How to obtain this differential equation? 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!
 A: $\newcommand{\pd}[2]{\frac{\partial#1}{\partial#2}}$
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.
