Velocity as a function of position, given velocity as a function of time

Question

Knowing the expression for the acceleration as a function of time: $$ \frac{dv}{dt} = - c v^n$$ (for some constant c >0 and n >1), one needs to find the velocity as a function of time and as a function of position. Solving for the velocity as a function of time is pretty straightforward and has lead me to the following: $$ v(t) = [ (n-1)ct + v_{O}^{(1-n)}]^{\frac{1}{1-n}} $$ Given the initial condition that $ v(0) = v_0$. Now, to find $v(x)$, I tried to use
$$ \frac{dv}{dx} = \frac{dv}{dt} \frac{dt}{dx} = a(t) \frac{1}{v(t)} $$ This leads to $$ \int \frac{dv}{v^n} = \int \frac{-c}{ [ (n-1)ct + v_{O}^{(1-n)}]^{\frac{1}{1-n}}} dx = -ct $$ Where the last equality is obtained from the first DE after separating variables. I don't see how this could possibly give me $v(x)$. How could one change variables in the integral without being in the possession of a relationship between $t$ and $x$ (Or am I? Am I overseeing something?), which if I had I could simply insert into the RHS of the equation, without having to worry about the integral at all. Any push into the right direction is appreciated.

Answer 1

I want to emphasize that the '$t$' in the fourth equation should be treated as a function of $x$(displacement).

Let me just rewrite your third formula with a input $\tilde{x}$ (You probably have applied the inverse function theorem.) : $$ \frac{dv}{dx} (\tilde{x})= \frac{dv}{dt}(\,t(\tilde{x})\,)\,\, \frac{dt}{dx} (\tilde{x}) = a( t(\tilde{x}) ) \frac{1}{\frac{dx}{dt} (t (\tilde{x}))} = \frac{a( t(\tilde{x}) )}{v (t (\tilde{x}))} $$

, where $t(\tilde{x})$ is the inverse function of the time-displacement function.

This slightly different notation lets us understand the integral more precisely. Indeed, your integral is precisely $$ \int_{v_0}^{v(\tilde{x})} \frac{dv}{v^n} = \int_{0}^{\tilde{x}} \frac{-c}{ [\, (n-1)c\,\,t(x) + v_{O}^{(1-n)}\,]^{\frac{1}{1-n}}} dx \quad \text{}. $$ i.e. $$ \frac{1}{1-n} \left( v(\tilde{x})\,^{1-n} \,-\,v_0\,^{1-n} \right) = \mathrm{D}(\tilde{x}). $$

( D is the result of our integral. (i.e our Dream function) )

Then our $v(x)$ can be derived as follows. $$ v(x) = \left[ {v_0}^{1-n} - (n - 1)\, D(\tilde{x}) \right]^{1 / {1-n}}\, . $$