1
$\begingroup$

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! :)

$\endgroup$
1
$\begingroup$

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:$$\mathbf M = \begin{bmatrix}\frac{\partial V_1}{\partial x_1} & \frac{\partial V_1}{\partial x_2} & \frac{\partial V_1}{\partial x_3}\\\frac{\partial V_2}{\partial x_1} & \frac{\partial V_2}{\partial x_2} & \frac{\partial V_2}{\partial x_3}\\\frac{\partial V_3}{\partial x_1} & \frac{\partial V_3}{\partial x_2} & \frac{\partial V_3}{\partial x_3}\end{bmatrix}$$

$\endgroup$
  • $\begingroup$ Oh ok. But what is R? And if I am correct, P is the coordinate, right? $\endgroup$ – MathCurious314 May 8 '16 at 12:19
  • $\begingroup$ If I get it, P is the coordinate (orts-vektor) we linearize at (where the derivatives and V(P) is evaluated) and R is the coordinate (orts-vektor) we want the linear approximation at. You could maybe compare it to a first order Taylor expansion. $\endgroup$ – mathreadler May 8 '16 at 13:01
  • $\begingroup$ Yes, $R$ is an arbitrary point where the linearized field is being evaluated. $\endgroup$ – Paul Sinclair May 8 '16 at 18:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.