1
$\begingroup$

Good morning,

I am studying for an exam, and in my notes I found a curious... note.

It is a differential equation:

$$\frac{\text{d}^2h}{\text{d}t^2} = -a \frac{\partial^2h}{\partial x^2}$$

Does anyone of you know this equation and how could it be derived?

There are no hints or other explanations in my notes, as if the professor just told it to us just because. I wrote it down but nothing else is explained.

Since we are studying the HEAT equation, why is there a second derivative in time??

Thank you so much!

$\endgroup$
8
  • 1
    $\begingroup$ Why in the LHS $h$ has single variable (t) and on RHS is not? How exactly $h$ is represented? $\endgroup$
    – Jingeon An-Lacroix
    May 18, 2020 at 11:41
  • 1
    $\begingroup$ What course is this for? $\endgroup$
    – user619894
    May 18, 2020 at 11:56
  • $\begingroup$ This seems to pretty obviously be a physics exam you're studying for, in which case you'd derive the equation by first describing the physical system it represents. Now, the $kT$ in the denominator of that $-B^5$ expression suggests some kind of thermodynamic (or maybe statistical mechanical) system. But beyond that, I've never seen anything even remotely resembling the equations you've written. So why don't you first tell us all about the physical system under consideration (or if you don't know, first go back and ask your teacher about it, and then tell us). And maybe post on physics.se $\endgroup$
    – John Forkosh
    May 18, 2020 at 11:58
  • 1
    $\begingroup$ You get the 4th derivative when you consider the curvature as "cost", and $h$ is small enough so that the curvature is represented by the second derivative. So the dynamic of this equation will reduce spikes in curvature towards an equal curvature over the full curve, as can be observed in surfaces of viscous media in zero or negligible gravity. This is similar to the heat equation that equalizes the gradient of the temperature curve. $\endgroup$
    – Lutz Lehmann
    May 18, 2020 at 12:30
  • 1
    $\begingroup$ That now is a wave equation, as long as $a>0$. I do not know what you are discussing, it might be some overview over elliptic, parabolic and hyperbolic second order PDE. // It would have been better if you had just added the new topic instead of replacing the old one. $\endgroup$
    – Lutz Lehmann
    May 18, 2020 at 14:34

1 Answer 1

Reset to default
2
$\begingroup$

$\newcommand{\pd}[2]{\frac{\partial#1}{\partial#2}}$ This is about the original version of the question about $\pd{h}{t}=-B^5\pd{^5h}{x^5}$. The heat equation $\pd{h}{t}=c\pd{^2h}{x^2}$ can be treated similarly starting from the functional $I[h]=\int_a^b(\pd{h}{x})^2dx$.

The curvature of some curve $y(x)$ is the change in slope relative to the curve length, thus $\frac{y''(x)}{\sqrt{1+y'(x)^2}}$. If $y'$ is small (which in turn implies that $y(x)$ varies little), then the quotient can be simplified to its numerator. Then the given equation can be obtained as the continuous gradient descent for the mean square curvature.

Consider the functional of the square "sum" of the second derivative $$ I[h]=\int_a^b\left(\pd{^2 h}{x^2}(x,t)\right)^2\,dx $$ Its variation is \begin{align} δI[h]&=2\int_a^b\pd{^2 h}{x^2}(x,t)\pd{^2 δh}{x^2}(x,t)\,dx \\ &=\left[\pd{^2 h}{x^2}(x,t)\pd{δh}{x}(x,t)\right]_{x=a}^{x=b}-\int_a^b\pd{^3 h}{x^3}(x,t)\pd{δh}{x}(x,t)\,dx \\ &=\left[\pd{^2 h}{x^2}(x,t)\pd{δh}{x}(x,t)-\pd{^3 h}{x^3}(x,t)δh(x,t)\right]_{x=a}^{x=b}+\int_a^b\pd{^4 h}{x^4}(x,t)δh(x,t)\,dx \end{align}

With suitable boundary conditions the first term can be set to zero, the remaining term gives the functional gradient as $$ \frac{δI[h]}{δh(x,t)}=\pd{^4 h}{x^4}(x,t) $$ Continuous gradient descent is now provided by the equation $$ \pd{h}{t}(x,t)=-\alpha \frac{δI[h]}{δh(x,t)}=-\alpha\pd{^4 h}{x^4}(x,t) $$ That $\alpha=B^5$ has to be derived from some scaling argument and physics, there is nothing obvious in that choice.

$\endgroup$
1
  • $\begingroup$ Sorry, I miswrote... No need of that answer though (by the way, I noticed a downvote but I ignore who has been). Could you provide another answer for me? I would accept ti! $\endgroup$
    – Numb3rs
    May 18, 2020 at 14:28

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .