# Flow curves in a vector field?

In $(x,y,z)$ space, we have the following vector field,

$$V(x,y,z)=(V_1(x,y,z), V_2(x,y,z), V_3(x,y,z))=\left(z^2+x+1, y^2 - yz, y + \frac{z^2}{2}+\frac{1}{2}\right)$$

Let us consider the points, $P=(1,1,1)$ and $Q=(-1,-1,-1)$.

a) Determine the divergence of V in P and Q.

b) We have to define a linearization of V from P such a way that: Search for $i = 1 . . 3$ approximating the first power polynomium $U_i (x, y, z)$ for $V_i (x, y, z)$ with the development point $P$. vector field $$U(x, y, z) = (U_1(x, y, z), U_2(x, y, z), U_3(x, y, z))$$ is the desired linearising of $V$.

c) Determine the flow curve $r(t)$ of $U$ with an arbitrary initial condition $R (0) = (x_0, y_0, z_0)$.

Approached/Tried

I have tried to solve this question for so long and have not really come very far :/ For (a) I found the divergence to be $4$ for $P(1,1,1)$ and $-2$ for $Q(-1,-1,-1)$. I have found the approximated first power polynomium $U(x,y,z)=(2z+x, y-z,z+y)$. But I am totally lost for (c), because I am not sure how to find the flow curve. Any Hint or Help will be grately appreciated. Thank You.

P.S I do not expect anyone to solve the complete question but any hint on how to proceed will be great! :)

Not at all sure what that phrasing in b) is supposed to mean. But the linearization of a vector field is just: $$U(R) = \mathbf M (R - P) + V(P)$$ where $\mathbf M$ is the derivative matrix:$$\mathbf M = \begin{bmatrix}\frac{\partial V_1}{\partial x_1} & \frac{\partial V_1}{\partial x_2} & \frac{\partial V_1}{\partial x_3}\\\frac{\partial V_2}{\partial x_1} & \frac{\partial V_2}{\partial x_2} & \frac{\partial V_2}{\partial x_3}\\\frac{\partial V_3}{\partial x_1} & \frac{\partial V_3}{\partial x_2} & \frac{\partial V_3}{\partial x_3}\end{bmatrix}$$
• Yes, $R$ is an arbitrary point where the linearized field is being evaluated. – Paul Sinclair May 8 '16 at 18:12