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How to solve $$\frac{dv}{dt} = av^2 + bv + c$$ to obtain $x(t)$, where $a$, $b$ and $c$ are constants, $v$ is velocity, $t$ is time and $x$ is position. Boundaries for the first integral are $v_0$, $v_t$ and $0$, $t$ and boundaries for the second integral are $0$, $x_{max}$ and $0$, $t$.

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I am guessing the second term should be $b v$? –  copper.hat May 15 '12 at 18:51
    
Why don't you ask on physics? You'll get better answers, I think. –  Gigili May 15 '12 at 19:01
    
Yes you'r right the second term is bv. I posted it on physics too but they proposed me to post it on maths. –  ACL May 15 '12 at 19:36
    
Am I right in assuming that the speed varies in your example, i.e., we need to solve a differential equation for speed and then integrate? –  Johannes Kloos May 15 '12 at 19:44
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Solve for $v(t)$ by separation of variables; then integrate to get $x(t)$. –  Arturo Magidin May 15 '12 at 19:51

2 Answers 2

As suggested by Arturo: $$ \begin{eqnarray} \int_{0}^{t} \frac{\dot{v}(t)}{a v(t)^2+ b v(t) + c} \mathrm{d} t &=& \int_0^t \mathrm{d} t \\ \int_{v_0}^{v_t} \frac{\mathrm{d} u}{a u^2 + b u + c} &=& t \\ \int_{v_0}^{v_t} \frac{4 a }{\left(2 a u + b\right)^2 + 4 a c - b^2} \mathrm{d} u &=& t \\ \left.\frac{2}{\sqrt{4 a c - b^2}} \arctan\left(\frac{2 a u + b}{\sqrt{4 c a - b^2}}\right)\right|_{v_0}^{v_t} &=& t \\ \frac{2}{\sqrt{4 a c - b^2}} \arctan\left(\frac{2 a v_t + b}{\sqrt{4 c a - b^2}}\right) &=& \frac{2}{\sqrt{4 a c - b^2}} \arctan\left(\frac{2 a v_0 + b}{\sqrt{4 c a - b^2}}\right) + t \\ \frac{2 a v_t + b}{\sqrt{4 c a - b^2}} &=& \tan\left( \arctan\left(\frac{2 a v_0 + b}{\sqrt{4 c a - b^2}}\right) + \frac{\sqrt{4 c a - b^2}}{2} t \right) \\ \frac{2 a v_t + b}{\sqrt{4 c a - b^2}} &=& \frac{ \frac{2 a v_0 + b}{\sqrt{4 c a - b^2}} + \tan\left(\frac{\sqrt{4 c a - b^2}}{2} t\right)}{1 - \frac{2 a v_0 + b}{\sqrt{4 c a - b^2}} \tan\left(\frac{\sqrt{4 c a - b^2}}{2} t\right)} \end{eqnarray} $$ Now one solves for $v_t$, and integrates over time: $$ x(t) = x(0) + \int_0^t v_t \mathrm{d} t $$

The integral is not likely to admit evaluation in closed form. Typically trajectories for such system are defined parametrically.

Added
As pointed out in @ChristianBlatter answer, the integral is indeed elementary: $$ x(t) = x(0) - \frac{b}{2a} t - \frac{1}{2a} \log\left(\frac{4 a \left(a v_0^2 + b v_0 + c\right)}{4 a c-b^2} \cos^2\left(\frac{t}{2} \sqrt{4 a c - b^2} + \arctan\left(\frac{2 a v_0 + b }{\sqrt{4 a c - b^2}}\right)\right)\right) $$ Here is a comparison of the exact solution to the numerical solution of the differential equation in Mathematica:

enter image description here

Several physical conclusions:

  • if $4 a c - b^2 < 0$, trigonometric cosine becomes hyperbolic cosine, and position $x(t)$ increases with time linearly, approximately linearly.
  • if $4 a c - b^2 > 0$, the position $x(t)$ plunges to $-\infty$ in finite time, determined by zero of the cosine function.
  • if $4 a c = b^2$, the solution is easy to re-derived. It also goes to $-\infty$ in finite time (if $2a v_0 + b > 0$), or goes $\pm\infty$ (if $2 a v_0 + b < 0$), depending on initial conditions: $$ x(t) = x(0) - \frac{ b t}{2a} - \frac{1}{a} \log\left( 1 - \frac{b t}{2} - a t v_0\right) $$
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Integrating over time gives me (log(tan(atan(b/(4*a*c - b^2)^(1/2) + (2*a*v0)/(4*a*c - b^2)^(1/2)) + (t*(4*a*c - b^2)^(1/2))/2)^2 + 1) - b*t)/(2*a) when I incorportate my a, b, c and v0 values x oscillates and goes to infinite when t = inf. and that doesn't make sense with the physics of my situation so I thought I had made ​​a mistake. –  ACL May 16 '12 at 19:49
    
@ACL I have extended my answer, with exact solution, comparison with simulation, and analysis of long-time behavior. –  Sasha May 16 '12 at 20:46

"Separating variables" means writing $${dv\over av^2+bv +c}=dt\ .\qquad(1)$$ The next step depends on the values of $a$, $b$, $c$. Assuming $a>0$ one has $$a v^2+bv +c={1\over a}\Bigl(\bigl(av +{b\over 2}\bigr)^2+{4ac -b^2\over 4}\Bigr)\ ,$$ so that after a linear substitution of the dependent variable $v$ the equation $(1)$ transforms into one of $${du\over 1+u^2}= p\ dt,\quad{du\over u^2}=p\ dt,\quad {du\over u^2-1}=p \ dt\ ,$$ depending on the value of $4ac-b^2$. Up to scalings and shifts the first form implies $$\arctan u = t\quad{\rm or}\quad u=\tan t\ .$$ It follows that (again neglecting scalings and shifts) $$x(t)=\int_{t_0}^t v(t)\ dt=\int _{t_0}^t \tan t\ dt=-\log(\cos t)\Bigr|_{t_0}^t\ .$$

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And the second form becomes $u = - \tanh t$ and $x(t) = \left. - \log(\cosh t) \right|_{t_0}^t$. –  Robert Israel May 16 '12 at 19:36
    
Putting back the scalings and shifts, it's $$ x(t) = x_0 -\dfrac {b(t-t_0)}{2a} - \dfrac{\log (\cos(\sqrt{ac-b^2/4}(t-t_0)))}{a}$$ (if $4 ac > b^2$), or $$ x(t) = x_0 -\dfrac {b(t-t_0)}{2a} - \dfrac{\log (\cosh(\sqrt{b^2/4-ac}(t-t_0)))}{a}$$ (if $4ac < b^2$) –  Robert Israel May 16 '12 at 19:52

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