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I came across the PDE

\begin{align*} u_{tt} - \operatorname{div}(a(Du)) = 0 \end{align*} where $a:\mathbb{R}^n \rightarrow \mathbb{R}^n$, $Du$ is the gradient of the unknown $u$ and $\operatorname{div}(\cdot) = \operatorname{trace}(D\,\cdot)$ and I'm not seeing why it should be quasi-linear. Any ideas?

Cheers

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  • $\begingroup$ I read it in some Uni notes, being rather a side note and not a proof; my understanding of quasi-linear is that the coefficient functions of the highest degree derivates only depend on $x,y,..$, on lower degree derivates of $u$ or on $u$ $\endgroup$ – MJimitater May 31 '20 at 15:22
  • $\begingroup$ OK, if we want to extend this definition to the present equation, then we need the coefficients of the highest derivatives also dependent on $Du$, cf. answer. $\endgroup$ – EditPiAf May 31 '20 at 15:25
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The chain rule gives \begin{aligned} \text{div}\big(a(Du)\big) &= [a_{i}(u_{,j}\,{\bf e}_j)]_{,i} \\ &= a_{i,k}(u_{,j}\,{\bf e}_j)\, u_{,ki}\\ & = \text{tr}\big( J_a(Du)\, DDu \big) \end{aligned} where $J_a$ is the Jacobian matrix of $a$ (note that Einstein notation was used). Because of the resemblance of $$ u_{tt} - \text{tr}\big(J_a(Du)\, DDu\big) = 0 $$ with the linear wave equation $u_{tt} - \Delta u = 0$ where $\Delta$ denotes the Laplace operator, we may call quasi-linear this form of the present nonlinear wave equation. Note that the linear wave equation $u_{tt} - \Delta u = 0$ is recovered if $a = \text{id} + c$ where $c$ is an arbitrary constant vector.

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  • $\begingroup$ thanks for your answer! Do you mean by $u_{,j} = u_{x_j} = \frac{\partial u}{\partial x_j}$? $\endgroup$ – MJimitater May 31 '20 at 15:33
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    $\begingroup$ @MJimitater Yes, forgot to mention it. Summation on repeated indices is also omitted (Einstein notation) $\endgroup$ – EditPiAf May 31 '20 at 15:36
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    $\begingroup$ Thanks @EditPiAf! My difficulty was to simply apply the chain rule consistently, but thanks to your help, all is clear! Cheers! $\endgroup$ – MJimitater May 31 '20 at 15:41

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