Solution to "Heat Equation" with Fractional Laplacian in 2 Dimensions

Question

Statement of the Problem

We consider the equation:

$ \partial_t u + (- \Delta)^{1/2}u = 0 $

for $ u : \mathbb{R}^2 \rightarrow \mathbb{R} $.

I would like to find a non-trivial solution to this equation, using the Fourier Transform.

I believe I have used the right methods to find the answer, but think there must be a mistake. I would love for someone to check my results for me.

My Attempt

We first take the FT of the equation with respect to the space variable $x$:

$ \partial_t \hat{u} + |\xi| \hat{u} = 0 $.

The solution to this equation is obviously $ \hat{u}(t,\xi) = e^{-t |\xi|} $.

Then the solution $u$ to our original equation is:

$ u(t,x) = \mathcal{F}[e^{-t |\xi|}](x) $

$ = \int_{\mathbb{R}^2} e^{-t |\xi|} e^{2 \pi i x \cdot \xi} \text{d}\xi $.

We note that this equation is radially symmetric with respect to $x$. That is,

$ \mathcal{F}[e^{-t |\xi|}](O_2 x) = \mathcal{F}[e^{-t |\xi|}](x) $, where $O_2$ is a rotation in 2 dimensions.

Then we can write:

$ \mathcal{F}[e^{-t |\xi|}](x) = \mathcal{F}[e^{-t |\xi|}](|x|) = \int_{\mathbb{R}^2} e^{-t |\xi|} e^{2 \pi i |x| |\xi|} \text{d}\xi $, which we then rewrite in polar coordinates as:

$ \int_{0}^{2 \pi} \int^{\infty}_{0} e^{-t \rho} e^{2 \pi i |x| \rho} \rho \ \text{d}\rho \text{d}\theta = 2 \pi \int^{\infty}_{0} e^{-t \rho} e^{2 \pi i |x| \rho} \rho \ \text{d}\rho $

$ = 2 \pi \large ( \frac{-(4 \pi^2 x^2 - t^2)}{16 \pi^{16} x^4 + 8 \pi^2 t^2 x^2 + t^4 } + \frac{4 \pi i t |x|}{16 \pi^4 x^4 + 8 \pi^2 t^2 x^2 + t^4} ) $.

This last value was calculated by splitting $ e^{2 \pi i |x| \rho} = \cos(2 \pi i |x| \rho) + i \sin(2 \pi i |x| \rho) $, and then plugging the functions into an integral calculator.

The reason why I suspect that this answer is wrong is that it is a complex number. I have seen here that, since the function $e^{-t|\xi|}$ is real and even with respect to $\xi$, the FT should be real as well. Does this not also apply to the inverse FT?

Have I made a mistake somewhere? Please let me know. Thank you.

Answer 1

Here's a solution 'by hand', copied from my handwritten class notes. Oddly, the radial symmetry isn't used. We first use a tiny bit of complex analysis and the magic of Fubini's theorem to arrive at the following lemma:

Lemma(Subordination Principle.) $$e^{-\beta}=\frac{1}{\sqrt{\pi}} \int_{0}^{\infty} \frac{e^{-u}}{\sqrt{u}} e^{-i \beta^2 / 4 u} d u.$$ Proof: First note that \begin{align} \frac{2}{\pi} \int_{0}^{\infty} \frac{\cos \beta x}{1+x^{2}} d x &=\frac{1}{\pi} \int_{-\infty}^{\infty} \frac{e^{i \beta x}}{1+x^2} d x \\ &=2 \pi i \operatorname{Res}_{z=i}\left(\frac{1}{\pi} \frac{e^{i \beta x}}{1+x^{2}}\right) \\ &= e^{-\beta}. \end{align} Therefore: \begin{align} e^{-\beta} &=\frac{2}{\pi } \int_{0}^{\infty} \color{red}{\frac{1}{1+x^2}} \cos \beta x d x \\ &=\frac{2}{\pi }\int_0^\infty \cos\beta x \color{red}{\int_0^\infty e^{-u} e^{-x^2 u} du} dx \\ &\overset{\smash{Fubini}}{=}\frac{2}{\pi }\int_{0}^{\infty} e^{-u} \int_{0}^{\infty} \cos \beta x e^{-x^{2} u} d x d u \\ &=\frac{2}{\pi }\int_{0}^{\infty} e^{-u} \frac{1}{2} \int_{-\infty}^{\infty} e^{-x^{2} u} e^{i \beta x} d x d u \\ &\overset{x=2\pi y}= 2\int_{0}^{\infty} e^{-u} \int_{-\infty}^{\infty} e^{-4 \pi^{2} u y^{2}} e^{-2 \pi i \beta y} d y d u \\ &\overset{\substack{\text{FT of}\\ \text{Gaussian}}}=2\int_{0}^{\infty} e^{-u} \frac{1}{2\sqrt{\pi u}} e^{-\beta^{2} / 4 u} d u \\ &=\frac{1}{\sqrt{\pi}} \int_{0}^{\infty} \frac{e^{-u}}{\sqrt{u}} e^{-\beta^{2} / 4 u} d u . \end{align}

