Applications of the shift operator

„And then a step to the right“

In the last post I promised some more examples of using shift operators, so here we are. To begin, consider the simple trigonometric integral equation

$\displaystyle \int_0^\pi\mathrm{d}\theta\,\sin(\theta)u(x-\alpha\cos(\theta))=f(x)$ .

We can isolate the function $u(x)$ using the shift operator and arrive at

$\displaystyle \int_0^\pi\mathrm{d}\theta\,\sin(\theta)\mathrm{e}^{-\alpha\cos(\theta)\hat{D}_x}u(x)=\frac{2\sinh(\alpha\hat{D}_x)}{\alpha\hat{D}_x}u(x)=f(x)$ ,

from which the formal solution follows:

$\displaystyle u(x)=\frac{\alpha\hat{D}_x}{2\sinh(\alpha\hat{D}_x)}f(x)$ .

To proceed, we need to find a suitable representation of the pseudo-differential operator on the right. There are a number of possibilities, one obvious choice is the following Fourier transform:

$\displaystyle \frac{\alpha k}{2\sinh(\alpha k)}=\frac{\pi}{2\alpha}\int_{-\infty}^\infty\mathrm{d}\xi\,\frac{\mathrm{e}^{-\frac{\pi}{\alpha}\xi}}{(1+\mathrm{e}^{-\frac{\pi}{\alpha}\xi})^2}\mathrm{e}^{-\mathrm{i}\xi k}$ .

Replacing $k\to\hat{D}_x$ gives the solution for $u(x)$ in the form of an integral transformation of $f(x)$ :

$\displaystyle u(x)=\frac{\pi}{2\alpha}\int_{-\infty}^\infty\mathrm{d}\xi\,\frac{\mathrm{e}^{-\frac{\pi}{\alpha}\xi}}{(1+\mathrm{e}^{-\frac{\pi}{\alpha}\xi})^2}f(x-\mathrm{i}\xi)=\frac{\pi}{8\alpha}\int_{-\infty}^\infty\mathrm{d}\xi\,\mathrm{sech}^2\Bigl(\frac{\pi\xi}{2\alpha}\Bigr)f{(x-\mathrm{i}\xi)}$ .

For polynomial $f(x)$ , you can easily check that this is correct. However, except for some well-behaved functions, this integral will most likely not converge due to the complex shift. Another way to represent the pseudo-differential operator is the more complicated

$\displaystyle \frac{\alpha s}{2\sinh(\alpha s)}=\frac{\alpha^2}{\pi}s\Bigl(s^2+\Bigl(\frac{\pi}{2\alpha}\Bigr)^2\Bigr)\int_0^\infty\mathrm{d}\xi\,\Bigl(\Bigl\lvert\cos\Bigl(\frac{\pi}{2\alpha}\xi\Bigr)\Bigr\rvert-\cos\Bigl(\frac{\pi}{2\alpha}\xi\Bigr)\!\Bigr)\mathrm{e}^{-s\xi}$ ,

from which follows

$\displaystyle {u(x)=\frac{\alpha^2}{\pi}\int_0^\infty\mathrm{d}\xi\,\Bigl(\Bigl\lvert\cos\Bigl(\frac{\pi}{2\alpha}\xi\Bigr)\Bigr\rvert-\cos\Bigl(\frac{\pi}{2\alpha}\xi\Bigr)\!\Bigr)\Bigl(f'''(x-\xi)+\Bigl(\frac{\pi}{2\alpha}\Bigr)^2f'(x-\xi)\!\Bigr)}$ .

Which is just beautiful, don’t you think? A rigorous proof in general is probably more difficult.

The next example is to find the solution of the so-called polyharmonic equation of order $k$ :

$\displaystyle (-\triangle_x)^ku(\vec{x})=f(\vec{x})$

with the Laplace operator in $n$ dimensions $\triangle_x=\frac{\partial^2}{\partial x_1^2}+\frac{\partial^2}{\partial x_2^2}+\dotsm+\frac{\partial^2}{\partial x_n^2}$ . Note that the special case $k=1$ is the well-known Poisson’s equations:

$\displaystyle \triangle_x u(\vec{x})=-f(\vec{x})$ .

