# Find the extremal of $\int_0^d \ddot{x}^2-\alpha x \: dt$

We are asked to find the extremal of $$\displaystyle \int_0^d \ddot{x}^2-\alpha x \: dt$$ where $$x=x(t),\: \alpha$$ constant, $$d>0$$, $$x(0)=0, \: x(d)=0$$, $$\dot x(0) = 0, \: \dot x(d) = 0$$,

by considering: $$\displaystyle 0 = \frac{\partial f}{\partial x}- \frac{d}{dt}\frac{\partial f}{\partial\dot x} + \frac{d^2}{dt^2}\frac{\partial f}{\partial \ddot x}$$

What I have is: $$f(t,x,\dot x, \ddot x) = \ddot{x}^2-\alpha x$$

and solving: \begin{align*} 0 &= \frac{\partial f}{\partial x}- \frac{d}{dt}\frac{\partial f}{\partial \dot x} + \frac{d^2}{dt^2}\frac{\partial f}{\partial\ddot x}\\ &= -\alpha + 0 + \frac{d^2}{dt^2}2\ddot x\\ x^{(4)}&=\frac{\alpha}{2}\\ \Longrightarrow x &= \frac{\alpha}{48}t^4 + \beta t^3 + \gamma t^2 + \delta t + \varepsilon, \quad \beta, \gamma, \delta, \varepsilon \in \mathbb R \end{align*}

From $$x(0)=0 \Longrightarrow \varepsilon=0$$, $$\dot x(0)=0 \Longrightarrow \delta = 0$$.

$$x(d)=0 \Longrightarrow 0 = \frac{\alpha}{48} d^4 + \beta d^3 + \gamma d^2 \Longrightarrow 0 = \frac{\alpha}{48} d^2 + \beta d + \gamma \Longrightarrow \gamma = - \frac{\alpha}{48} d^2- \beta d$$

$$\dot x(d)=0 \Longrightarrow 0 = \frac{\alpha}{12} d^3 + 3\beta d^2 + 2\gamma d$$

$$\Longrightarrow 0 = \frac{\alpha}{12} d^2 + 3\beta d + 2\gamma = \frac{\alpha}{12} d^2 + 3\beta d + 2(- \frac{\alpha}{48} d^2- \beta d)= \frac{\alpha}{24} d^2 + \beta d \Longrightarrow 0 =\frac{\alpha}{24} d^2 + \beta d$$

$$\Longrightarrow \beta = - \frac{\alpha}{24} d$$

$$\Longrightarrow \gamma = - \frac{\alpha}{48} d^2- \beta d = - \frac{\alpha}{48} d^2 - (-\frac{\alpha}{24} d)d =\frac{\alpha}{48} d^2$$

So, $$\displaystyle x(t) = \frac{\alpha}{48}t^4 -\frac{\alpha}{24} d t^3 + \frac{\alpha}{48} d^2t^2$$ is the extremal.

I'm just checking if this is correct?

• try taking the derivatives and plugging into integral to verify Sep 4, 2019 at 19:08
• $\dddot{x}-\alpha x$=? Sep 4, 2019 at 19:09
• I think he means for you to plug the solution into the EL equations to check that it works and satisfies the boundary conditions. Sep 4, 2019 at 19:23

## 1 Answer

For a critical value of $$\int_0^d\left(\ddot x^2-\alpha x\right)\mathrm{d}t\tag1$$ we need \begin{align} 0 &=\delta\int_0^d\left(\ddot x^2-\alpha x\right)\mathrm{d}t\\ &=\int_0^d\left(2\ddot x\,\delta\ddot x-\alpha\,\delta x\right)\mathrm{d}t\\ &=\int_0^d\left(-2\dddot x\,\delta\dot x-\alpha\,\delta x\right)\mathrm{d}t\\ &=\int_0^d\left(2\,\ddddot x\,\delta x-\alpha\,\delta x\right)\mathrm{d}t\tag2 \end{align} Thus, the critical point is when $$\ddddot x=\frac\alpha2$$. That would mean $$x=\frac\alpha{48}t^4+a_3t^3+a_2t^2+a_1t+a_0\tag3$$ Computing $$a_k$$ so that $$x(0)=x(d)=\dot x(0)=\dot x(d)=0$$ and gives $$x=\frac\alpha{48}t^4-\frac{\alpha d}{24}t^3+\frac{\alpha d^2}{48}t^2\tag4$$ So, your answer looks correct.