I used the following two properties of the inverse Fourier transform, which I state for $n$ dimensions (which depends on your convention)- \begin{align} \mathcal F^{-1}(e^{-\pi |x|^2}) (\xi) = e^{-\pi |\xi|^2}\\ \mathcal F^{-1}(f(\lambda x))(\xi) = \frac1{\lambda^{n}} (\mathcal F^{-1}f)\left(\frac\xi\lambda\right) \end{align} Now lets compute the inverse Fourier transform of $e^{-2\pi |\xi|}$ (I think you actually want this factor of $2\pi$), in dimension $n$: \begin{align} \int_{\mathbb R^n} \color{blue}{e^{-2\pi |\xi|} }e^{2\pi i x \xi} d\xi &= \frac1{\sqrt\pi}\int_{\mathbb R^n} \color{blue}{\int_0^\infty \frac{e^{-u}}{\sqrt{u}} e^{-4\pi^2|\xi|^{2} / 4 u} d u } e^{2\pi i x \xi} d\xi \\ &\overset{Fubini}= \frac1{\sqrt\pi} \int_0^\infty e^{-u}{\sqrt u} \int_{\mathbb R^n} e^{-\pi ^2 |\xi|^2/u} e^{2\pi i x\xi} d\xi du \\ &\overset{\substack{\text{FT of}\\ \text{Gaussian}}}= \frac1{\sqrt\pi} \int_0^\infty \frac{e^{-u}}{\sqrt u} \frac{u^{n/2}}{\pi^{n/2}} e^{-u |x|^2} du \\ &\overset{v = (1+|x|^2)u}= \frac1{\pi^{\frac{n+1}2}}\int_0^\infty v^{\frac{n-1}2} e^{-v} dv \frac1{(1+|x|^2)^{\frac{n-1}2}} \frac1{1+|x|^2} \\ &= \frac{\Gamma\left({\frac{n+1}2}\right)}{\pi^{\frac{n+1}2}}\frac1{(1+|x|^2)^{\frac{n+1}2}} \end{align} Having solved the problem for $t=1$, we now note that \begin{align} \mathcal F^{-1}(e^{-2\pi t|\xi|})(x) &= \int_{\mathbb R^n} e^{-2\pi t|\xi|} e^{2\pi i x \xi} d\xi \\ &\overset{\eta = t\xi}= \int_{\mathbb R^n} e^{-2\pi |\eta|} e^{2\pi i (x/t) \eta} d\eta t^{-n} \\ &= \frac1{t^n}\mathcal F^{-1}(e^{-2\pi|\xi|})(x/t) \\ &=\frac{\Gamma\left({\frac{n+1}2}\right) t}{\pi^{\frac{n+1}2}(t^2+|x|^2)^{\frac{n+1}2}} \end{align} For $n=2$ we have $\Gamma(3/2)=\frac{\sqrt\pi}2$, so

$$ \mathcal F^{-1}(e^{-2\pi t|\xi|})(x) = \frac{t}{2\pi (t^2+|x|^2)^{3/2}}$$ If you want $ \mathcal F^{-1}(e^{-t|\xi|})(x)$ instead here it is for reference- we simply sub in $t=T/2\pi$, $$ \mathcal F^{-1}(e^{- T|\xi|})(x) =\frac{T}{(2\pi)^2(\frac{T^2}{(2\pi)^2}+|x|^2)^{3/2}} = \frac{T}{(2\pi)^{2-3}(T^2+|2\pi x|^2)^{3/2}} = \frac{2\pi T}{(T^2+|2\pi x|^2)^{3/2}} $$ and I seem to have discovered that one of us is off by a factor of 2...

