Heat Equation derivative in terms of Laplace If the heat equation is 
$ \frac{\partial u}{\partial t} - \alpha \nabla^2 u=0$
Is the second derivative of u w.r.t t is the laplacian of the lapacian?
 A: Yes. Writing $\Delta$ instead of $\nabla^2$ one has
$$u_{tt}={\partial\over\partial t}(u_t)={\partial\over\partial t}\bigl(\alpha\>\Delta u\bigr)=\alpha\>\Delta(u_t)=\alpha\>\Delta\bigl(\alpha\>\Delta u\bigr)=\alpha^2\>\Delta^2 u\ \ .$$
A: Working intuitively, it seems to me that if
$\dfrac{\partial u}{\partial t} - \alpha \nabla^2 u=0, \tag{1}$
or
$\dfrac{\partial u}{\partial t} = \alpha \nabla^2 u, \tag{1}$
and we are working in an open set $U \in \Bbb R \times \Bbb R^n$ on which $u(t, \vec x)$, $\vec x = (x_1, x_2, \ldots, x_n)$, and enough of its derivatives are "nice", that is, defined and continuous, then writing
$\nabla^2 = \dfrac{\partial^2}{\partial x_1^2} + \dfrac{\partial^2}{\partial x_2^2} + \ldots + \dfrac{\partial^2}{\partial x_n^2} = \sum_1^n \dfrac{\partial^2}{\partial x_i^2} \tag{3}$
and using the fact that for such "nice" $u(t, \vec x)$ the partial derivatives with respect to $t$ and the $x_i$ all "commute" with each other, then we may write
$\dfrac{\partial^2 u}{\partial t^2} = \dfrac{\partial}{\partial t}\dfrac{\partial u}{\partial t} = \dfrac{\partial}{\partial t}(\alpha \nabla^2 u) =\alpha \nabla^2 \dfrac{\partial u}{\partial t} = \alpha^2 \nabla^2 \nabla^2 u; \tag{4}$
so yeah, $\partial^2 u / \partial t^2$ given by the laplacian of the laplacian of $u(t, \vec x)$, up to the factor of $\alpha^2$.  And if $u(t, \vec x)$ is nice enough, we can just keep going:
$\dfrac{\partial^3 u}{\partial t^3} = \alpha^3 \nabla^2 \nabla^2 \nabla^2 u = \alpha^3 (\nabla^2)^3 u,  \tag{5}$
and so forth:
$\dfrac{\partial^n u}{\partial t^n} = \alpha^n (\nabla^2)^n u.  \tag{6}$
In fact, expressions such as (6) more lead to more advanced formulas such as
$u(t, \vec x) = e^{(\alpha (t - t_0)\nabla^2)}u(t_0, \vec x), \tag{7}$
for if we expand $u(t, \vec x)$ in a power series in $t$ around $t_0$, we see that, writing informally,
$u(t, \vec x) = \sum_0^\infty \dfrac{(t - t_0)^n}{n!} \dfrac{\partial^n u(t_0, \vec x)}{\partial t^n}$
$= \sum_0^\infty \dfrac{\alpha^n (t - t_0)^n}{n!} (\nabla^2)^n u(t_0, \vec x) = e^{(\alpha(t - t_0)\nabla^2)} u(t_0, \vec x). \tag{8}$
Of course, the above is a heuristic derivation and more careful, refined technique and arguments are necessary to put it on a rigorous footing.
In any event, the answer to the OP's stated question is, basically,
YES!
Hope this helps.  Cheers,
and as always,
Fiat Lux!!!
