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I'm having a lot of trouble with this particular differential equation, mainly because I don't think I've ever done a nonlinear second order ODE. But here's the equation:

$$y''(t)+2\beta [y'(t)]^2+c^2 y(t)=0 \quad \text{where $\beta$ and $c^2$ are constants}$$ $$\text{initial conditions:}\quad y(0)=0, \quad y'(0)=v_0$$

My first thought was to do a Laplace Transform, but I'm pretty sure those only work on linear differential equations, so I'm not sure how to proceed..

Like I'm just genuinely stuck on how to even start. If possible it would be preferable to find an analytical solution. Any help is appreciated, thanks in advance.

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    $\begingroup$ Autonomous equation: use substitution $v(y)=y'(t)$. Then multiply by an integrating factor and find $v(y)$. In principle this can be integrated again (since $v=dy/dt$) then inverted to find $y(t)$. $\endgroup$
    – Sal
    Sep 5, 2021 at 22:04
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    $\begingroup$ What's the source of this problem? Do you have any reason to believe that it can be solved explicitly? $\endgroup$ Sep 6, 2021 at 2:24
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    $\begingroup$ If this is for an oscillator (mechanical spring or something) with air friction, then the second term should be $2β|y'(t)|y'(t)$, so that energy gets dissipated in any movement direction. $\endgroup$ Sep 6, 2021 at 15:26
  • $\begingroup$ @LutzLehmann Haha you're spot on about the premise, it's supposed to be an oscillator with quadratic drag. I was able to solve the same problem with linear drag, but the force for the quadratic air resistance is written as $-km\dot q^2$, so i figured it was literally the derivative squared.. $\endgroup$
    – pyat
    Sep 6, 2021 at 19:43
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    $\begingroup$ Then you can do some perturbation calculations if $β$ or $c$ are small, or do some averaging calculus to capture the general decline in amplitude. But I'm quite certain that no exact solution for the general case exists. $\endgroup$ Sep 6, 2021 at 20:35

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Let $u=y'$ and $u'=\frac{du}{dy}u$.
Then $\frac{du}{dy}u+2\beta u^2+c^2y=0 \Rightarrow (2\beta u^2+c^2y)dy+udu=0 \cdots (a)$.
Let's denote $M=2\beta u^2+c^2y$ and $N=u$.
Since $\frac{\partial M}{\partial u}=4\beta u$, $\frac{\partial N}{\partial y}=0$, we have $\frac{M_u-N_y}{N}=4\beta$.
So by multiplying $\mu(y)=e^{\int \frac{M_u-N_y}{N}dy}=e^{4\beta y}$ to the both side of $(a)$, we can make it exact.
Thus, $e^{4\beta y}(2\beta u^2+c^2y)dy+e^{4\beta y}udu=M^*dy+N^*du=0$.
So there exists a function $f$ satisfying $\frac{\partial f}{\partial y}=M^*$ and $\frac{\partial f}{\partial u}=N^*$, where $f=0$ is the solution of ODE.
Since $\frac{\partial f}{\partial u}=e^{4\beta y}u$, we have $f=\frac{1}{2}e^{4\beta y}u^2+g(y)$.
Now $\frac{\partial f}{\partial y}=2\beta e^{4\beta y}u^2+g'(y)=2\beta e^{4\beta y}u^2+c^2e^{4\beta y}y \Rightarrow g'(y)=c^2e^{4\beta y}y$.
By integration, $g(y)=\frac{c^2}{4\beta}\left(ye^{4\beta y}-\frac{1}{4\beta}e^{4\beta y}\right)+C_1$ where $C_1$ is general constant.
As a conclusion, we obtain the solution $f=\frac{1}{2}e^{4\beta y}u^2+\frac{c^2}{4\beta}\left(ye^{4\beta y}-\frac{1}{4\beta}e^{4\beta y}\right)+C_1=0$.
Now substitute the initial condition $y(0)=0$ and $y'(0)=u(0)=v_0$.
Then $C_1=\frac{c^2}{16\beta^2}-\frac{1}{2}v_0^2$ so that $\frac{1}{2}e^{4\beta y}u^2+\frac{c^2}{4\beta}\left(ye^{4\beta y}+\frac{1}{4\beta}-\frac{1}{4\beta}e^{4\beta y}\right)-\frac{1}{2}v_0^2=0$.
Here, we have $u^2=(y')^2=\frac{c^2}{2\beta}\left(y-\frac{1}{4\beta}\right)+\left(v_0^2-\frac{c^2}{8\beta^2}\right)e^{-4\beta y}=F(y)$.
By assuming $y'(0)=v_0>0$, $y'=\frac{dy}{dt}=\sqrt{F(y)}$.
Since it is a seperable equation, we can solve it by integrating both sides: $t=\int_0^y\frac{1}{\sqrt{F(w)}}dw$.
Unfortunately, I do not think $\frac{1}{\sqrt{F(y)}}=\frac{1}{\sqrt{\frac{c^2}{2\beta}\left(y-\frac{1}{4\beta}\right)+\left(v_0^2-\frac{c^2}{8\beta^2}\right)e^{-4\beta y}}}$ is integrable.
But it would be fine to leave it in the integral form to write down the solution.

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