_x + (1-v_x) u_y - u - 1= 0.$$(For certain purposes it may be more fruitful to do something else with the first-order term - it's not immediately obvious what the "correct" choice is to me. For the purpose of establishing the maximum principle it doesn't matter.)

A function $u$ is then a solution of the original PDE if and only if $F_u(u) = 0$; i.e. iff it is a solution of the linear PDE $F_u (\cdot)= 0$. In some cases this could give a way to find solutions: if the linear PDE $F_v(u) = 0$ has unique solution given by $u = G(v)$ then solving the non-linear PDE is just solving the fixed-point problem $G(u)=u$.

Whether we establish the existence of solutions in this manner or another, the most useful application comes in analysing these solutions - since $u$ is a solution of the linear PDE $F_u(\cdot)=0$, you can apply all the same a priori estimates you're familiar with from the linear elliptic theory. All you have to do is control the properties of the solution enough for the operator $F_u$ to be elliptic.

In this case, you are given that $u$ is non-negative, so you should be able to verify that $F_u$ is indeed elliptic. Thus the maximum principle holds for $u$.

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  • $\begingroup$ Why is the new PDE $F_v(u)$ a linear one? $\endgroup$ – math12 Nov 4 '13 at 12:42
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    $\begingroup$ @math12: The coefficients of $u, u_i, u_{ij}$ depend only on $x$ and $v(x)$. $\endgroup$ – Anthony Carapetis Nov 4 '13 at 12:47
  • $\begingroup$ In order to use the maximum principle for linear elliptic PDE, I have to verify if (1) $F_u(\cdot)$ is uniformly elliptic in $\Omega$ and that (2) $a_{11}(x,y)=1+x^2, a_{12}(x,y)=a_{21}(x,y)=-x, a_{22}(x,y)=1+u(x,y), c=-1, f=1$ are in $C(\overline{\Omega})$. (2) is clear because they are all continious on $\mathbb{R}^2$ resp. $u$ is by definition in $C(\overline{\Omega})$. But how can I show (1)? Is it enough to show that the coefficient matrix of the main part is positive definite, i.e. only has positive eigenvalues? $\endgroup$ – math12 Nov 4 '13 at 13:42
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    $\begingroup$ @math12: for uniform ellipticity you need a positive uniform lower bound on the eigenvalues. If you work it through, this should be true because $u\ge 0$ and $\Omega$ (and thus the domain of $x$) is bounded. $\endgroup$ – Anthony Carapetis Nov 4 '13 at 14:22
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    $\begingroup$ @math12: it's a quadratic equation in $\lambda$, so e.g. apply the quadratic formula. $\endgroup$ – Anthony Carapetis Nov 4 '13 at 14:49

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