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.

To communicate my experience level and intent: I'm an undergraduate, this is not homework, I'm trying to write a physical simulation for fun and xp and am stuck just before (what looks to me like) the end of the tunnel.

I am attempting to solve a system of ordinary nonlinear differential equations and have either made a mistake that I can't see or have run out of tools to hit this thing with.

$v'(t) = \alpha r^2 v^2$

$r'(t) = \beta v^2$

In my attempt to solve it, I assume $v(t) = c_v e^{st}$, which implies:

$v'(t) = sce^{st} = \alpha r^2 c_v^2 e^{2st}$

$r'(t) = \beta c_v^2 e^{2st}$.

I solve $v'(t)$ for $c_v$ and make the substitution $c_v = \dfrac{s}{\alpha r^2}e^{-st}$ to eliminate $c_v$ from $r'(t)$:

$r'(t) = \dfrac{\beta s^2}{\alpha^2r^4}$

Now I solve for $r(t)$ by the separation of variables $\int{r^{-4}dr} = \int{\dfrac{\beta s^2}{\alpha^2}dt}$:

$r(t) = \bigg(\dfrac{1}{-3\beta(\frac{s}{a})^2t-3c_r)}\bigg)^{1/3}$

Setting $r(0) = r_0$, I solve the IVP for $c_r = \dfrac{-1}{3r_0^3}$,

and in the interest of carpal health I make the substitutions:

$-3\beta\big(\dfrac{s}{a}\big)^2 = \gamma$

$-3c_r = \dfrac{1}{r_0^3} = \delta$

which rewrites $r(t)$ into

$r(t) = (\gamma t + \delta)^{-1/3}$

Okay, one down. Now back to $v(t)$... I expand $r(t)$ into its new form to give

$v'(t) = \alpha v^2 (\gamma t + \delta)^{-2/3}$

Now I use separation of variables again, though because $r'(t)$ is a function of $v'(t)$ and I already assumed $v = c_v e^{st}$ I'm not totally sure this is ok, but...

$\int{\dfrac{dv}{v^2}} = \int{\alpha (\gamma t + \delta)^{-2/3}dt}$

which gives

$v(t) = \dfrac{-\gamma}{\gamma c_v + 3(\gamma t + \delta)(\gamma t + \delta)^{-2/3}}$

Now I get stuck. I isolate $c_v$ by setting $v(0) = v_0$ and substitute back in all of my $\gamma$s and $\delta$s to get the full picture:

$c_v = -\dfrac{1}{v_0} + \dfrac{1}{r_0\beta(\frac{s}{\alpha})^2}$

I've still got that pesky $s$ in there from when I assumed $v(t) = c_v e^{st}$ and I'm out of equations to extract it with. I can't isolate $s$ in my equation for $r(t)$, I don't have the skill to begin to unravel that mess and I don't even know if its possible. Wolfram Alpha reports no solutions, for what its worth.

So, how do I get at that $s$, and if I can't, where did I go wrong?

share|improve this question

2 Answers 2

up vote 1 down vote accepted

Well, the Mathematica solution is horrifying, involving inverse arbitrary functions. I can at least verify that your initial ansatz is definitely wrong; $v$ is not an exponential function of $t$. The inverse functions will require numerical solutions, so you may be just as well off numerically solving your differential equations.

share|improve this answer
    
Could you tell me, how were you able to verify that the initial ansatz is wrong? Is it because the expression I wound up with for $v(t)$ is not exponential? I actually find it a bit puzzling that I wound up with that expression after guessing an exponential. –  evenex_code Jun 5 at 6:05
    
Actually, it's because the result Mathematica produces is not exponential. Your observation also works since your ansatz should be self-consistent and is not. –  Eric Towers Jun 5 at 6:09

Starting with $$ v'(t) = \alpha r^2 v^2,\\ r'(t) = \beta v^2. $$

we can write $$ v' = \dfrac{\alpha}{\beta}r^2r' = \dfrac{\alpha}{\beta}\frac{d}{dt}\dfrac{r^{3}}{3} $$ spo yeilding $$ v(t) = \dfrac{\alpha}{3\beta}r^3 + C_{1},\\ r(t) = \left(\dfrac{3\beta}{\alpha}\right)^{1/3}\left(v(t)-C_1\right)^{1/3} $$ now we have $$ v' = \alpha\left(\dfrac{3\beta}{\alpha}\right)^{2/3}\left(v(t)-C_1\right)^{2/3}v^2 = \alpha\left(\dfrac{3\beta}{\alpha}\right)^{2/3}\left[v\left(v-C_1\right)^{1/3}\right]^2,\\ r'=\beta \left[\dfrac{\alpha}{3\beta}r^3 + C_{1}\right]^2 $$

Looking at $v'$ first

$$ \int\frac{dv}{\left[v\left(v-C_1\right)^{1/3}\right]^2} = \lambda_0 t + C_2 $$ where $\lambda_0 = \alpha\left(\dfrac{3\beta}{\alpha}\right)^{2/3}$

using the change of variable $X = (v-C_{1})^{1/3} \rightarrow dX = \frac{1}{3}(v-C_{1})^{-2/3}dv$

we obtain $$ \int \dfrac{dX}{\left(X^3+C_1\right)^2} = \lambda_1 t + C_2 $$

and $\lambda_1 = 3\lambda_0$

focus on $r'$

$$ \int \frac{dr}{\left(r^3 + \lambda_2\right)^2} = \lambda_3 t + C_3 $$ where $$ \lambda_2 = \dfrac{3\beta}{\alpha}C_1,\\ \lambda_3 = \dfrac{9\beta}{\alpha^2} $$

So basically the analytic solution will be found $\textbf{if}$ one can compute the integral $$ \int \frac{dy}{\left(y^3+b\right)^2}? $$

Though this rather extended comment might all be for nought :).

$\textbf{update:}$ I will not claim to solve this integral myself (wolfram ;)) but apparently the integral above equates to $$ \frac{1}{9b^{5/3}}\left[\frac{3b^{2/3}y}{b+y^3} - \log\left(b^{2/3}-b^{1/3}y+y^2\right) + 2\log\left(b^{1/3} + y\right) - 2\sqrt{3}\tan^{-1}\left(\frac{1-\frac{2y}{b^{1/3}}}{\sqrt{3}}\right)\right] $$

..apparently. Hopefully someone can verify this in the meantime.

share|improve this answer

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.