Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

In one physics problem, I got the following system of non-linear equations that I need to solve for $v_e$, $v_s$ and $\omega$:

$$m_p v_a = m_s v_s + m_p v_e$$

$$\frac{l}{2} m_p v_a = \frac{l}{2} m_p v_e + I \omega$$

$$\frac{1}{2} m_p v_a^2 = \frac{1}{2} m_p v_e^2 + \frac{1}{2} m_s v_s^2 + \frac{1}{2} I \omega^2$$

I know how to solve this with substitution and a lot of scratch paper.

With linear systems, one can just derive the matrix and use gauss-jordan / reduced row echolon form and then the solution is directly apparent.

Is there something handy for non-linear equations as well?

share|cite|improve this question
Looks quite like something you can use Gröbner bases for, in the general case. – J. M. Dec 31 '11 at 0:26
Gröbner bases can be viewed as a generalization of Gaussian Elimination for systems of polynomial equations. – user2468 Dec 31 '11 at 0:35
It looks like what I need, I just think it is a little over my head currently. I'll read into it. – Martin Ueding Dec 31 '11 at 15:42
up vote 3 down vote accepted

For your example, you have a "nearly" linear system, because only one equation is quadratic. In that case, you can express all solutions of the linear part of the system in the form $x_0+\alpha x_h$, substitute that expression into the quadratic equation and solve the resulting equation for $\alpha$.

If you apply this technique to your system, you get $v_e=v_a-\alpha$, $v_s=\frac{m_p}{m_s}\alpha$ and $\omega=\frac{lm_p}{2I}\alpha$. If you substitute this into the quadratic equation $$m_p (v_a^2-v_e^2) = m_s v_s^2 + I \omega^2$$ you get

$$m_p (2v_a-\alpha)\alpha = \frac{m_p^2}{m_s}\alpha^2 + \frac{l^2m_p^2}{4I}\alpha^2$$ One solution of this quadratic equation is obviously $\alpha=0$, but I guess you are more interested in the other solution. Assuming $m_p\alpha\neq0$, we can divide by $m_p\alpha$ to get $2v_a-\alpha = \frac{m_p}{m_s}\alpha + \frac{l^2m_p}{4I}\alpha$. It's easy to solve this equation for $\alpha$.

share|cite|improve this answer
Thanks a lot, this made the whole problem painless. – Martin Ueding Dec 31 '11 at 15:41

Your Answer


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.