# Converting Newton's Second Law Differential Form to Integral Form and Integration Limits

I am reading a report titled Definition of safe separation criteria for external store and pilot escape capsules'' in which the author uses following expressions.

From Newton's second law: $$z \left( t\right) = \frac{1}{m} \int \int\limits_{0}^{t}F \left( \tau \right)d\tau dt$$

From Taylor Series,

$$F \left(\tau \right) = F \left( 0 \right) + \sum\limits_{n=1}^{\infty} \frac{d^{n}F}{d\tau^n} \Bigg\rvert_{\tau = 0} \frac{\tau^{n}}{n!}$$

Substituting the Taylor series expansion in first equation, we get

$$z \left( t\right) = \frac{1}{m} \left[ \frac{F \left( 0 \right) t^{2}}{2!} + \sum\limits_{n=1}^{\infty} \frac{d^{n}F}{d\tau^n} \Bigg\rvert_{\tau = 0} \frac{\tau^{n+2}}{\left(n + 2 \right)!}\right] + w \left( 0 \right) t + z \left( 0 \right)$$

Questions:

1. I am familiar with differntial equation form of Newton's Second law, i.e., $$F = m \frac{d^{2}z}{dt^{2}}$$ but I don't know how the author obtained integral form, especially the limits of integration.
2. Since the integral with respect to $$\tau$$ is a definite integral, why the author has added constant of integration when integrating with respect to $$\tau$$.

### Review of Terms

w(t) -> v(t) -> velocity at time t

z(t) -> x(t) -> position at time t

w(0) -> initial velocity

### Derivation

We can compute $$w(t)$$ from force, since $$\frac{dw}{dt} = F(t) / m$$, so $$w(t) - w(0) = \int_{0}^{t}\left(\frac{1}{m}F(\tau)d\tau\right)$$

Which means $$w(t) = w(0) + \int_{0}^{t}\left(\frac{1}{m}F(\tau)d\tau\right)$$

We can compute $$z(t)$$ from velocity: $$z(t) - z(0) = \int_{0}^{t} \left( w(\tau) d\tau \right)$$

Where we substitute the above integral expression for $$w(t)$$: $$z(t) - z(0) = \int_{0}^{t}\left( w(0) + \int_{0}^{t}\frac{1}{m}F(\tau) d\tau\right)d\tau$$

Since $$w(0)$$ is constant, we can compute its integral in closed form: $$w(0) * t$$.

$$z(t) = z(0) + w(0)t + \int_{0}^{t}\left(\int_{0}^{t}\frac{1}{m}F(\tau) d\tau\right)d\tau$$

And you can use that Taylor expansion to write an expression for this integral.

### Conclusion

1. The integral form is the double-integral of the differential equation.

2. The integration constants you are seeing come from the fundamental theorem of calculus — they’re the initial conditions. Recall the classic kinematic equation: $$z(t) = z(0) + w_0t + \frac{a_0t^2}{2}$$ for the same expression, where acceleration is known to be constant. $$w_0$$ is $$w(0)$$. The body’s initial velocity applies a linear offset to its motion

I have never seen such an expression of the Newton second law. From the first equation you provide, it seems that there are two different time variables, $$t$$ and $$\tau$$, which is obviously not the case. From what I get this notation is just a translation of a double integration in time. I agree with you that the limits of the integration are quite confusing. To my mind both integral limits should be 0 and t, and I would rewrite the whole equation this way: $$z \left( t\right) = \frac{1}{m} \int\limits_{0}^{t} (\int\limits_{0}^{t}F \left( \tau \right)d\tau) d\tau$$ As for your second question, the constants of integration just reflect the physical initial conditions. $$w(0)$$ is the velocity and $$z(0)$$ the position at time $$t=0$$, I do not think it has something to do with the integral being definite.

• I think integration constants are not added when evaluating definite integrals. Thanks. Apr 14 '20 at 9:37