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
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.