# Computing the inverse of Laplacian operator.

I am considering the following equation:

$$f(t):=\int_\Omega[(I-t\Delta)^{-1}\Delta(I-t\Delta)^{-1}u]\cdot u\,dx$$ where $u\in C_c^\infty(\Omega)$ and $t\geq 0$ a real number. $I$ is the identity operator and by $-1$ we denote the inverse of operator. For example, if $v=\Delta w$, we may write $w=\Delta^{-1}v$.

I am trying to study the positiveness of $f$. Clearly, when $t=0$, I have $$f(0)=\int_\Omega\Delta u\cdot u = -\int_\Omega |\nabla u|^2<0$$ My question: I wish to prove $f(t)\leq 0$ for $t>0$ and $\lim_{t\to\infty}f(t)=0$ as well. I am confused at as $t$ gets larger, how $(I-t\Delta)^{-1}u$ changes?

Update:

We have \begin{align*} f(t)&=\int_\Omega[(I-t\Delta)^{-1}\Delta(I-t\Delta)^{-1}u]\cdot u=\\ &=\int_\Omega(\Delta[(I-t\Delta)^{-1}(I-t\Delta)^{-1}u])\cdot u\\ &=-\int_\Omega(\nabla[(I-t\Delta)^{-1}(I-t\Delta)^{-1}u])\cdot\nabla u\\ &=-\int_\Omega((I-t\Delta)^{-1}(I-t\Delta)^{-1}\nabla u])\cdot\nabla u \end{align*}

But why I have $$\int_\Omega((I-t\Delta)^{-1}(I-t\Delta)^{-1}\nabla u])\cdot\nabla u = \int_\Omega((I-t\Delta)^{-1}\nabla u])\cdot((I-t\Delta)^{-1}\nabla u) \tag 1$$ I am not sure I can move $(I-t\Delta)^{-1}$ like this.

• Do you consider $I + t \,\Delta : H_0^1(\Omega) \to H_0^1(\Omega)^*$? By the way, you have $f(t) = -|(I-t\,\Delta)^{-1}u|^2_{H^1(\Omega)}$, with the $H^1$-seminorm, similar to your case $t = 0$.
– gerw
Jan 26, 2016 at 18:31
• @gerw Yes, can you expand your answer a bit? Thank you! Jan 26, 2016 at 20:04
• This expression for $f(t)$ is obtained using integration by parts, as in the case $f(0)$ and b using $(I-t\,\Delta)^-1u = 0$ on $\partial\Omega$.
– gerw
Jan 26, 2016 at 20:12
• @gerw Hi I updated my post. Can you have a look? I am not so sure how you have $H^1$ seminorm in the end... Thank you! Jan 26, 2016 at 21:05
• Doeesn't $\Delta$ commute with $(I-t\Delta)^{-1}$? Jan 26, 2016 at 21:08

To answer your edit: once you have found the correct domain of definition for the Laplacian, you have that $(I-t\Delta)^{-1} = \sum \limits _{k=0} ^\infty t^k \Delta^k$. Since $\int (\Delta u) v = \int u (\Delta v)$ (on test functions, at least), you may use induction and prove that $\int (\Delta^k u) v = \int u (\Delta^k v)$, so $\int (\sum \limits _{k=0} ^\infty t^k \Delta^k u) v = \int u (\sum \limits _{k=0} ^\infty t^k \Delta^k v)$, which means $\int [(I-t\Delta)^{-1} u] v = \int u [(I-t\Delta)^{-1} v]$.
• I have one more question. What do you think about the continuity of function $f(t)$? I think it is continuous, right? Maybe even differentiable? Jan 27, 2016 at 15:24
• Also, could you provide a reference for $(I-t\Delta)^{-1}=\sum_{k=0}^\infty t^k \Delta^k$? Jan 27, 2016 at 23:29
• We could similarly define $\exp (it \frac{d}{dx})$ by a series and prove that $(u, exp (i t\frac{d}{dx})v) = 0$ when $u(x),v(x)$ have disjoint support. But we expect $\exp (it \frac{d}{dx})$ to be a translation operator. The flaw is that the series for $\exp (it \frac{d}{dx})v$ does not converge. Dec 23, 2018 at 19:24