The formal solution of the polyharmonic equation is easily found as

$\displaystyle u(\vec{x})=(-\triangle_x)^{-k}f(\vec{x})$ ,

where we can treat the resulting operator as usual using the identity

$\displaystyle \frac{1}{a^n}=\frac{1}{\Gamma(n)}\int_0^\infty\mathrm{d}s\,s^{n-1}\mathrm{e}^{-as}$

giving

$\displaystyle u(\vec{x})=\frac{1}{\Gamma(k)}\int_0^\infty\mathrm{d}s\,s^{k-1}\mathrm{e}^{-s(-\triangle_x)}f(\vec{x})$ .

How do we deal with the operator $\mathrm{e}^{s\triangle_x}f(\vec{x})$ ?* It is similar to the shift operator, but now with a second-order derivative instead of a first-order derivative. Again, we can use an integral transformation, namely

$\displaystyle \mathrm{e}^{sx^2}=\frac{1}{\sqrt{4\pi s}}\int_0^\infty\mathrm{d}\xi\,\mathrm{e}^{-\frac{\xi^2}{4s}}\mathrm{e}^{\xi x}$ ,

to transform the second-order derivative into a first-order derivative, yielding

$\displaystyle \mathrm{e}^{s\triangle_x}f(\vec{x})=\frac{1}{(4\pi s)^{\frac{n}{2}}}\int_{\mathbb{R}^n}\mathrm{d}^n\xi\,\mathrm{e}^{-\frac{\lVert\vec{\xi}\rVert^2}{4s}}\mathrm{e}^{\vec{\xi}\cdot\nabla_x}f(\vec{x})$ .

Combining this with the original integral expression, we get

$\displaystyle u(\vec{x})=\frac{1}{\Gamma(k)(4\pi)^{\frac{n}{2}}}\int_0^\infty\mathrm{d}s\int_{\mathbb{R}^n}\mathrm{d}^n\xi\,s^{k-\frac{n}{2}-1}\mathrm{e}^{-\frac{\lVert\vec{\xi}\rVert^2}{4s}}\mathrm{e}^{\vec{\xi}\cdot\nabla_x}f(\vec{x})$ ,

which after integrating over $s$ gives the familiar integral

$\displaystyle u(\vec{x})=\frac{\Gamma(\frac{n}{2}-k)}{\Gamma(k)(4\pi)^{\frac{n}{2}}}\int_{\mathbb{R}^n}\mathrm{d}^n\xi\,f(\vec{x}+\vec{\xi})\lVert\vec{\xi}\rVert^{2k-n}=\frac{\Gamma(\frac{n}{2}-k)}{\Gamma(k)(4\pi)^{\frac{n}{2}}}\int_{\mathbb{R}^n}\mathrm{d}^n\xi\,\frac{f(\vec{\xi})}{\lVert\vec{x}-\vec{\xi}\rVert^{n-2k}}$ .

Note that there are some important conditions on the dimension $n$ and order $k$ for this expression to exist and be correct (for a detailed analysis of polyharmonic equations, see this paper and the references therein).

The third example touches on a topic of particular interest to me: finding the operators corresponding to integral transforms⁺. To demonstrate this, we take the spherical mean of a function defined by

$\displaystyle \bar{f}(r,\vec{x})=\frac{1}{S_{n-1}}\int_{\lVert\vec{\xi}\rVert=1}\mathrm{d}^n\xi\,f(\vec{x}+r\vec{\xi})$ ,

where $S_{n-1}$ is the area of the $n$ -dimensional unit sphere and $r$ is the radius of the sphere we want to average over. The form of the integral suggests the use of the shift operator. Therefore, we find

$\displaystyle \bar{f}(r,\vec{x})=\frac{1}{S_{n-1}}\int_{\lVert\vec{\xi}\rVert=1}\mathrm{d}^n\xi\,\mathrm{e}^{r\vec{\xi}\cdot\nabla_x}f(\vec{x})$ ,

and after computing the integral, we arrive at the expression

$\displaystyle \bar{f}(r,\vec{x})=\Gamma\left(\frac{n}{2}\right)\biggl(\frac{2}{r\lVert\nabla_x\rVert}\biggr)^{\frac{n}{2}-1}I_{\frac{n}{2}-1}(r\lVert\nabla_x\rVert)f(\Vec{x})$ ,

where $I_\alpha(x)$ is the modified Bessel function of the first kind. We can write this expression even more compact by using generalized hypergeometric functions

