Optimizing a Path

A particle of mass m m travels along a one-dimensional path from ( x s , t s ) = ( 0 , 0 ) (x_s, t_s) = (0,0) to ( x f , t f ) = ( D , 2 T ) (x_f, t_f) = (D, 2 T) . Here, x x and t t represent space and time coordinates. From t = 0 t = 0 to t = T t = T , the particle has constant velocity v 1 v_1 , and from t = T t = T to t = 2 T t = 2T , the particle has constant velocity v 2 v_2 .

Define the quantity S S as follows:

S = 1 2 m v 1 2 T + 1 2 m v 2 2 T S = \frac{1}{2} m v_1^2 T + \frac{1}{2} m v_2^2 T

The quantity S S is a function of variables v 1 v_1 and v 2 v_2 . Subject to the constraint v 1 T + v 2 T = D v_1 T + v_2 T = D , determine the values of v 1 v_1 and v 2 v_2 such that the following is true.

S v 1 = S v 2 = 0 \frac{\partial{S}}{\partial{v_1}} = \frac{\partial{S}}{\partial{v_2}} = 0

What is the relationship between the resulting v 1 v_1 and v 2 v_2 values.

Note: To get the most out of this problem, solve it on its own terms without applying any Newtonian intuition. What is the physical significance of this result?

v 1 > v 2 v_1 > v_2 v 1 = v 2 v_1 = v_2 v 1 < v 2 v_1 < v_2

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1 solution

Karan Chatrath
Feb 13, 2021

We are asked to minimize the convex function:

S = m T 2 ( v 1 2 + v 2 2 ) S = \frac{mT}{2}\left(v_1^2 + v_2^2\right)

Subject to:

v 1 + v 2 = D T v_1+v_2 = \frac{D}{T}

Introducing a Lagrange multiplier and augmenting the cost function: F = S + λ ( v 1 + v 2 D T ) F = S + \lambda\left(v_1+v_2 - \frac{D}{T}\right)

Now:

F v 1 = m T v 1 + λ = 0 \frac{\partial F}{ \partial v_1} = mTv_1 + \lambda = 0 F v 2 = m T v 2 + λ = 0 \frac{\partial F}{ \partial v_2} = mTv_2 + \lambda = 0

Solving for v 1 v_1 and v 2 v_2 in terms of λ \lambda leads to the conclusion that:

v 1 = v 2 v_1=v_2

Now, coming to the physics of this problem. The quantity S S can be thought of as the action integral - The time integral of the Lagrangian:

S = 0 2 T ( T V ) d t S =\int_{0}^{2T} (\mathcal{T} - \mathcal{V}) \ dt

In the absence of any external forces or potentials, the action integral becomes:

S = 0 2 T ( T ) d t S =\int_{0}^{2T} (\mathcal{T} ) \ dt S = 0 T m v 1 2 2 d t + T 2 T m v 2 2 2 d t S =\int_{0}^{T} \frac{mv_1^2}{2} \ dt + \int_{T}^{2T} \frac{mv_2^2}{2} \ dt S = m T 2 ( v 1 2 + v 2 2 ) S = \frac{mT}{2}\left(v_1^2 + v_2^2\right)

So essentially, according to the problem statement, at time t = T t = T , there is an abrupt change in the speed of the particle. From the laws of motion, we know that this is an impossibility in the absence of a force or potential energy that tends to drive the particle. Therefore, our intuition tells us that the speed must not change at any instant. This is in agreement with the result of the optimization problem.

Another very neat observation that we see from the optimization problem is that the particle's linear momentum is always constant, which is also in agreement with classical intuition that momentum is conserved in the absence of external forces.

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