# The inverse of Laplacian operator for different orders

I post this question in MSE couple of days before and get no response. So I repost it here for better luck. Thank you!

Let $u,v\in C_c^\infty(\Omega)$ and $\Omega\subset \mathbb R^N$ is open bounded with smooth boundary. Let $\Delta$ denote the Laplacian operator, $I$ denotes the identity operator and $t\in\mathbb R^+$ is a positive real number.

Let $$f(t):=\|(I-t\Delta)^{-1}\nabla u\|_{L^2}^2-\|(I-t\Delta)^{-\frac32}\nabla v\|_{L^2}^2.$$ It is given that $\|\nabla u\|_{L^2}^2<\|\nabla v\|_{L^2}^2$, that is, $f(0)<0$; and I know that there exists $t_0>0$ such that $f(t_0)=0$. I also know that $$\int u = \int v.$$

My question: Can I prove that $f(t)<0$ for $0<t<t_0$ and $f(t)>0$ for $t>t_0$?

My try: I compute that $$\frac{d}{dt}(\|(I-t\Delta)^{-1}\nabla u\|_{L^2}^2)=-\|(I-t\Delta)^{-\frac32}\Delta u\|_{L^2}^2<0$$ so I know $\|(I-t\Delta)^{-1} \nabla u\|_{L^2}^2$ is decreasing as $t$ increasing, so is $\|(I-t\Delta)^{-\frac32} \nabla v\|_{L^2}^2$. But I can't prove that the later one decreasing faster... I guess the order $3/2$ would do sth but I am not sure...

Also, I am wondering that how may I write

$$\|(I-t\Delta)^{-s}u\|_{L^2}^2=\sum_{k=0}??$$ where $t$ and $s$ are real numbers, based on Fourier transform.

I was trying to look for Bessel potential but had no luck...

• I guess going to the frequency space wouldn't hurt you. – Fan Zheng Feb 1 '16 at 16:10
• @FanZheng You mean by using Fourier basis? Yea, I actually have trouble about it... – JumpJump Feb 1 '16 at 16:15
• For your final box: you can expand $u$ in terms of sum of eigenfunctions of Laplacian (with fixed, say Dirichlet, boundary conditions). The eigenfunctions are orthogonal to each other in $L^2$. The $\int u = \int v$ condition can be trivially satisfied by assuming $\int u = \int v = 0$ which is the same as saying that they have no $0$-frequency components. – Willie Wong Feb 1 '16 at 17:40
• Just write $u,v$ as linear combinations of the eigenfunctions of $-\Delta$ and see what happens. – Christian Remling Feb 1 '16 at 17:48
• (@ChristianRemling: sorry to have wasted your time; I hid my answers while I was editing since the OP changed the question after I posted a computation along the lines of what you suggest.) – Willie Wong Feb 1 '16 at 17:51

Disclaimer

The below argument applies directly to the previous version of the question. For the current version the basic argument still hold, using that $$\| \nabla u\|_2 = \langle u, -\Delta u\rangle$$ and so behaves nicely for eigenfunctions of the Laplacian.

To make the argument also work for $C^\infty_c$ functions simply apply a cut-off near the boundary. This is a small perturbation and so you just need to add some small $\epsilon$ terms to the whole thing.

The added requirement that $\int u = \int v$ is of no consequence since the constructed example below has no zero-frequency component, and that the insertion of $\nabla$ in the new version of the question kills this component anyway.

End disclaimer

With the given information it seems unlikely.

Let $b,g,h$ be three eigenfunctions of $-\Delta$ with eigenvalues $\lambda_b \gg \lambda_g \gg \lambda_h$, normalized so $\|b\|_2 = \|g\|_2 = \|h\|_2 = 1$. Let $\tau_0 = 0$, $\tau_1 = 1/\lambda_g$, and $\tau_2 = 1 / \lambda_h$.

Let $u = a_g g$ and $v = b + a_h h$ where $a_g$ and $a_h$ are scalars. We have that \begin{align} f(\tau_0) &= a_g^2 - (1 + a_h^2) \\ f(\tau_1) &= \frac{a_g^2}{4} - \frac{1}{(1 + \frac{\lambda_b}{\lambda_g})^3} - \frac{a_h^2}{(1 + \frac{\lambda_h}{\lambda_g})^3} > \frac{a_g^2}{4} - \frac{\lambda_g^3}{\lambda_b^3} - a_h^2 \\ f(\tau_2) &= - \frac{a_h^2}{8} - \frac{1}{(1 + \frac{\lambda_b}{\lambda_h})^3} + \frac{a_g^2}{(1 + \frac{\lambda_g}{\lambda_h})^2} < - \frac{a_h^2}{8} + \frac{a_g^2 \lambda_h^2}{\lambda_g^2} \end{align} It is clear that for very large $t$, $f(t) > 0$ (since $t^{-1}$ decays slower than $t^{-3/2}$). So to produce an almost counterexample (only thing missing is that $u$ and $v$ are not in $C^\infty_0(\Omega)$), we just need to arrange

\begin{align} a_g^2 &< a_h^2 + 1 \\ a_g^2 & \geq 4 a_h^2 + \frac{4 \lambda_g^3}{\lambda_b^3} \\ a_g^2 & \leq \frac{\lambda_g^2}{8 \lambda_h^2} a_h^2 \end{align}

This can be satisfied by, for example, $a_h^2 = 1/6$ and $a_g^2 = 1$ provided the ratio of eigenvalues $24\lambda_g^3 < \lambda_b^3$ and $48 \lambda_h^2 < \lambda_g^2$.

• Can you write more details about how you obtain $f(\tau_1$? Thank you! – JumpJump Feb 1 '16 at 17:10
• and $C_0^\infty$ is kind of important to me since I need to do integration by parts to obtain $f(t)$. – JumpJump Feb 1 '16 at 17:11
• and I do apologize that I changed the question a bit by changing $u$ to $\nabla u$ and add one more assumption. But your example really gives me some ideas. Thank you! – JumpJump Feb 1 '16 at 17:29
• @tankonetoone: for your first comment: use that eigenfunctions of the Laplacian with different eigenvalues are orthogonal. And just compute. – Willie Wong Feb 1 '16 at 17:50
• and I am still confused at how you compute fraction order without using Fourier transform? – JumpJump Feb 1 '16 at 18:36