# Non-divergence form of a 2nd order PDE

This might be a trivial question but I'm very rusty with regards to calculus and am new to PDEs. How would you write the following second order quasilinear equation in it's non-divergence form:

The equation is: $$-\nabla\cdot\big(a(u,\nabla u)\big)+c(u,\nabla u) = g.$$

Based on the definition of a quasilinear second order pde defined in Lawrence Evans book it should be of the form: $$- \sum_{i,j =1}^{n}a_{ij}(x,y,\nabla u) \frac{\partial^{2}u }{\partial x_{i}\partial x_{j}} + c(x,u,\nabla u) = g.$$ I think it is an application of the chain rule but I'm having difficulty getting a good form. What steps would you use to get this result?

Thanks.

I am using the notation $\nabla:=\mathrm{div}$. Apparently $a=a(u,\nabla u)$ is a vector, if "div" is applied on it, i.e., $a=(a_1,\ldots,a_n)$. Then $$-\nabla\cdot a(u,\nabla u)=-\sum_{k=1}^n \frac{\partial a_k}{\partial x_k},$$ and if we set $a_k=a_k(u,v_1,\ldots,v_n)$, we have $$\frac{\partial a_k}{\partial x_k}=\frac{\partial a_k}{\partial u}\frac{\partial u}{\partial x_k}+\sum_{j=1}^n\frac{\partial a_k}{\partial v_j}\frac{\partial^2 u}{\partial x_k\partial x_j}.$$ Thus $$\nabla\cdot a(u,\nabla u)=\sum_{k=1}^n\left(\frac{\partial a_k}{\partial u}\frac{\partial u}{\partial x_k}+\sum_{j=1}^n\frac{\partial a_k}{\partial v_j}\frac{\partial^2 u}{\partial x_k\partial x_j}\right)=\sum_{k,j=1}^n\frac{\partial a_k}{\partial v_j}\frac{\partial^2 u}{\partial x_k\partial x_j}+\sum_{k=1}^n \frac{\partial a_k}{\partial u}\frac{\partial u}{\partial x_k}.$$
Therefore $$-\text{div}(a(u,\nabla u))+c(u,\nabla u) = g$$ can be written also as $$-\sum_{k,j=1}^n\frac{\partial a_k}{\partial v_j}\frac{\partial^2 u}{\partial x_k\partial x_j}-\sum_{k=1}^n \frac{\partial a_k}{\partial u}\frac{\partial u}{\partial x_k}+c(u,\nabla u) = g.$$
• Why do you get $$\frac{\partial a_k}{\partial x_k}=\frac{\partial a_k}{\partial u}\frac{\partial u}{\partial x_k}+\sum_{j=1}^n\frac{\partial a_k}{\partial v_j}\frac{\partial^2 u}{\partial x_k\partial x_j}.$$ is it not simply $$\frac{\partial a_k}{\partial x_k}=\frac{\partial a_k}{\partial u}\frac{\partial u}{\partial x_k}+\sum_{j=1}^n\frac{\partial a_k}{\partial v_j}\frac{\partial v_{j}}{\partial x_{k}}.$$ ? – Alex May 9 '14 at 14:39
• But $v=\nabla u$, so $v_j=\partial u/\partial x_j$. – Ted Shifrin Jun 24 '14 at 20:50