4
$\begingroup$

Let $u$ be a solution of the heat equation $$u_t - u_{xx} = 0, \quad t>0, x \in \mathbb{R}$$ with initial data $u(0,\cdot) = u_0$. Fix $\alpha >0$. How can I estimate (without using explicitly the heat kernel) $$\sup_{t>0}\int_{\mathbb{R}} t^\alpha |u_x|^2 \ dx,$$ in terms of the initial data? Could you point out a reference where such an estimate is obtained?

Is it fair to call what we obtain a decay estimate?

$\endgroup$
  • $\begingroup$ Using the Fourier transform with respect to the spatial variable it is not difficult to show the (possibly non-optimal) estimate $\sup_{t > 0} \int_{\mathbb{R}} t^\alpha \lvert u_x\rvert^2 dx \le (\alpha/e)^\alpha \int_{\mathbb{R}} \lvert x\rvert^{2-2\alpha} \lvert \hat u_0(x)\rvert^2 dx$; for $\alpha \in (0,1)$ the integral on the right is closely related to a Sobolev norm of $u_0$. If this is of any help for you I can post the details. $\endgroup$ – Jochen Glueck Jul 29 '18 at 18:35
  • $\begingroup$ @JochenGlueck Yes, please, I'd like to know the details. Especially about the meaning of that integral. $\endgroup$ – Riku Jul 29 '18 at 19:10
  • $\begingroup$ I posted the details in an answer (and it seems that there was a small computational error in the constant in front of the integral in my above comment). $\endgroup$ – Jochen Glueck Jul 30 '18 at 19:38
3
$\begingroup$

I'm sorry for the late answer, but joined MathOverflow just this week. The Fourier Splitting method, developed by María Elena Schonbek in the 80's asserts that "decay is determined by the low frequencies of the solutions" for many dissipative linear and nonlinear equations (heat, fractional heat, Navier-Stokes, dissipative quase-geostrophic, amongst many others). For $v_0 \in L^2 (\mathbb{R} ^n)$, assume there is a unique $-\frac{n}{2} < r^{\ast} (v_0) < \infty$, called the decay character, such that

\begin{equation} r^{\ast} (v_0) = \lim _{\rho \to 0} \rho ^{-2r-n} \int _{B(\rho)} \large|\widehat{v}_0 (\xi) \large|^2 \, d \xi \end{equation}

i.e., the decay character measures the "order" of $v_0$ at the origin in frequency space (see Bjorland and Schonbek, Adv. Diff. Eq. 2009; Niche and M.E. Schonbek, J. London Math. Soc. 2015). Then, for a large family of dissipative operators, the decay character provides sharp estimates for decay of linear (and some nonlinear) equations. In the specific case of the heat equation in $\mathbb{R}^n$, we have that for the solution $v$ with initial data $v_0$

\begin{equation} C_1 (1 + t)^{- \left( \frac{n}{2} + r^{\ast} \right)} \leq \Vert v(t) \Vert _{L^2} ^2 \leq C_2 (1 + t)^{- \left(\frac{n}{2} + r^{\ast} \right)}, \end{equation} for some absolute constants $C_1, C_2 > 0$, see Theorem 6.5 in Bjorland and M.E. Schonbek (op.cit.) and Theorem 2.10 in Niche and M.E. Schonbek (op. cit.).

Now take $u_0 \in \dot{H} ^1 (\mathbb{R})$, i.e. $\partial_x u_0 \in L^2 (\mathbb{R})$. For $r^{\ast} = r^{\ast} (\partial_x u_0)$ the estimate above implies

\begin{equation} C_1 (1 + t)^{- \left( \frac{1}{2} + r^{\ast} \right)} \leq \Vert u(t) \Vert _{\dot{H}^1} ^2 \leq C_2 (1 + t)^{- \left( \frac{1}{2} + r^{\ast} \right)}. \end{equation} If $u_0 \in H^s (\mathbb{R}), s > 0$, then $u_0 \in L^2 (R)$ as well and from Theorem 2.11 in Niche and M.E. Schonbek we have $r^{\ast} (\partial_x u_0) = s + r^{\ast} (u_0)$ so

\begin{equation} C_1 (1 + t)^{- \left( \frac{1}{2} + s + r^{\ast} (u_0) \right)} \leq \Vert u(t) \Vert _{\dot{H}^1} ^2 \leq C_2 (1 + t)^{- \left( \frac{1}{2} + s + r^{\ast} (u_0) \right)}. \end{equation}