Answer 2

Two points:

1. We see that your original Fourier integrand $$e^{-t|\xi|}e^{2\pi i x\cdot\xi}$$ mapping $\xi\to-\xi$ is equivalent to taking a complex conjugate, so the integral must be real. However, when it is manipulated into $$e^{-t|\xi|}e^{2\pi i |x||\xi|}$$ it has lost this property, and something has gone wrong. I see know reason that the second exponential can be simplified to $e^{2\pi i|x||\xi|}$; perhaps a factor of $\cos(\theta)$ went missing when simplifying the dot product?

2. Your use of Euler's identity $e^{ix}=\cos(x)+i\sin(x)$ seems to have imaginary arguments on both sides. In the original identity they differ by a factor of $i$.

Answer 3

The problem is that a radially symmetric function doesn't allow you to ignore the dependence on the angle between $x$ and $\xi$. The integral can be solved without this assumption. Finding the radial symmetry is very cumbersome, but it is possible (full solution below):

We know that

$$ \mathcal{F}^{-1}[e^{-t|\xi|}] = \int_{\mathbb{R}^2} e^{-t |\xi|} e^{2 \pi i x\cdot \xi} \text{d}\xi$$

From there, we can then use the definition of dot product $x \cdot \xi = |x||\xi|cos(\phi_{\xi}-\phi_x)$ where $\phi_\xi$ and $\phi_x$ are the angles of the vectors $x$ and $\xi$. Then we have that, in polar coordinates,

$$ \mathcal{F}^{-1}[e^{-t|\xi|}] = \int_0^{2\pi} \int_0^{\infty} e^{-|\xi|(t-2\pi i |x|cos(\phi_{\xi}-\phi_x))}|\xi| d|\xi|d\phi_\xi .$$

Solving the improper integral we get

$$ \mathcal{F}^{-1}[e^{-t|\xi|}] = \int_0^{2\pi} \frac{1}{[t-2\pi i |x|cos(\phi_{\xi}-\phi_x)]^2} d\phi_\xi ,$$

where $t>0$. Making a change of coordinates $\phi_\xi' = \phi_\xi - \phi_x$, the integral becomes then

$$ \mathcal{F}^{-1}[e^{-t|\xi|}] = \int_{-\phi_x}^{2\pi-\phi_x} \frac{1}{[t-2\pi i |x|cos(\phi_{\xi}')]^2} d\phi_\xi' $$

which, by solving on Maxima and considering $0\leq \phi_x \leq 2\pi$, we get, after some simplification,

$$ \mathcal{F}^{-1}[e^{-t|\xi|}] = \frac{-\sqrt{4\pi^2 |x|^2+t^2}(4\pi^3 cos^2(\phi_x)t|x|^2 + \pi t^3)}{-64 \pi^6 cos^2(\phi_x)|x|^6 - (32cos^2(\phi_x) + 16)\pi^4 t^2 |x|^4 - (4 cos^2(\phi_x)+8)\pi^2 t^4 |x|^2 - t^6} $$

Changing to cartesian coordinates where $x = (x_1,x_2)$ and simplifying further, we have

$$ \mathcal{F}^{-1}[e^{-t|\xi|}] = \frac{-\sqrt{4\pi^2 |x|^2+t^2}(4\pi^2 x_1^2 + t^2) \pi t}{-64 \pi^6 x_1^2 |x|^4 - (32 x_1^2 |x|^2 + 16 |x|^4) \pi^4 t^2 - (4 x_1^2 + 8 |x|^2) \pi^2 t^4 - t^6} $$

Finally, the denominator can be factored:

$$ \mathcal{F}^{-1}[e^{-t|\xi|}] = \frac{-\sqrt{4\pi^2 |x|^2+t^2}(4\pi^2 x_1^2 + t^2) \pi t}{-(4 \pi^2 x_1^2 + t^2)(4 \pi |x|^2 + t^2)^2} $$

Simplifying the fraction we see finally the radial symmetry:

$$ \mathcal{F}^{-1}[e^{-t|\xi|}](x) = \frac{\pi t}{(4 \pi |x|^2 + t^2)^{3/2}} $$