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I would like to draw a curve between two points that minimizes the square of the second derivative integrated along the curve.

$J(y) = \int_{1}^{0} {y}''^2 dx $

The first derivative for the start and end point are known and must be preserved, and all values on the curve between the start and end point must fall within some range l < e < u.

$y(0) = a, y(1) = b, y'(0) = c, y'(1) = d$

$y(t) = e, \forall t \in [0, 1] \implies l < e < u$

Additionally the curve should be continuous in the interval.

Sincere thanks to Wadim Zudilin for recommending posting an example, I know realize my first post was not even asking the right thing :(

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In this vague formulation (without any example!), it's hard to be helpful. The only related optimization problem is Lagrange multipliers ( and its variational generalization, Pontryagin's principle. – Wadim Zudilin Jun 29 '10 at 9:32
Perhaps you can set the potential energy to infinity outside the box, like they do in billiards. – muad Jun 30 '10 at 17:25
@Wadim - Yes it is vague. This is very much outside my area of expertise and I don't really know the correct terminology to express the problem, for that I apologize. @muad - Thank you for the suggestion. Glancing at the formulation of dynamically billiards it looks like a promising route, although admittedly I am unsure of how to translate a problem expressed in terms of energy to a problem where the conditions are expressed in velocity and position. – Jonathan Fischoff Jul 5 '10 at 8:15
A bit confused: the box is one dimensional? (Since you write l < y(t) < u) Then can't you use the fundamental theorem and conclude that for any path, J(y) = c - d? Or do you actually want to minimize |y''| instead? – Willie Wong Jul 5 '10 at 10:59
@muad: billiard is probably not the right potential to use, at least by itself. If the solution bounces off the walls then the second derivative of y cannot be square integrable (as it'd contain a delta function). So you probably need condition that whenever $y(t) = l \implies y'(t) = 0, y''(t) \geq 0$ and similarly for the upper bound. – Willie Wong Jul 5 '10 at 18:32
up vote 0 down vote accepted

Just to elaborate a bit on what Rahul and I mentioned in the comments.

Take the action functional to be $\int_0^1 (y'')^2 dt$, with prescribed boundary conditions $y(0) = a$, $y(1) = b$, $y'(0) = c$, $y'(1) = d$. For finding the free evolution, take the variation of the function relative to $y$ and set it to zero. Immediately this gives $$\int_0^1 y'' (\delta y)'' dt = 0 $$ for any perturbation. The boundary conditions prescribed implies that $\delta y(0) = \delta y(1) = \delta y'(0) = \delta y'(1) = 0$. So we are allowed to integrate by parts twice (assuming the solution is $C^4$) and obtain $y'''' = 0$, which implies that $y(t)$ is a cubic polynomial in time and thus has 4 free parameters, which we can fix by the boundary values.

The intuition for the bounded case is that, until the evolution hits the wall, locally the equation of motion should be identical to the free evolution. So the solution should be composed piecewise of cubic polynomials. Every time it hits the wall it should receive a hard impact, which suggests that $y''''(\tau_i) = c\delta_{\tau_i}$, the Dirac delta. (The same way that for a hard billiard which away from the walls travel via $x'' = 0$ receive a delta function impact in the second derivative when it hits a wall.) This suggests that $y'''$ is a step function of $t$, and $y''$ is continuous.

For an example, let $l = -1, u = 1$, let the initial time be $t = 0$, and final time be $t = 2$. Let $a = b = 0$, and $c = d = 6$.

First solve the free problem: we want $$ k_3 t^3 + k_2 t^2 + k_1 t + k_0 = y(t) $$ Plugging in the values for the four points we find by solving the linear system that the equation should be $$ y(t) = 3t^3 - 9 t^2 + 6t $$ which achieves its local max and min in $[0,2]$ at $1\pm \sqrt{1/3}$, at which points $|y| = |\pm \sqrt{4/3}| > 1$. So the free evolution is no go.

There should be two break points, but for illustration we try first with one break point. Assume the break point is at $\tau$ where $y(\tau) = 1$. So we have a system of equations using the data at the points $2,0,\tau$ and the cubic ansatz $$ y |_{[0,\tau]} = \sum \alpha_kt^k,\qquad y|_{[\tau,2]} = \sum \beta_k t^k$$ which leads to $$\alpha_0 = 0, \alpha_1 = 6, \alpha_3 \tau^3 + \alpha_2\tau^2 + 6 \tau = 1, 3\alpha_3\tau^2 + 2\alpha_2\tau + 6 = 0$$ and $$8\beta_3 + 4\beta_2 + 2\beta_1 + \beta_0 = 0, 12\beta_3 + 4\beta_2 + \beta_1 = 6, \beta_3\tau^3 + \beta_2\tau^2 + \beta_1\tau + \beta_0 = 1, 3\beta_3\tau^2 + 2\beta_2\tau + \beta_1 = 0$$ which gives $$\alpha_0 = 0, \alpha_1 = 6, \alpha_2 = -12 / \tau, \alpha_3 = 6 / \tau^2$$ $$\beta_3 = (6\tau - 2)/(\tau - 2)^3, \beta_2 = -3(5\tau^2 - 3\tau + 6) / 2(\tau - 2)^3 $$ Continuity of $y''$ then implies $$ 6\tau \alpha_3 + 2\alpha_2 = 6 \tau\beta_3 + 2\beta_2 $$ or $$ 3\tau^2 + 23 \tau^2 - 42 \tau - 32 = 0$$ which has two negative roots and one positive one at $\tau = 2$ which we can throw out. And thus 1 break point it not enough.

A direct computation (if I did it right, which is not guaranteed) with 2 break points $\sigma\in [1,2]$ and $\tau \in[0,1]$ yields that $\tau = 1/2$ and $\sigma = 3/2$ an admissible pair. It is just linear algebra in the end.

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This is very cool. I'm still go over your work, but the process makes sense to me. Thanks a lot :) – Jonathan Fischoff Jul 6 '10 at 8:59
Hum, the last paragraph is almost surely wrong. I shouldn't do computation that late at night. – Willie Wong Jul 6 '10 at 10:19
Solving the equation, you should have that $\sigma$ is the real root of $4\sigma^3 + \sigma^2 - 4\sigma - 10$, which Wolfram Alpha tells me is $\left( -1 + \sqrt[3]{2087 + 36\sqrt{3270}} + \sqrt[3]{2087 - 36\sqrt{3270}}\right) / 12$ which is roughly 1.51 – Willie Wong Jul 6 '10 at 10:55

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