$\displaystyle {}_pF_q(a_1,\dots,a_p;b_1,\dots,b_q;x)=\sum_{n=0}^\infty\prod_{i=1}^p\frac{\Gamma(a_i+n)}{\Gamma(a_i)}\prod_{j=1}^q\frac{\Gamma(b_j)}{\Gamma(b_j+n)}\frac{x^n}{n!}$ ,

as the operator

$\displaystyle \Bar{f}_S(r,\Vec{x})={}_0F_1\Bigl(;\frac{n}{2};\frac{r^2\triangle_x}{4}\Bigr)f(\Vec{x})$ .

This operator is a function of the Laplacian and contains all relevant information about the spherical mean of that function. To make sense of the operator, we interpret it in terms of the power series of ${}_0F_1(;\frac{n}{2};x)$ at $x=0$ :

$\displaystyle {}_0F_1\Bigl(;\frac{n}{2};\frac{r^2\triangle_x}{4}\Bigr)f(\Vec{x})=\Bigl(1+\frac{r^2}{2n}\triangle_x+\frac{r^4}{8n(n+2)}\triangle_x^2+\dotsm\Bigr)f(\Vec{x})$ .

This already gives us a way to efficiently approximate the spherical mean for small radii $r$ . This is certainly nice, but I guess you’re wondering: What’s the use of an associated operator of an integral transformation? Why bother looking for one? In short, it simplifies a lot of things. For instance, we can write down the inverse operation of spherical averaging right away; it’s just a matter of dividing by the operator:

$\displaystyle f(\Vec{x})=\Bigl[{}_0F_1\Bigl(;\frac{n}{2};\frac{r^2\triangle_x}{4}\Bigr)\Bigr]^{-1}\Bar{f}_S(r,\Vec{x})$ .

Again, we interpret the operator by its power series. More interesting, however, is the ability to use the operator to give meaning to fractional spherical means, such as the $\frac{1}{2}$ -th spherical mean:

$\displaystyle \Bigl[{}_0F_1\Bigl(;\frac{n}{2};\frac{r^2\triangle_x}{4}\Bigr)\Bigr]^{\frac{1}{2}}f(\Vec{x})=\Bigl(1+\frac{r^2}{4n}\triangle_x+\frac{(n-2)r^4}{8n^2(n+2)}\triangle_x^2+\dotsm\Bigr)f(\Vec{x})$ .

It’s easy to check that applying the square root of the operator to the function again returns the usual spherical mean:

$\displaystyle \Bigl[{}_0F_1\Bigl(;\frac{n}{2};\frac{r^2\triangle_x}{4}\Bigr)\Bigr]^{\frac{1}{2}}\biggl(\Bigl[{}_0F_1\Bigl(;\frac{n}{2};\frac{r^2\triangle_x}{4}\Bigr)\Bigr]^{\frac{1}{2}}f(\Vec{x})\biggr)={}_0F_1\Bigl(;\frac{n}{2};\frac{r^2\triangle_x}{4}\Bigr)f(\Vec{x})$ .

Generalizations of this kind are certainly not easy to do just from the integral definition. So, in this case, operators help us to do things that would be hard to do otherwise.

An important area where one frequently encounters shift-type operators is quantum mechanics in the form of time evolution operators. They form an essential part of the formalism known as Schrödinger picture, in which the wave function evolves in time according to the Hamiltonian:

$\displaystyle \mathrm{i}\hbar\frac{\partial}{\partial t}\psi(t,x)=\hat{H}\psi(t,x)$ .

Formally, by treating the Hamiltonian as a number, the solution of this differential equation for some initial condition $\psi_0(x)$ is

$\displaystyle \psi(t,x)=\mathrm{e}^{-\frac{\mathrm{i}}{\hbar}t\hat{H}}\psi_0(x)$ .

