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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

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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
up vote 8 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 $C = z_0$.

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