Note that the decay character does not always exists, there are $v_0 \in L^2 (\mathbb{R}^n)$ which oscillate a lot near the origin (in frequency space) for which $r^{\ast} (v_0)$ is not defined, see the article by Brandolese in SIAM J. Math. Anal. 2016 to find a precise characterization in terms of Besov spaces for when the decay character exists and for other results.

$\endgroup$
3
$\begingroup$

For every $t > 0$ and every $\alpha \in (0,1)$ we have \begin{align*} t^\alpha \lVert u_x \rVert_{L^2(\mathbb{R})}^2 \le \big( \frac{\alpha}{2e} \big)^\alpha \lVert u_0 \rVert_{H^{1-\alpha}}^2, \end{align*} where $\lVert \cdot\rVert_{H^{1-\alpha}(\mathbb{R})}$ denotes a fractional Sobolev norm over $\mathbb{R}$ (for a few details about fractional Sobolev spaces see the end of this post).

Proof. For each $v \in L^2(\mathbb{R})$ we denote by $\mathcal{F}v = \hat v$ the Fourier transform of $v$; by $\xi$ we denote the variable of Fourier transforms of functions. By taking the Fourier transform of the heat equation we obtain \begin{align*} \frac{d}{dt} \hat u(t) & = -\xi^2\hat u(t), \\ \hat u(0) & = \hat u_0. \end{align*} This problem has the unique solution (more precisely: the unique mild solution, which is the usual definition of a solution within the theory of operator semigroups) \begin{align*} \hat u(t) = e^{-t\xi^2} \hat u_0. \end{align*} Hence, by Plancherel's theorem, \begin{align*} t^\alpha \lVert u_x(t) \rVert_{L^2(\mathbb{R})}^2 = t^\alpha \lVert \widehat{u_x}(t) \rVert_{L^2(\mathbb{R})}^2 = t^\alpha \lVert \xi \hat u(t) \rVert_{L^2(\mathbb{R})}^2 = \int_{\mathbb{R}} \xi^2 \; t^\alpha e^{-2t\xi^2} \, \lvert\hat u_0(\xi)\rvert^2 \, d\xi, \end{align*} If we fix $\xi \in \mathbb{R} \setminus \{0\}$ for a moment, a bit of elementary calculus shows that the function $(0,\infty) \ni t \mapsto t^\alpha e^{-2t\xi^2}$ has the maximal value $\big(\frac{\alpha}{2e}\big)^\alpha\xi^{-2\alpha}$. Hence, we obtain from the above estimate \begin{align*} & t^\alpha \lVert u_x(t) \rVert_{L^2(\mathbb{R})}^2 \le \int_{\mathbb{R}} \xi^2 \; \big(\frac{\alpha}{2e}\big)^\alpha\xi^{-2\alpha} \lvert\hat u_0(\xi)\rvert^2 \, d\xi \\ & \le \big(\frac{\alpha}{2e}\big)^\alpha \int_{\mathbb{R}} (1 + \xi^2)^{1-\alpha} \lvert \hat u_0(\xi)\rvert^2 \, d\xi = \big(\frac{\alpha}{2e}\big)^\alpha \lVert u_0 \rVert_{H^{1-\alpha}(\mathbb{R})}^2. \end{align*} This proves the claim.

