April 26th, 2018, 12:46 am

This post is a bit long, because I present a full derivation of my solution mentioned in my last post and clear up the confusion and misconception in several previous posts. In the following [$]\Theta[$] is the Heaviside step function.

The spatial domain is already compact. Saying a function is compact supported in the space dimension is a bit leading. Usually saying a function has compact support means the closure of the point of the domain where the function value is nonzero is a compact strict subset of the whole domain. If it is truly a strict compact subset and that function is analytic within the whole domain, that function can only be zero by the analytic continuation. The partial sum or truncated Fourier series

[$]f_{N}(x)=\Theta(|x-\frac12|)\big(\sum_{m=1}^{N}\frac{8}{m^2\pi^2}e^{-m^{2}\pi^{2}t}\sin\frac{m\pi}{2}\sin m\pi x\big)[$]

is only infinitely differentiable in [$][0,1][$] and not in [$][-\delta, 1+\delta][$] for some positive [$]\delta[$]. So saying [$]f_{N}(x)[$] is compact-supported and infinitely differentiable is misleading.

According to Cuchulainn, the problem to be solved is a parabolic (heat) PDE [$]u[$] define on [$](t,x)\in[0,\infty)\times [0,1][$] with Dirichlet boundary condition

[$]u(t,x=0)=u(t,x=1)=0[$]. If the equation is [$]\Big(\frac{\partial }{\partial t}-\frac12\frac{\partial^2}{\partial x^2}\Big)u(t,x)=0[$] We can solve it two ways. One is the Fourier series method as ppauper used to obtain

[$]u(t,x)=\sum_{m=1}^{\infty}\frac{8}{m^2\pi^2}e^{-m^{2}\pi^{2}t}\sin\frac{m\pi}{2}\sin m\pi x[$]

For [$]t=0[$], the Fourier series can be shown to converge uniformly [$]\sim\frac1m[$]. The proof is a bit complicated. I may put it up later though. Thus the convergence is very slow for very small [$]t[$] at least for the first partial sums but fast for large [$]t[$]. Besides, the partial sum is going to oscillate, which is not desirable to represent a heat equation solution with a concave initial condition and a vanishing boundary.

The other is the image (Green's function) method or the Poisson summation formula Alan and I used to obtain the following solution

[$]u(t,x)=\frac1{\sqrt{2\pi t}}\int_{-\infty}^\infty dy\, e^{-\frac{(x-y)^2}{2t}} \sum_{m=-\infty}^\infty (u(t=0,y+2m)-u(t=0,-y+2m)),[$]

where in this setting [$]u(t,y):=u(t,y)\Theta(|\frac12-y|)[$] i.e., the extension of [$]u(t,\cdot)[$] to the whole real axis with zero values assigned outside of [$][0,1][$]. This is extends the initial value function to an odd function of period [$]2[$]. This series converges very fast for small [$]t[$] at the rate of [$]\sim e^{-\frac{m^2}{2t}}[$] but slowly for large [$]t[$]. So we use this series.

The fast convergence allows us to just take the first term. We would like to enforce as much as we can without complicating the mathematics too much the vanishing boundary condition, so we take, say the quadratic Taylor expansion at the boundary and make an odd function extension with respect to that boundary point. In our particular case, it is even simpler. We just extend the wedge (not the hat) to infinity. We have

[$]u_{\text{l}}(t,x)=\frac1{\sqrt{2\pi t}}\int_{-\infty}^\infty dy\, e^{-\frac{(x-y)^2}{2t}}(1-2|y-\frac12|)=1-\frac2{\sqrt{2\pi t}}\int_{-\infty}^\infty dy\, e^{-\frac{y^2}{2t}}|x-\frac12-y|[$]

[$]=2\Big(\frac12-\big(\frac12-x\big)\text{erf}(v) - \sqrt{\frac {2t}\pi}\,e^{-v^2} \Big),[$]

where [$]v:=\frac{x-\frac12}{\sqrt{2t}}[$]. We see that the above [$]u_{\text{l}}(t,x)[$] is the positively and analytically weighted [$]y[$]integral of concave functions [$]-|x-\frac12-y|[$] which is then concave and analytic in [$]x[$]. Note the exponential function reflects the effect of [$]y[$] term or the slope at each point, particularly the broken slope at the vertex, of the initial condition function. The concavity is provided by the exponential function. So it can not be dropped. The error function gives the base line shape and rounds off (making the resulting function analytic) the sharp corner of the vertex.

In fact, the error of the approximation is

[$]e\big(t,x+\frac12\big) = (1-2|x|)-u_{\text{l}}(t,x+\frac12)=\frac2{\sqrt{2\pi t}}\int_{-\infty}^\infty dy\, e^{-\frac{y^2}{2t}}(|x|-|x-y|).[$]

We can easily see from the graph of the integrand in the parenthesis that [$]e\big(t,x+\frac12\big)>0[$] and its maximum occurs at [$]x=0[$] i.e. the vertex with value [$]e\big(t,\frac12\big)=2\sqrt{\frac{2t}\pi}[$]. I have to emphasize that this expression particularly [$]\sqrt t[$] is to give the functional dependency between [$]\max_x(e(t,x))[$] and the parameterization of [$]u_{\text l}(t,x)[$]. One may just pick [$]\max_x(e(t,x))[$] as the parameter and re-parameterize [$]u_{\text l}(\cdot,x)[$]. The approximation is always below the target function. This lowering is the effect of the exponential function. and drops the lowest at the vertex. So we know not only the convergence rate but the exact location of the error and its exact value. This is contributed solely by the exponential term. This gives yet another reason the exponential function cannot be dropped.

It is only natural that this approximation should possess all the properties of a heat equation because it **IS** a solution of a heat equation, while dropping one without the exponential function is **NOT**.

I have now derived my formula and its desired properties.

Last edited by

lovenatalya on April 26th, 2018, 4:42 pm, edited 4 times in total.