Harmonic oscillator with squared damping term

Question

Does a solution exist for a harmonic oscillator with a squared damping term?

$$m\ddot{u}+c\dot{u}^2+ku=0$$

Answer 1

Yes, you can look at Smith (2012) who discusses how you might go about getting solutions to this non-linear differential equation. The results are mathematically very challenging to say the least. They are also not exact, but work as close approximations to what you would get from a brute force numerical technique.

The outcome appears to be that the quadratically damped oscillator undergoes a decaying oscillation, but that the decay rate is slower than exponential (see Fig.6). A second important conclusion is that such an oscillator can never be overdamped or critically damped.

Such systems are of interest when the damping is provided by (for instance) turbulence in a fluid or a damper where there is an opening that allows fluid to escape.

Answer 2

When I first learned about harmonic oscillators, I played with Mathematica and learned different solutions for different cases. I tried to solve this equation very long time ago, but unfortunately, I couldn't find any analytical solution for it.

So I have good news and bad news.

The bad news: I don't think analytical solutions exist. If such a model equation has a solution, I think Mathematica would have had it, and it doesn't, so most likely no analytical solutions exist.

The good news is: Analytical solutions are not the only way to solve such a problem. Depending on your application, you can solve this differential equation numerically. Software like Matlab, Mathematica and Maple, and even low programming languages like C/C++ and Fortran can very, very easily solve this equation numerically for any set of parameters you like. Then you can remodel the cases you need with algebraic equations, depending on the complexity of the numerical solutions.

More about such equations:

It may seem attractive to students to solve such equations just for the hell of it, but in reality, it's not important at all. In the physics of the real world, systems are usually linear with a very good approximation. Normally when such crazy, nonlinear terms show up, we don't try to solve their equations analytically (unless it's really easy). The way to go is by using something similar to what we do in quantum mechanics: Perturbation Theory. So we learn the analytical solution of the simple case, and then we add small nonlinear perturbations that make us understand our system better. So in your case, you could find a way to write this equation in a linear way, then make the quadratic term a small perturbation, which is solvable for some approximation.

The bigger challenge on differential equations is not such nonlinear equations, which don't occur in physics that often. Actually it's orders of magnitude more difficult to solve systems of differential equations that occur very often in nature, such as coupled systems of Bloch equations. Although such a system is linear, it's still not possible to generally get an analytical solution for it due to a theorem called the Abel-Ruffini theorem. Notice that the guys in that paper still found a solution to the problem they have in their lab. They simplified the system with an approximation from 6 to 4 equations. That worked well for them!

So the bottom line is: Depending on your application, you may find approximations that will get you the information you need from this differential equation.

Answer 3

Assuming we want to study the differential equation

$$\ddot{y} + \alpha\dot{y}^2 + \beta y = 0$$

Then we can easily get rid of $\alpha$ and $\beta$.

$$ \frac{d^2 y}{dt^2}+\alpha \left(\frac{d y}{dt} \right)^2+\beta y=0 $$

Replacing:

$$ t \rightarrow \frac{t}{\sqrt{ \beta }}~~~~~y \rightarrow \alpha y $$

We will get.

$$ \frac{d^2 y}{dt^2}+\left(\frac{d y}{dt} \right)^2+y=0 $$

Now we make a replacement

$$ w(y)=\left(\frac{d y}{dt} \right)^2 $$

Then:

$$ \frac{d^2 y}{dt^2}=\frac{d \sqrt{w}}{dt} = \frac{d \sqrt{w}}{d y} \frac{d y}{d t} = \frac{d \sqrt{w}}{dy} \sqrt{w}=\frac{d w}{2dy} $$

The equation transforms to:

$$ \frac{d w}{dy}+2 w+2 y=0 $$

We only need to solve two first order ODEs. The solution for $w$ can be found easily (I used Wolframalpha):

$$ w(y)=C_1 e^{-2y}-y+\frac{1}{2} $$

Then we need to solve first order nonlinear equation:

$$ \left(\frac{d y}{dt} \right)^2=C_1 e^{-2y}-y+\frac{1}{2} $$

$$ \frac{d y}{dt}=\pm \sqrt{C_1 e^{-2y}-y+\frac{1}{2}} $$

Since this equation is fully separable, we can try to approximate the integral. For $C_1=0$ this ODE is exactly solvable.

$$ y=\frac{1}{2}-\left(\frac{t}{2}+C_2 \right)^2 $$

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As for the best way to approximate the solution - it heavily depends on the context - if it was a simple pendulum in air for example, then you have harmonic oscillator motion with resistance proportional to velocity squared, and the second term can be interpreted as perturbation. However, since $\alpha$ is positive, it's apparently some other case. There is also the case of nonlinear Schrödinger, and there is a number of methods developed to solve this problem also.

My point is - there can be no 'general' approximate solutions, the appropriate approximation depends on the context of the original problem.