The case of an initial velocity greater than final velocity with a drag force proportional to the velocity squared

Question

consider an object with a mass $m$ falling in a fluid with a drag force proportional to its velocity squared $(f=kv^2)$. the governing differential equation can be found using Newton's second law of motion as $$ A \frac{dv}{dt} + v^2 = v_{lim}^2 ~,~v(0)=v_0$$ where $A=\frac{m}{k}$ and $v_{lim}=\sqrt{\frac{mg}{k}}$ is the final velocity with the initial condition $v(0)=v_0$.
the solution of the equation is $$v(t)=v_{lim} \tanh \left(\frac{v_{lim}t}{A}+ \tanh^{-1} \left(\frac{v_0}{v_{lim}} \right) \right)$$ where $\tanh^{-1}$ is the inverse hyperbolic tangent.

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The given solution does reflect the physical phenomenon in the case of $(v_0<v_{lim})$ ie the velocity increases from $v_0$ to $v_{lim}$.

In the case of $(v_0>v_{lim})$, the expected physics "behavior" of the solution is that the velocity decreases over time (from $v_0$ to $v_{lim}$) and yet we never see such decrease when plotting the function due to the fact that the $\tanh^{-1}(x)$ function is only defined for $x<1~ie~(v_0<v_{lim})$.
So, is there maybe another formula that better describe the second case?

Answer 1

Using the software Maple 2022 to solve the differential equation, I end up with the solution

$$v(t) = v_\mathrm{lim} \frac{\frac{v_0}{v_\mathrm{lim}} \cosh(x) + \sinh(x)}{\frac{v_0}{v_\mathrm{lim}} \sinh(x) + \cosh(x)},$$

where $x = t \frac{v_\mathrm{lim}}{A}$.

This solution agrees with yours for $v_0 < v_\mathrm{lim}$ but is also valid when $v_0 \geq v_\mathrm{lim}$.

It seems to me that the solution given to you is simply only valid in the case $v_0 < v_\mathrm{lim}$.

Answer 2

You were quite right to worry that the $\tanh(x)$ function seems not to give the solution to your problem but there was a function of time involved.

To keep things simple assume that $v_{\rm lim}=1$ and $A=1$.

$v(t)=1\cdot \tanh \left(\frac{1}{1}\cdot t+ \tanh^{-1} \left(\frac{v_0}{1} \right) \right)=\tanh \left(t+ \tanh^{-1} \left(v_0 \right) \right)$

Lets see what happens as the initial velocity, $v(0)$, changes relative to the terminal velocity.

If $v(0)=1$ then you get that $v(t) = \tanh \left(t+ \tanh^{-1} \left(1 \right) \right)=1$, as expected?

Now an example where the body will speed up with $v_0 = \frac 12$.

What you did not like about your solution was that it did not seem to give a sensible view of what happened when the initial velocity was greater than the terminal velocity.
So with $v_0=2$ your solution is as follows.

$\LARGE !!!!!$

That final solution can be written in many alternative ways and here is a sample.

But this form of the solution might be the easiest to comprehend?

$v(t) = \dfrac{3e^{2t}+1}{3e^{2t}-1}$

and this form of the solution only seems to be generated by applications like WolframAlpha and Maple when the constants are given numerical values.

Answer 3

It's not strictly true that the inverse hyperbolic tangent function is undefined for $|x| > 1$. You can, it turns out, extend the function to the entire complex plane; and for real $x > 1$ (or $x < -1$), the function is complex-valued.

Under this definition, the $\tanh^{-1}$ function satisfies the identity $$ \tanh^{-1} (z) = \tanh^{-1} \left( \frac{1}{z} \right) \pm \frac{i \pi}{2} $$ where the plus-or-minus depends on the value of $z$. For real $z > 1$, the minus sign applies, and so $$ v(t)=v_{lim} \tanh \left(\frac{v_{lim}t}{A}+ \tanh^{-1} \left(\frac{v_{lim}}{v_0} \right) - \frac{i \pi}{2} \right) $$ Moreover, we also have $$ \tanh( x - i \pi/2) = \coth(x) $$ and so the actual value yielded by the function is still real.