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My book says, "PDE is said to be quasi-linear if it is linear in the highest-ordered derivative of the unknown function." For example, $$u_xu_{xx} + xuu_y = \sin y$$ is quasi-linear PDE. Now, what is the meaning of 'linear in highest ordered derivative of unknown function'? I read first answer of this question: Linear vs nonlinear differential equation. But it doesn't answer my question. How can we find out linear operator in this case (or where is linearity here)? On which variable should we apply linear operator here? Some sources defines quasilinear PDE as the PDE in which the coefficients of the highest derivatives of $u$ depend only on lower derivatives of $u$. How does this make PDE linear in highest order derivative? Any help would be appreciated. Thank you.

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In short, you have \begin{align} u_{xx}+xu\frac{u_y}{u_x}=\frac{\sin y}{u_x} \end{align} which is linear in the highest derivative.

On the other hand \begin{align} u_{xx}^2+xu\frac{u_y}{u_x}=\frac{\sin y}{u_x} \end{align} is not quasi-linear.

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  • $\begingroup$ But why first one is linear? By definition it should be operation preserving under addition of function and scalar multiplication. Then how this definition applies here? $\endgroup$
    – ramanujan
    Commented Nov 20, 2018 at 21:20
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    $\begingroup$ @ramanujan It's not linear, but only linear in the highest derivative. In short, you can view the problem as $Lu=F$ where $F$ has lower order terms. $\endgroup$ Commented Nov 20, 2018 at 21:22
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    $\begingroup$ @ramanujan Generally, the highest order terms have the greatest effect on the behavior of solutions. Quasilinear equations are important because even though they are strongly nonlinear, the highest order terms are linear, so solutions will still behave somewhat similarly to the linear case. $\endgroup$
    – Alex Jones
    Commented Nov 20, 2018 at 21:25
  • $\begingroup$ @Jacky Chong In $u_xu_{xx} + xuu_y = \sin y$, $L(u)= u_{xx}$ and $F= \frac{\sin y - xuu_y}{u_x}$. So, $L$ is linear operator. Is this true? $\endgroup$
    – ramanujan
    Commented Nov 20, 2018 at 21:29
  • $\begingroup$ Yes. You are correct. $\endgroup$ Commented Nov 20, 2018 at 21:30

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