# Why is the non-linear wave equation $u_{tt} = \operatorname{div}(a(Du))$ quasi-linear?

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

• 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 $u$ – MJimitater May 31 '20 at 15:22
• 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. – EditPiAf May 31 '20 at 15:25

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.
• thanks for your answer! Do you mean by $u_{,j} = u_{x_j} = \frac{\partial u}{\partial x_j}$? – MJimitater May 31 '20 at 15:33