Impulsive force and simple integration

[Trying2Learn]

TL;DR

Is this process, correct?

May I ask if the following process is correct?

Given: F=ma

Apply an impulsive force using the dirac delta near 0 (with F nearly constant over the tiny impulsive interval)

ma = Fδ(t)

This is a second order differential equation with a forcing function. However, I cannot readily integrate this differential equation.

Instead, I turn to Linear Momentum:

Initial momentum + (integral of force over time) = Final momentum

mv⁺=FΔt + mv^-

With zero initial velocity, I now have:

v⁺ = F/(mΔt)

And now I turn my original differential equation to this

ma = 0

With these two initial conditions:
x(0) = 0
v(0) = F/(mΔt)

The solution is:

x(t) = (F/(mΔt)) * t

This seems strange to me.

Is this process correct?

Can someone explain in words (sorry, I am embarrassed) what I am doing (if this is correct)?

Would I get the same results by solving the original equation with a convolution or numerical method?

It seems so strange to me: as if I skirted the complexity of a nonhomogeneous differential equation (I cheated).

 

[ergospherical]

It's essentially the linear 2nd order ODE:\begin{align*}
\dfrac{d^2 x}{dt^2} = \delta(t)
\end{align*}modulo constant factors. In each of $t<0$ and $t>0$ the forcing term (right-hand side) vanishes so the general solution is a linear combination of $1$ and $t$,\begin{align*}
x = \begin{cases}
a+bt & t<0 \\
c+dt & t>0 \\
\end{cases}
\end{align*}What happens if you integrate the original ODE between $t = -\epsilon$ and $t= \epsilon$? \begin{align*}
\int_{-\epsilon}^{\epsilon} \dfrac{d^2 x}{dt^2} dt &= \int_{-\epsilon}^{\epsilon} \delta(t) dt \ \implies \ \dfrac{dx}{dt} \bigg{|}_{\epsilon} - \dfrac{dx}{dt} \bigg{|}_{-\epsilon} = 1
\end{align*}i.e. the jump conditions at $t=0$ are $\left[\frac{dx}{dt} \right] = 1$ and $[x] = 0$, which imply that $c=a$ and $d=b+1$, i.e.\begin{align*}
x = \begin{cases}
a+bt & t<0 \\
a+(b+1)t & t>0 \\
\end{cases}
\end{align*}

 

[Trying2Learn]

It's essentially the linear 2nd order ODE:\begin{align*}
\dfrac{d^2 x}{dt^2} = \delta(t)
\end{align*}modulo constant factors. In each of $t<0$ and $t>0$ the forcing term (right-hand side) vanishes so the general solution is a linear combination of $1$ and $t$,\begin{align*}
x = \begin{cases}
a+bt & t<0 \\
c+dt & t>0 \\
\end{cases}
\end{align*}What happens if you integrate the original ODE between $x = -\epsilon$ and $x= \epsilon$? \begin{align*}
\int_{-\epsilon}^{\epsilon} \dfrac{d^2 x}{dt^2} dt &= \int_{-\epsilon}^{\epsilon} \delta(t) dt \ \implies \ \dfrac{dx}{dt} \bigg{|}_{\epsilon} - \dfrac{dx}{dt} \bigg{|}_{-\epsilon} = 1
\end{align*}i.e. the jump conditions at $t=0$ are $\left[\frac{dx}{dt} \right] = 1$ and $[x] = 0$, which imply that $c=a$ and $d=b+1$, i.e.\begin{align*}
x = \begin{cases}
a+bt & t<0 \\
a+(b+1)t & t>0 \\
\end{cases}
\end{align*}

I am sorry. I am dense. Would you be so kind as to run that by me again, but explain this a, b, c, d, in terms of initial quiescent conditions?

Yes, I get that in theory you have cases BEFORE and AFTER, but I am only interested in (to ensconce this) what happens AFTER you slug the ball.

 

[ergospherical]

If the particle is at rest at the origin before you hit it, then $a=b=0$. Then you have $x = t$ for $t>0$, which is interpreted as the particle having gained a speed of $1$.

 

[Trying2Learn]

OK, so

If the particle is at rest at the origin before you hit it, then $a=b=0$. Then you have $x = t$ for $t>0$, which is interpreted as the particle having gained a speed of $1$.

OK then, we get the same answer. So what did I do in my solution that is different from what you did?

Is the application of linear momentum the same as your integrating the dirac?

 

[ergospherical]

You didn't arrive at the same answer because you assumed a force $F$ acting over a time $\Delta t$, giving rise to an impulse $F\Delta t$ which depends on $\Delta t$. If on the other hand you consider a delta-function force $F\delta(t)$, then this provides a fixed impulse $F$.

 

[Trying2Learn]

OK then... in a practical engineering problem, where a ball is hit with a bat (ignoring the lift of the ball, and deformation, etc -- just assume the force is transmitted in a "small" interval, ) what is wrong with taking my approach?

Namely, assume the constant contact force, F, happens in Δt?

 

[ergospherical]

Nothing wrong with it, but you're solving a different problem to the one you started with.

 

[Trying2Learn]

Oh I see. You are telling me that my original (first equation) is NOT related to my engineering application?

OK... I get that.

Ignoring then, my very first equation.

In YOUR mathematical (which is pure) understanding, what am I doing when I:

1. use linear momentum to get an initial velocity and, with that
2. solve an homongeneous differential equation

Is it "so engineering" that all mathematical precision is lost?

 

[ergospherical]

I don't understand the question, they are of course related. You can model the delta function $\delta(t)$ in several ways, for example with a top hat function $\delta_{\epsilon}(t)$ defined by
\begin{align*}
\delta_{\epsilon}(t) = \begin{cases}
1/\epsilon & t \in [- \frac{\epsilon}{2}, \frac{\epsilon}{2}]\\
0 & \mathrm{otherwise} \\
\end{cases}
\end{align*}With such approaches you guarantee that the impulse is still of fixed size unity, no matter how small you make the time interval $\epsilon$.

 

[Trying2Learn]

I don't understand the question, they are of course related. You can model the delta function $\delta(t)$ in several ways, for example with a top hat function $\delta_{\epsilon}(t)$ defined by
\begin{align*}
\delta_{\epsilon}(t) = \begin{cases}
1/\epsilon & t \in [- \frac{\epsilon}{2}, \frac{\epsilon}{2}]\\
0 & \mathrm{otherwise} \\
\end{cases}
\end{align*}With such approaches you guarantee that the impulse is still of fixed size unity, no matter how small you make the time interval $\epsilon$.

OK, your answer helps.

But I still get the feeling that I "cheated" to get my answer.
You used the precise definition of the dirac.
I abused it.

But it seems that what I did was legit. (It just seems as if what I did was sneaky).

 

[ergospherical]

What I'm trying to convey is that when you write:

The solution is:
x(t) = (F/(mΔt)) * t

then if F is a constant, the resulting velocity F/(mΔt) diverges as you make Δt smaller and smaller. If you want to model a fixed impulse (which is the more physically reasonable thing to do), you need to ensure that F varies with Δt in such a way that FΔt is a fixed number.

 

[Trying2Learn]

Yes, yes... I am aware of that. Yes, in engineering we do make the Δt small, but not infinitely small.

My issue concerned that bridge between an engineering approximation and the mathematical precision.

Now, with your words, I can see the connection.

Thank you so much for your patience with me!