Let $u$ be a solution of the heat equation
$$u\_t - \Delta u = 0, \qquad t >0, \ x \in \mathbb T^d$$
and $v$ be a solution of the bi-harmonic heat equation
$$v\_t +\Delta^2 v = 0, \qquad t >0, \ x \in \mathbb T^d$$
with the same initial data $f$.
Is it true that, for every fixed time $T >0$,
$$\|v(T,\cdot)\|\_{L^2} \le \|u(T,\cdot)\|\_{L^2}$$
holds? That is, in some sense, $v$ is "more dissipated"?



$L^2$ norm for solutions of evolution equations driven by different elliptic operators
Not necessarily. I mean, it depends upon the torus you consider. Notice that in the case of the standard one ${\mathbb T}^d={\mathbb R}^d/{\mathbb Z}^d$, the answer is positive. But if you torus is ${\mathbb R}^d/a{\mathbb Z}^d$, then it is positive if $a\le2\pi$ and negative otherwise. The reason is that both semi-groups $H\_t$ and $B\_t$ are co-diagonal with an orthogonal eigenbasis, but they have eigenvalues $e^{-t\mu}$ and $e^{-t\mu^2}$, where $\mu\_0=0<\mu\_1,\ldots$ are the eigenvalues of $-\Delta$. To have the required inequality is equivalent to having $\mu^2\ge\mu$ for every eigenvalue, that is $\mu\_1\ge1$. Whence the condition on the size of the torus.


More generally, if ${\mathbb T}^d={\mathbb R}^d/\Lambda$ where $\Lambda$ is a lattice, the inequality amounts to the fact that every non-zero element of the dual lattice $\Lambda^\*$ has norm $\ge\frac1{2\pi}$. Here $\Lambda^\*$ is the set of points $\alpha\in{\mathbb R}^d$ such that $\langle \alpha, p\rangle\in{\mathbb Z}$ for every $p\in\Lambda$.


**Edit**. Here are the details. Let me consider a general torus $T={\mathbb R}^d/\Lambda$ where $\Lambda$ is a lattice. Both the Laplacian and its square are diagonal in the Fourier basis of exponentials
$$\phi\_\alpha(x):=c\_\alpha\exp(2i\pi\alpha\cdot x),\qquad \alpha\in\Lambda^\*,$$
where the constant $c\_\alpha$ normalizes: $\|\phi\_\alpha\|\_{L^2}=1$.


The corresponding eigenvalues are $4\pi^2|\alpha|^2$, respectively $(4\pi^2|\alpha|^2)^2$. Decomposing an arbitrary data
$$u(0,\cdot)=\sum\_{\alpha\in\Lambda^\*}a\_\alpha\phi\_\alpha,$$
we have
$$u(T,\cdot)=\sum\_{\alpha\in\Lambda^\*}e^{4\pi^2|\alpha|^2T}a\_\alpha\phi\_\alpha,\qquad v(T,\cdot)=\sum\_{\alpha\in\Lambda^\*}e^{(4\pi^2|\alpha|^2)^2T}a\_\alpha\phi\_\alpha.$$
Using the fact that the Fourier basis is $L^2$-orthogonal, we find
$$\|u(T)\|\_{L^2}^2=\sum\_{\alpha\in\Lambda^\*}e^{8\pi^2|\alpha|^2T}|a\_\alpha|^2,\qquad \|v(T)\|\_{L^2}^2=\sum\_{\alpha\in\Lambda^\*}e^{32(\pi^2|\alpha|^2)^2T}|a\_\alpha|^2.$$


The required inequality is equivalent to
$$e^{32(\pi^2|\alpha|^2)^2T}\le e^{8\pi^2|\alpha|^2T}$$
for $T>0$ and every $\alpha$, that is to $2\pi|\alpha|\ge1$ for every non-zero element of $\Lambda^\*$.


