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I am trying to minimize the following integral

$$A(x)=\frac{1}{T}\int_0^T\left(\dfrac{\partial h}{\partial t}+U\dfrac{\partial h}{\partial x}\right)^2dt.$$

The parameter $U$ is constant and $h=h(x,t)$. So I thought that I could introduce $L=\left(\dfrac{\partial h}{\partial t}\right)^2+2U\dfrac{\partial h}{\partial t}\dfrac{\partial h}{\partial x}+U^2\left(\dfrac{\partial h}{\partial x}\right)^2$ as a Lagrangian.

I looked up the Euler-Lagrange equations for this type of problem: $$\dfrac{\partial L}{\partial h}-\dfrac{\partial}{\partial x}\left(\dfrac{\partial L}{\partial h'} \right)-\dfrac{\partial}{\partial t}\left(\dfrac{\partial L}{\partial \dot{h}} \right).$$

I use $h'=\dfrac{\partial h}{\partial x}$ and $\dot{h}=\dfrac{\partial h}{\partial t}$ to simplify notation.

Using this equation and applying it to my Lagrangian $L$ results in the follwing equation:

$$\dfrac{\partial^2 h}{\partial t^2}+2U\dfrac{\partial^2 h}{\partial x\partial t}+U^2\dfrac{\partial^2 h}{\partial x^2}=0.$$

Solving this PDE will give $h(x,t)=g(Ut-x)+xf(Ut-x)$, in which $g$ and $f$ are arbitrary functions.

Is my procedure correct (solution of PDE is not important, but rather the way I set up the PDE)?

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  • $\begingroup$ Just to make sure, if $x = x(t)$? $\endgroup$ – user99914 Apr 1 '17 at 10:50
  • $\begingroup$ Hmm ... that is really a good question. I actually think that $x$ is not a function of time. So if it was, I would have to be careful with the derivative with respect to time? Are you pointing this out? $\endgroup$ – MrYouMath Apr 1 '17 at 11:14
  • $\begingroup$ @JohnMa: What happens if $x=x(t)$? $\endgroup$ – MrYouMath Apr 1 '17 at 14:43
  • $\begingroup$ Why is better expression for the general solution? $\endgroup$ – Rafa Budría Apr 3 '17 at 20:00
  • $\begingroup$ @Qmechanic: $$g(Ut-x)+(Ut-x+2x)f(Ut-x)=g(Ut-x)+(Ut-x)f(Ut-x)+x(2f(Ut-x))=h(Ut-x)+xz(Ut-x)$$ The solutions are the same. What are you pointing out? But your solution for the easy case somehow looks different. It would be great if you could explain in further detail. $\endgroup$ – MrYouMath Apr 3 '17 at 21:14
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  1. It seems relevant to point out that there is a shortcut, see Section 3 below. No need to deal with the second-order PDE from the Euler-Lagrange (EL) equation.

  2. In fact more troublesome: the EL equation relies on boundary conditions (BC), which seem absent here. Without BC, the procedure/approach by OP is flawed/unjustified, cf. OP's main question in yellow (v1).

  3. Instead it is evident from the $A$-functional, that a minimizing solution $h$ should satisfy the first-order PDE $$\frac{\partial h}{\partial t}+U\frac{\partial h}{\partial x}~=~0 ,$$ with complete solution of the form $$ h(x,t)~=~g(Ut-x). $$

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  • $\begingroup$ The solutions aren't the same. $\endgroup$ – Rafa Budría Apr 3 '17 at 16:36
  • $\begingroup$ @Rafa Budría: OP's solution method (v1) has not been properly justified, as mentioned in my answer. $\endgroup$ – Qmechanic Apr 3 '17 at 19:41

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