What it means that a function depends on $x,y,u,u_x,u_y$?

Question

Definition: A PDE equation is quasilinear if $Au_{xx}+Bu_{xy}+Cu_{yy}+Du_x+Eu_y+Fu+G+\Phi(x,y,u,u_x,u_y)=0,$ where $A,B,C,D,E,F,G$ are functions of $x,y,u.$

Is this equation quasilinear? $$(x^2+u^2)u_x-xyu_y=u^3x+y^2$$

The answer is yes it is because $D=(x^2+u^2),E=-xy,F=u^2x, G=y^2$.

However in this pdf http://nptel.ac.in/courses/Webcourse-contents/IIT-%20Guwahati/maths3/module_13/pdenotes.pdf pag. [2] says that $xu_x+yu_y+u^2=0$ is not linear, semilinear nor quasilinear. And I don't understand why, according to definition it's quasilinear.

So maybe I'm not understanding correctly the meaning of function of.. or maybe the pdf file is wrong.

Your question is not about function. – Apr 01 '18 at 20:58 — , Apr 01 '18 at 20:58

score 0 · Answer 1 · answered Apr 01 '18 at 20:57

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Using the definition in the note:

A PDE is said to be quasilinear if it is linear in its highest derivative. That is, the coefficient of highest order does not depend on any highest order partial derivative.

So the PDE $xu_x + yu_y +u^2 = 0$ IS quasilinear. Indeed in the note it is not claimed that the above is NOT quasilinear. It said that it is "nonlinear".

The $xu_x + yu_y +u^2 = 0$ is not linear because of that $u^2$ term. Thus it is nonlinear. (So it is both quasilinear and nonlinear).

answered Apr 01 '18 at 20:57

Did you read definition 1.7? It says PDE is nonlinear if it doesn't belong to any of the three types. – Apr 01 '18 at 21:00
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The only possible conclusion is that the notes have it wrong: $xu_x + yu_y +u^2 = 0$ is, according to Def. 1.5, semilinear, so it is not, according to Def. 1.7, nonlinear. – user539887 Apr 01 '18 at 21:05
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See Simple PDE classification question for another definition. – user539887 Apr 01 '18 at 21:09

What it means that a function depends on $x,y,u,u_x,u_y$?

1 Answers1