# Genaral PDE for scalar and vectors

1. Apr 1, 2014

### Jhenrique

I realized that a PDE of 2nd order can written like: $$Af+\vec{b}\cdot\vec{\nabla}f+cf=0$$
$$\begin{bmatrix} a_{11} & a_{12}\\ a_{21} & a_{22}\\ \end{bmatrix}:\begin{bmatrix} \partial_{xx} & \partial_{xy}\\ \partial_{yx} & \partial_{yy}\\ \end{bmatrix}f+\begin{bmatrix} b_1\\ b_2\\ \end{bmatrix}\cdot\begin{bmatrix} \partial_x\\ \partial_y\\ \end{bmatrix}f+cf=0$$
But and if f is a vector, how would be a general PDE of 2nd order for a vector?

EDIT: Let me be more clear with my ask:

An ODE of 2nd order for a vector is: $$A\frac{d^2\vec{r}}{dt^2} + B\frac{d\vec{r}}{dt} + C\vec{r}=0$$ and a PDE is: $$a \frac{\partial^2 \vec{\rho}}{\partial u^2} + 2b \frac{\partial^2 \vec{\rho}}{\partial u \partial v} + c\frac{\partial^2 \vec{\rho}}{\partial v^2} + d\frac{\partial \vec{\rho}}{\partial u} + e\frac{\partial \vec{\rho}}{\partial v} + f\vec{\rho}=0$$ But how write this PDE in the correct way? Because if I replace the scalar $f$ (in the 1st equation above) by vector position $\vec{\rho}$, I'll have a direct product between the hessian $H$ and $\vec{\rho}$, and another between $\vec{\nabla}$ and $\vec{\rho}$, thing that I never saw.

Last edited: Apr 1, 2014