Fractional differential operator

Question

We consider the operator $$\eqalign{ & T: = \left(I - {{{\partial ^2}} \over {\partial {x^2}}}\right):H_0^1(0,L) \cap {H^2}(0,L) \to {L^2}(0,L) \cr & \cr} $$ i'm wondering, is this true : $$ (Tu,u) = ({T^{1/2}}u,{T^{1/2}}u)$$ ? Is there an explicit formula of $ {T^{1/2}} $ ? Thank you.

Answer 1

It is true since $T$ is symmetric and has positive eigenvalues. In fact, it is easy to check $$ (Tu,v)=(u,Tv), \forall u,v\in H_0^1(0,L)\cap H^2(0,L) $$ which means that $T$ is symmetric. Since $\sin(\frac{n\pi x}{L})\}_{n=1}^\infty$ is dense in $H_0^1(0,L)\cap H^2(0,L)$ and $$ T\sin(\frac{n\pi x}{L})=(1+\frac{n^2\pi^2}{L^2})\sin(\frac{n\pi x}{L}), n=1,2,\cdots, $$ namely $T$ has positive eigenvalues $1+\frac{n^2\pi^2}{L^2}$ with the corresponding eigenfunctions $\sin(\frac{n\pi x}{L})$, $n=1,2,\cdots$. For $u\in H_0^1(0,L)\cap H^2(0,L)$, let $$ u=\sum_{n=1}^\infty u_n \sin(\frac{n\pi x}{L}). $$ Clearly $$ Tu(x)=\sum_{n=1}^\infty u_n (1+\frac{n^2\pi^2}{L^2})\sin(\frac{n\pi x}{L}). $$ Define $$ (T^{\frac12}u)(x)=\sum_{n=1}^\infty u_n\sqrt{1+\frac{n^2\pi^2}{L^2}}\sin(\frac{n\pi x}{L}).$$ It is easy to check that $$ (T^{\frac12}u, T^{\frac12}u)=(Tu,u).$$

Answer 2

Another way which is useful if you need discrete approximations and if you accept using complex numbers is

$$\cases{D^2 = \frac{1}{k} [1,0,-2,0,1]\\I = [0,0,1,0,0]}$$

If we now consider $$I-iD = [0,0,1,0,0] - i[0,1,0,-1,0]$$

Now performing the self-convolution $(I-iD)*(I-iD)$ the complex numbers will give us a real part which is $[-1,0,3,0,-1]$, which is correct for $k=1$, but we need to expand to use complex numbers and throw the imaginary part away.

$$(I-D)^2 = \mathcal{Re}((I-iD)*(I-iD))$$