Short supplement on fractional Sobolev spaces. Fix $s \in (0,\infty)$. For each $f \in L^2(\mathbb{R})$ we set \begin{align*} \lVert f \rVert_{H^s(\mathbb{R})}^2 := \int_{\mathbb{R}} (1 + \xi^2)^s \lvert \hat f(\xi)\rvert^2 \, d\xi \in [0,\infty]. \end{align*} By definition, the space $H^s(\mathbb{R})$ consists of all functions $f \in L^2(\mathbb{R})$ for which this number is finite. The mapping $\lVert\cdot\rVert_{H^s(\mathbb{R})}$ is a norm on $H^s(\mathbb{R})$ which renders this space a Banach space (in fact, even a Hilbert space with respect to an appropriate inner product).

Note that fractional Sobolev spaces (and the norms on them) can also be defined without using the Fourier transform; this approach is for instance explained in these lecture notes, and for the relation to the above definition via the Fourier transform we refer to Section 4 (and in particular to Proposition 4.2) in these notes. Note however that there are various ways in the literature to define (equivalent but not necessarily equal) norms on Sobolev spaces.

$\endgroup$
2
+25
$\begingroup$

Write the heat equation in the abstract form $u'+Au=0$, where $A$ is a maximal dissipative operator over the Hilbert space $H=L^2({\mathbb R})$, with domain the Sobolev space $D(A)=H^2({\mathbb R})$. Because $A$ is self-adjoint, it is a classical result that if $u_0\in H$, then $u(t)\in D(A)$ for every $t>0$, with an inequality $$\|Au(t)\|\le\frac{\|u_0\|}t\,.$$ Here this gives $$t^2\int_{\mathbb R}u_{xx}^2dx\le\int_{\mathbb R}u_0^2dx.$$ Now, because of $$\int_{\mathbb R}u_{x}^2dx\le\|u\|\cdot\|u_{xx}\|$$ and the decay of $\|u(t)\|$, we obtain $$t\int_{\mathbb R}u_{x}^2dx\le\int_{\mathbb R}u_0^2dx.$$

Reference : H. Brézis, Functional analysis, Sobolev spaces and partial differential equations.

Edit. Another technique is the following. Denote $\|\cdot\|_p$ the $L^p$-norm. We start from the energy estimate, in the form $$\frac{d}{dt}\|u\|_2^2+2\|u_x\|_2^2=0.$$ Apply Gagliardo-Nirenberg inequality $\|v\|_2^3\le C\|v\|_1^2\|v_x\|_2$.We have $\|u(t)\|_2^3\le C\|u_0\|_1^2\|u_x\|_2$. Therefore $$C^2\|u_0\|_1^4\|u\|_2^{-6}\frac{d}{dt}\|u\|_2^2+2\le0.$$ By a Gronwall argument, we infer a dispersion inequality $$\|u(t)\|_2\le c\,\frac{\|u_0\|_1}{t^{1/4}}\,.$$ Combining with our first estimate, we have $$t^{3/2}\int_{\mathbb R}u_{x}^2dx\le\|u_0\|_1^2dx.$$ Using the Riesz-Thorin interpolation theorem, we obtain $$t^{\frac12+\frac1p}\int_{\mathbb R}u_{x}^2dx\le\|u_0\|_p^2dx$$ for every $p\in[1,2]$.

There is no such inequality with $p\in(2,\infty)$. Because a translation invariant operator, such as $u_0\longmapsto \partial_xu(t)$, cannot be a bounded operator from $L^p$ to $L^q$ when $q<p<\infty$ (Hörmander).

$\endgroup$
  • $\begingroup$ Could you ellaborate a bit on your last sentence? I have difficulties to see why $t^\alpha \int_{\mathbb{R}} u_x^2 \, dx$ should be bounded by an $L^p$-norm of $u_0$ rather than by a Sobolev norm. $\endgroup$ – Jochen Glueck Jul 30 '18 at 18:36
  • $\begingroup$ @DennisSerre: Thanks for your edit, now I see your point. $\endgroup$ – Jochen Glueck Jul 30 '18 at 21:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.