Smooth ride with the subway (Optimization with Lagrange multiplier)

Question

I have following problem and im struggeling with it.

I need to find be a twice differentiable function $ h:[0,1] \rightarrow \mathbb{R}$ which fullfills the conditions:

• $h(0)=0$ , $h(1)=1$ (the position of the subway train)
• $h'(0)=h'(1)=0$ (the velocity)
• $\vert h''(t)\vert \leq c\in \mathbb{R}^+$

Such that the cost function $U(h):=\int_{0}^{1}(h''(t))^2 dt$ is minimal.(Also: for which c exists a solution at all?) I tried with the Lagrange-Ansatz, but i fail already there, as i am not sure how to express the second condition in the lagrange equation.(since the function is "only" twice differentiable so the i cannot express the first derivative as an integral with the second derivative, can I?), what i got so far:

$\Lambda(h,\lambda_1,\lambda_2)=U(h)+\lambda_1(\int_{0}^1h'(t)dt-1)+ \lambda_2(...)$.

Im glad for every shred of help. Thanks in advance.

Answer 1

The post from Christian Blatter and the methods for finding the catenoid-equation gave some new ideas to tackle the problem:

First:

$$h(1)-h(0)=\int\limits_0^1h'(t)dt=\int\limits_0^1\int\limits_0^t h'(s) ds\ dt = \int\limits_0^1\int\limits_s^1h'(s)dt\ ds=\int\limits_0^1h'(s)(1-s)ds=1$$ Second (Ignoring that $h'(t)$ might not be continously differentiable): $$h'(1)-h'(0)=\int\limits_0^1 h''(t)dt=\int\limits_0^1\int\limits_0^th''(s)ds \ dt=\int\limits_0^1\int\limits_s^1h''(s)dt\ ds=\int\limits_0^1h''(s)(1-s)ds=0$$

Henze our Langrange equation looks like this: $$\Lambda(h'',h',\lambda_1,\lambda_2)=\int\limits_0^1(h''(s))^2ds+\lambda_1(\int\limits_0^1h'(s)(1-s)ds-1)+\lambda_2(\int\limits_0^1h''(s)(1-s)ds)$$ and therefore:

$$\frac{\partial}{\partial s}\frac{\partial \Lambda}{\partial h''}-\frac{\partial \Lambda}{\partial h'}=0 \Leftrightarrow 2 \frac{d^3 h}{ds^3}-\lambda_1-\lambda_2(1-s)=0 \Leftrightarrow \frac{d^3 h}{ds^3}=\frac{1}{2}(\lambda_1+\lambda_2(1-s))$$

Integration gives us:

$$h(x)=\frac{-\lambda_2}{48}x^4+\frac{(\lambda_1-\lambda_2)}{12}x^3+\frac{c_1}{2}x^2+c_2x+c_3; \ c_i\in \mathbb{R}$$

With the initial conditions one gets:

• $c_2=c_3=0$
• $\lambda_1=\frac{-1}{2}(48+\lambda_2)$
• $c_1=\frac{-1}{12}(3\lambda_1+2\lambda_2)=6-\frac{\lambda_2}{24}$

This delivers:

$$h_{\lambda_2}(x)=\frac{-\lambda_2}{48}x^4+\frac{1}{12}x^3(\lambda_2-\frac{1}{2}(48+\lambda_2))+\frac{1}{24}x^2(\frac{3}{2}(48+\lambda_2)-2\lambda_2)$$

So i got an equation which still contains a unknown variable $\lambda_2$.

Calculating the integral $\int_{0}^{1} (h_{\lambda_2}'')^2 dx=12+\frac{\lambda_2^2}{2880}$. Now we can differentiate for $\lambda_2$ and set it equal to $0$ which gives us:

$\frac{\lambda_2}{1440}=0 \Leftrightarrow \lambda_2=0$

Hence: $h(t)=3 t^2 - 2 t^3$ (and therefore a Solution exists only if $c\geq 6$ otherwise the train will overshoot or won't arrive at all)

I welcome any feedback and i thank uniquesolution and Christian Blatter for their answers.

Answer 2

Put $f(t)=h'(t)$. Then you want to minimize the functional $$\int_0^1 L(t,f(t),f'(t))\,dt\enspace\enspace (1)$$ where $L(t,f,f')=(f')^2$. The corresponding Euler-Lagrange equation is: $$\frac{\partial L}{\partial f}-\frac{d}{dt}\frac{\partial L}{\partial f'}=0$$ that is: $$\frac{d}{dt}(2f')=0$$ which implies $f''=0$. It follows that $f$ is a linear function, say, $f(t)=at+b$. Since $f=h'$, we deduce that $$h(t)=\frac{1}{2}at^2+bt+d$$ Correction Unfortunately, there does not exist a polynomial of order $2$ satisfying the initial conditions. Since the Euler-Lagrance equation must be satisfied by a stationary function for the functional $(1)$, and since a twice-differentiable function has second derivative identically zero if and only if it is linear, the argument above effectively proves that the general solution must be a polynomial of degree $2$. It follows that your problem has no solution with the initial conditions given.

Answer 3

If you allow discontinuities of $h''$ you can argue as follows: By symmetry it is sufficient to solve the problem on $\bigl[0,{1\over2}\bigr]$ with conditions $$h(0)=h'(0)=0,\quad h\left({1\over2}\right)={1\over2},\quad 0\leq a(t):=h''(t)\leq c\ .$$ One has $$h'(t)=\int_0^t a(\tau)\>d\tau,\quad h(t)=\int_0^ta(\tau)(t-\tau)\>d\tau\ ,$$ so that we have to minimize $\int_0^{1/2}a^2(t)\>dt$ under the conditions $$0\leq a(t)\leq c,\quad \int_0^{1/2}a(t)\left({1\over2}-t\right)\>dt={1\over2}\ .$$ This easily implies $c\geq4$ as a necessary condition, whereby in the case $c=4$ the train has to be maximally accelerated during the whole interval $\bigl[0,{1\over2}\bigr]$.

Forgetting about the condition $a(t)\leq c$ for the moment we have to minimize the $L^2$-norm of $a$ under the condition $\langle a,u\rangle={1\over2}$ for a given vector $u$. The minimum is attained for $a_*=\lambda u$ for a certain scalar $\lambda$. Calculation gives $$a_*(t)=12\left({1\over2}-t\right)\ ,\tag{1}$$ so that $a_*(0)=6$. We therefore can say the following: If $c\geq6$ then $(1)$ is the solution to the problem for the interval $\bigl[0,{1\over2}\bigr]$, and a suitable reflection will take care of the interval $\bigl[{1\over2},1\bigr]$. Since $a_*\bigl({1\over2}\bigr)=0$ the extended $a_*$ will even be continuous on $[0,1]$.

If $4\leq c<6$ the condition $a(t)\leq c$ really enters the picture, and I don't know how to deal with that.