This is a very prominent appearance of the time evolution operator. In fact, we can write the solution to any partial differential equation with Cauchy boundary conditions of the type

$\displaystyle \frac{\partial}{\partial t}f(t,x)=\hat{\mathbb{D}}f(t,x)$ ,

for any time-independent differential operator $\hat{\mathbb{D}}$ , in the form

$\displaystyle f(t,x)=\mathrm{e}^{t\hat{\mathbb{D}}}f_0(x)$ ,

where $f_0(x)$ is the initial condition. This is useful in two ways: first, we can take the time evolution operator and use operational methods to determine its effect on the initial condition to find the solution, or second, we can interpret an expression of the form „exponential of an operator“ as a time evolution operator and give it meaning by means of the corresponding partial differential equation. For example, suppose we don’t know what $f(x,t)=\mathrm{e}^{t\hat{D}_x}f(x)$ produces. The corresponding differential equation is

$\displaystyle \frac{\partial}{\partial t}f(t,x)=\hat{D}_xf(t,x)$ ,

whose solution is given by

$\displaystyle f(t,x)=g(x+t)$ ,

for some function determined by the initial condition. However, we know that $f(x,0)=\mathrm{e}^{0\hat{D}_x}f(x)=f(x)$ , so $g(x)=f(x)$ and

$\displaystyle f(t,x)=\mathrm{e}^{t\hat{D}_x}f(x)=f(x+t)$ ,

which indeed is the correct answer. As an example of the first case of using time evolution operators in conjunction with operational methods to find solutions to PDEs, consider the differential equation

$\displaystyle \frac{\partial}{\partial t}f(t,x)=(x^3\hat{D}_x^3+3x^2\hat{D}_x^2+x\hat{D}_x)f(t,x)=(x\hat{D}_x)^3f(t,x)$ .

As outlined above, the formal solution is

$\displaystyle f(t,x)=\mathrm{e}^{t(x\hat{D}_x)^3}f_0(x)$ .

So we need a way to transform the third-order derivative into something manageable like the first-order derivative. It turns out that there is a suitable identity, namely

$\displaystyle \mathrm{e}^{tx^3}=\int_{-\infty}^\infty\mathrm{d}\xi\,\mathrm{Ai}(\xi)\mathrm{e}^{\sqrt[3]{3t}\xi x}$ ,

where $\text{Ai}(x)$ is the Airy function. Together with the shift operator, we find

$\displaystyle f(t,x)=\int_{-\infty}^\infty\mathrm{d}\xi\,\mathrm{Ai}(\xi)f_0\bigl(x\mathrm{e}^{\sqrt[3]{3t}\xi}\bigr)$ ,

provided the integral converges. A second example of using operational methods with the time evolution operator is the following PDE:

$\displaystyle \frac{\partial}{\partial t}f(t,x)=\sqrt{1-x^2}\hat{D}_xf(t,x)$ .

The corresponding time evolution operator is

$\displaystyle f(t,x)=\mathrm{e}^{t\sqrt{1-x^2}\hat{D}_x}f_0(x)$ .

However, since $\hat{D}_x\arcsin(x)=\frac{1}{\sqrt{1-x^2}}$ , we have

$\displaystyle f(t,x)=\mathrm{e}^{t\hat{D}_{\arcsin(x)}}f_0(x)=f_0\bigl(\sqrt{1-x^2}\sin(t)+x\cos(t)\bigr)$ ,

which can be easily verified to be the solution to the original PDE. Solving this differential equation took two sentences, which again shows how simple certain problems become when using operational calculus.

I hope these examples give a good impression of the possibilities of the shift operator in solving complicated mathematical problems. Now imagine what is possible with operators in general, and you will perhaps reach the same conclusion as I did some time ago: „Why didn’t I hear about this sooner?“ Well, the answer is quite mundane: lack of rigor. Basically, all solutions like I have presented here should be taken with a grain of salt and really verified before making any high-stakes decisions with them. However, as I have pointed out in previous posts, they allow us to gain valuable insight into the solution set of the problem, thus simplifying the entire solution-finding process. But that’s it for now!

See you next time, Cheers!

If you’re still interested in what’s possible with operators, take another look at the work of D. Babusci and G. Dattoli (I highly recommend it)

*If we interpret the expression as a time evolution operator, it corresponds to the heat (or diffusion) kernel
⁺The following example is taken from my paper on spherical means

Pseudo-Riemannian Manifold of Random Ideas