Second order quasilinear PDE

Question

Some quick question about PDE's. Only recently started studying PDE's so this might be trivial.

The second-order quasilinear elliptic equation is given by:

$ -\sum_{i=1}^{n} \frac{\partial}{\partial x_{i}}a_{i}(x,u(x),\nabla u(x)) + c(x,u(x),\nabla u(x)) = g(x)$

where $u : U \rightarrow \mathbb{R}$ and $U \subset \mathbb{R}^{n}$

Since $\nabla u(x)$ is included as an independent variable of $a_{i}(x,u(x),\nabla u(x))$, does it then follow that $a_{i}(x,u(x),\nabla u(x))$ is then a vector valued function?

If we had $\nabla u(x)$ as a term would it follow that $\frac{\partial}{\partial x_{i}}\nabla u(x) = $ $ (\frac{\partial \nabla u}{\partial x_{i}},...,\frac{\partial \nabla u}{\partial x_{n}})$?

Is there a simple example of such a PDE?

Thanks

Answer 1

No. Defining $a(x,u(x),\nabla u(x)) := |\nabla u(x)|$ takes a vector-valued argument to a real number.

I would say $$\frac{\partial}{\partial x_i}\nabla u(x) = (u_{x_1x_i}, u_{x_2x_i}, ..., u_{x_nx_i})$$ but the covention used in the book might be different.

Answer 2

No, $a_i$ is a scalar valued function that takes a vector among its arguments. Just because the input is a vector, the output need not be.

But the collection of all $a_i$, $(a_1,\dots,a_n)$, can be considered as a vector-valued function, and often it is convenient to do so. Then the PDE takes the form $$-\operatorname{div} A(x,u(x),\nabla u(x)) + c(x,u(x),\nabla u(x)) = g(x)$$ A classical example is the $p$-harmonic equation, in which $c=g=0$ and $A(x,u(x),\nabla u(x))= |\nabla u(x)|^{p-2} \nabla u(x)$.