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As the general form of a linear PDE of degree 2 we wrote $$ (Lu)(x):=\sum_{i,j=1}^{n}a_{ij}(x)\frac{\partial^2 u}{\partial x_i\partial x_j}+\sum_{i=1}^{n}b_i(x)\frac{\partial u}{\partial x_i}+c(x)u=f(x) $$ Now I have the PDE $$ (1+x^2)\frac{\partial^2 u}{\partial x^2}-2x\frac{\partial^2 u}{\partial y^2}-(1+u^2)\frac{\partial u}{\partial x}+(1+\frac{\partial u}{\partial x})\frac{\partial u}{\partial y}-u=1 $$

I try to transfer the general form to this example.

What I see is:

$$ a_{11}(x)=1+x^2,~~a_{12}(x)=-2x,~~c(x)=-1,~~f(x)=1 $$

But what are the other coefficients?

1 Answers1

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Since you have $\partial u/\partial x$ multiplied by $\partial u/\partial y$, it is not linear.

Empy2
  • 50,853
  • I have to apply a maximum principle here to show that for the PDE above and the condition $u(x,y)=\frac{\sin^2(x)}{1+y^2}$ on the border it is for a nonnegative solution $u(x,y)\leq 1$. In our reading we had a maximum principle for linear elliptic PDE. So I thought it must be linear and elliptic... hm! –  Nov 02 '13 at 13:05
  • I'll ask a separate question for that. –  Nov 02 '13 at 13:05
  • This is a quasi-linear PDE, right? –  Nov 02 '13 at 13:36