# Is Jacobi Theta function same as Heat Kernel ? How to derive Jacobi Theta from Heat Kernel?

My understanding is that the Jacobi Theta function is fundamental solution of heat equation:

$\displaystyle \vartheta (x,it)=1+2\sum _{n=1}^{\infty }\exp \left(-\pi n^{2}t\right)\cos(2\pi nx)$

The following heat kernel is also fundamental solution of heat equation:

$\Phi (x,t)={\frac {1}{\sqrt {4\pi kt}}}\exp \left(-{\frac {x^{2}}{4kt}}\right)$

But I do not see how to expand above heat kernel to get Jacobi Theta function.

Can anyone provide some hints on how to expand above heat kernel to get Jacobi Theta function ?

Thank you.

• Don't you have to take the convolution of the theta function when using it as a "fundamental solution" type entity (c.f., these notes)? – Alex Nelson Jul 26 '17 at 16:38
• @AlexNelson Thank you for the reply. How is the heat kernel $\Phi (x,t)$ related to Jacobi Theta function ? – david Jul 26 '17 at 16:51
• You would want to consider a solution periodic in $x$ with period $\ell$, this gives you $\theta(x/\ell, it)$. Taking the limit as $\ell\to\infty$ might "give" you $\Phi(x,t)$ but periodicity might screw things up...such at least is the heuristic approach I would take. – Alex Nelson Jul 26 '17 at 17:56
• @AlexNelson, thank you. This approach will not "give" $\Phi(x, t)$ – david Jul 26 '17 at 18:47
• What we call a fundamental solution to a PDE depends on more than just the PDE itself: we need to specify the domain and the boundary conditions and different choices leads to different solutions. When the domain is $\mathbb{R}$ then the fundamental solution is the usual heat kernel. When the domain is a finite interval then we get a theta function. See e.g. en.wikipedia.org/wiki/Heat_kernel – Winther Jul 26 '17 at 22:16

I recommend the book A Brief Introduction to Theta Functions by Richard Bellman reprinted by Dover Publications. The first expansion you wrote is the Fourier series of the theta function. Using the heat kernel you sum over a lattice of periods to match the periodicity of the theta function. Something like $\sum_{k\in \mathbb{Z}} \Phi (x+2\pi k,t)$. Notice the Fourier transform connection between the two exponentials in your two equations. The summation over a period lattice is equivalent to convolution by a periodic sum of shifted Dirac delta functions (a Dirac comb), and the Fourier transform of this is the point-wise product of the Fourier transform of the heat kernel with another Dirac comb.

• Thank you for the reply. You said: "Notice the Fourier transform connection between the two exponentials in your two equations." Just want to confirm that I understand what you said. Use symbols in my original post, $\vartheta (x,it)$ is just Fourier expansion of $\Phi (x,t)$, am I right ? – david Jul 26 '17 at 22:35
• No, Fourier series is not the same as Fourier transform although they are related. The Fourier transform of $\exp(-ax^2)$ is $\sqrt{\pi/a}\exp(-\pi^2k^2/a)$. See the MathWorld article Fourier Transform -- Gaussian – Somos Jul 26 '17 at 23:14
• @david One way to think of the theta function is the heat flow of a circular ring which is subjected to a heat spike (heat kernel) at one point. The $x$ represents space coordinates around the circular ring and hence the theta (heat) function is periodic in space, and $t$ represents time and the heat from the spike flows out to the rest of the ring and approaches uniform distribution of heat as $t\to\infty$.. – Somos Jul 26 '17 at 23:41
• The theta function $\vartheta(x,t) = \sum_n e^{-\pi n^2 x}e^{2i \pi n t}$ is the periodic fundamental solution (or Green function) of the heat equation, ie. if we have a $1$-periodic initial condition $f(x)$, the solution to the heat equation is $F(x,t) = \int_0^1 f(y) \vartheta(x-y,t)dt$. In term of distributions, $\vartheta(x,t)$ is the solution for the initial condition $\sum_n \delta(x-n)$ @david – reuns Jul 27 '17 at 3:59

HINT :

To derive Jacobi Theta from Heat Kernel just compare the respective PDE.

$$\vartheta_3 (z,q)=1+2\sum _{n=1}^{\infty }q^{n^{2}}\cos(2 nz)$$ This Jacobi Theta function is a solution of : $$4\,q\,\frac{\partial\,\vartheta_3}{\partial q}+\frac{\partial^2\,\vartheta_3}{\partial z\;^2}=0$$

It looks like the heat equation, isn't it ?

http://functions.wolfram.com/EllipticFunctions/EllipticTheta3/13/01/

With $z=\pi x$ and $q=e^{-\pi t}$ , I suppose that you can take it from here.

• Thank you. Just want to confirm that I understand what you said. Use symbols in my original post, $\vartheta (x,it)$ is just Fourier expansion of $\Phi (x,t)$, am I right ? – david Jul 26 '17 at 22:33
• The wording of your question doesn't give sufficient information on what exactly is your problem. The heat equation doesn't appears explicitly in your question. So, it isn't possible to give you the exact related change of variables to go from the elliptic theta PDE to your heat PDE. So, you have to do it by yourself. Sorry, I have not more time to spent. Good luck for your work. – JJacquelin Jul 27 '17 at 6:41