Differential equation where forcing function depends on variable

Question

The force acting on a body depends on the displacement $x(t)$ of the body from the origin. This dependence is given by

$$ F(x) = \begin{cases} F_0 - mg, &-L < &x < 0,\\ -mg,& &x > 0. \end{cases} $$

where $L$ is a positive number, $F_0 > 0$ is a constant resting force and $g$ is the gravitational acceleration of Earth.

Newton's second law gives $m\ddot{x} = F(x)$ and we impose the initial conditions $\dot{x}(0) = 0, x(0) = -L.$

How does one solve an equation like this? My main problem is that if we restrict to the branch $x > 0$ we get $x = -g\frac{t^2}{2} + At + B$. I don't know if I am allowed to plug in the boundary conditions here, but if I were to, I would get $x = -g\frac{t^2}{2} - L$. But this function is clearly always negative, so then how could this satisfy the branch $x>0$?

Any tips or clarifications?

Answer 1

You just have to solve two ODE's. First solve the ODE for $x\in[-L,0)$ using your initial conditions $\dot{x}(0)=0$, $x(0)=-L$. That equation is valid until $x=0$. Then you can get values of $\dot{x}$ for when $x=0$. That is your new initial condition to use when solving the second ODE.

Answer 2

From

$$ m\ddot x = F(x) = \left(\theta(x+L)-\theta(x)\right)(F_0-m g)- m g\theta(x) $$ where $\theta(x)$ is the Heaviside step function, we have

$$ m\ddot x\dot x = F(x)\dot x $$

or equivalently with $v = \dot x$

$$ m v dv = F(x)dx $$

and then

$$ \frac 12 m v^2 = (x+L)(F_0-m g)\theta(x+L)-x F_0\theta(x)+C_0 $$

now assuming that $v(-L) = 0\Rightarrow C_0=0$ we have

$$ \frac{dx}{\sqrt{(x+L)(F_0-m g)\theta(x+L)-x F_0\theta(x)}} = \sqrt{\frac 2m}dt $$

now assuming $-L \le x \le 0$

$$ \int_{-L}^x\frac{du}{\sqrt{(u+L)(F_0- m g)}}= \sqrt{\frac 2m}t $$

giving

$$ x(t) = \frac{t^2 (F_0-g m)}{2 m}-L $$

and for $x > 0$

$$ \int_{-L}^0\frac{du}{\sqrt{(u+L)(F_0- m g)}}+\int_0^x\frac{du}{\sqrt{(u+L)(F_0- m g)-u F_0}}= \sqrt{\frac 2m}t $$

giving

$$ \frac{2L}{\sqrt{L(F_0-m g)}}+\frac{2 \left(\sqrt{L (F_0-m g)}-\sqrt{F_0 L-m g(L+x)}\right)}{m g} = \sqrt{\frac 2m}t $$

and now inverting we can find $x(t)$

$$ x(t) = \frac{F_0}{g m-F_0}-1+\frac{ F_0 L \sqrt{\frac{2}{m}}}{\sqrt{L (F_0-g m)}}t-\frac{g}{2}t^2 $$

Answer 3

The motion is a sequence of two uniformly accelerated movements. The first is immediately found to be

$$x(t)=(f_0-g)\frac{t^2}2-L$$ and $$\dot x(t)=(f_0-g)t.$$

$x$ cancels after

$$t^*=\sqrt{\frac{2L}{f_0-g}}$$ seconds, with speed $$v^*=\sqrt{2L(f_0-g)}.$$

Then the motion continues as

$$x(t)=-g\frac{(t-t^*)^2}2+v^*(t-t^*).$$

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$f_0:=\dfrac{F_0}m.$