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.

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

I was trying to simulate a physical system which lead me to this equation. I don't know if it has any solution or not, but I guess you guys can help me find the answer. $$v'(t) = a + s * \frac{v(t)}{|v(t)|}$$ in which t is an scalar variable, s is an scalar constant and a is a vector with same dimensions as v(t) (either 2D or 3D)

share|cite|improve this question
If $s$ and $v$ are vectors, what kind of a product does $*$ denote? The only product that produces a vector from two vectors is the cross product, but that's usually not denoted by $*$, and also that wouldn't work in two dimensions. – joriki Aug 27 '12 at 23:17
I believe $s$ is not a vector. – Tunococ Aug 27 '12 at 23:58
@joriki s was an scalar constant as tunococ suggested, I've fixed the question. – Ali.S Aug 28 '12 at 11:10

Interpretation 1: Perhaps we can write your problem as

$$ \frac{dv}{dt} - \frac{s}{|v|}v=a $$

Provided $s$ is in fact a scalar. Take the dot-product with $v$ to obtain:

$$ v \cdot \frac{dv}{dt} - \frac{s}{|v|}v\cdot v= v \cdot a $$


$$ \frac{1}{2}\frac{d}{dt} (v \cdot v )-s|v| = v \cdot a $$

One silly solution is $v(t)=v_o$ where $v_o$ is taken to fit the condition $v_o \cdot a=-s|v_o|$. I assume $s,a$ are given constants.

Interpretation 2: Another solution, suppose we seek a constant speed solution then $|v|= s_o$. We face

$$ \frac{dv}{dt} - \frac{s}{s_o}v=a $$

Suppose $s$ is a scalar function of $t$. We can use the obvious generalization of the usual integrating factor technique (note $v$ is a vector in contrast to the usual context where the typical DEqns student faces $\frac{dy}{dt}+py=q$). Construct $\mu = exp( -\int \frac{s}{s_o} dt)$. Multiply by this integrating factor,

$$ \mu\frac{dv}{dt} - \frac{s}{s_o}\mu v= \mu a $$

By the product rule for a scalar function multiplying a vector,

$$ \frac{d}{dt}\biggl[ \mu v \biggr] = \mu a $$

Integrating, we reduce the problem to quadrature:

$$ v = \frac{1}{\mu} \int \mu a \, dt $$

where $\mu = exp( -\int \frac{s}{s_o} dt)$ and we choose constants of integration such that $|v|=s_o$ (if that's even possible...)

share|cite|improve this answer
your answer makes sense but the problem is I don't know how can I use it in my application! as I said both a and s are constants given and I know initial value for $v_0$. – Ali.S Aug 28 '12 at 11:18

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.