Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

How do I caluclate the integral curves of a vector field, i.e. how would I go about calculating the integral curves of:

Define the vector field in $\mathbb{R}^3$ by:

$ u = x_1\displaystyle\frac{\partial}{\partial x_2} +x_2\frac{\partial}{\partial x_1} + x_3\frac{\partial}{\partial x_3}$

Thanks for any help

share|improve this question
    
Calculate a parametrization, an implicit equation, or a numerical approximation? These are all hard problems in general. Do you have a specific vector field in mind? –  lhf Jan 25 '12 at 13:11
    
Well yes, i have some questions (non-assessed) to do which give the vector fields and ask for the integral curve, so I gave one above –  hmmmm Jan 25 '12 at 13:15
add comment

1 Answer 1

up vote 5 down vote accepted

What lhf says in the comments is generally true: Most of the time, there is no hope of finding a closed form solution. However, in this case, we can do it.

Let $\gamma(t) = (\gamma_1(t), \gamma_2(t), \gamma_3(t)$ be an integral curve with initial point $\gamma(0) = (x_0, y_0, z_0)$. What does this mean?

It means that $u$ at the point $\gamma(t)$ is equal to $\gamma'(t)$. Let's write this out. I'm going to use $\partial_k$ for $\dfrac{\partial}{\partial x_k}$ to save typing.

$$u(\gamma(t)) = \gamma_1 \partial_2 + \gamma_2 \partial_1 + \gamma_3\partial_3$$ and $$\gamma'(t) = (\gamma_1'(t), \gamma_2'(t), \gamma_3'(t)) = \gamma_1' \partial_1 + \gamma_2'\partial_2+\gamma_3' \partial_3.$$

Setting these equal to each other and equating coefficients gives us a system of ODEs to solve:

$$\gamma_1' = \gamma_2$$ $$\gamma_2' = \gamma_1$$ $$\gamma_3' = \gamma_3$$

The first two equations are coupled but the third is not, so lets solve that one first. The solution to $\gamma_3' = \gamma$ is $\gamma(t) = Ce^t$ for some constant $C$.

There is a known process for solving coupled linear ODEs, but in this case, I think it's easier to just guess a solution. We want two functions so that if we start with one and take two derivatives, we get back where we started. This suggests we try $\gamma_1(t) = Ae^t + Be^{-t}$. Plugging this into the second equation gives $\gamma_2(t) = Ae^t-Be^{-t}$, and it's easy to check that this choice of $\gamma_1$ and $\gamma_2$ solves the second equation.

The upshot is we now know $\gamma(t) = (Ae^t + Be^{-t}, Ae^t -Be^{-t}, Ce^t)$. What are $A$, $B$, and $C$?

Well, $\gamma(0) = (A+B,A-B, C) = (x_0,y_0,z_0)$. So, $A = \frac{x_0+y_0}{2}$ and $B = \frac{x_0-y_0}{2}$ while $C = z_0$.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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