I am considering the following problem.
Let $\mathcal{P}$ the classical heat-diffusion problem:
$$\mathcal{P} : \left(\partial_t u (t,x)=\frac{1}{2}\partial_{xx}^2u(t,x)\text{ with }u(0,\cdot) = f(x)\right).$$
Suppose that the initial condition $f$ satisfies the conditions:
- $f\in\mathcal{C}^\infty(\mathbb{R},\mathbb{R})\cap L^\infty(\mathbb{R}, \mathbb{R})$ (smooth and bounded)
- $f$ is negative before $0$ and positive after (one zero-crossings).
We know that the function $u(t,\cdot)$ has a unique zero-crossing point denoted $x_t$ (I can add details on that).
My problem is the following: can one find a constant $K>0$ such that:
$$\forall t\in\mathbb{R}_+,~|x_t| \leq Kt \text{ ? }.$$
As well done by @Iosif Pinellis in this post, we have an asymptotic approximation of $x_t$ as $t$ tends to 0.
Does anyone have an idea on that ?
Thank you very much!
P.S.: feel free to make any remarks on MO guidelines
Best Answer
$\def\R{\mathbb R}$$\def\aha{{1/2}}$$\def\maha{{-1/2}}$ Edit. Now it looks correct.
I can prove that $x_t$ grows at most like $t$ for $t\geq 1$, up to a multiplicative constant.
For simplicity, assume that $\|f\|_{L^\infty}\leq 1$. Call $$ 0<\varepsilon:=\min_{x\in[1,2]} f(x). $$ Now write $u$ as a convolution with the heat kernel (I will use the kernel for the equation $u_t=u_{xx}$ and drop the factor $1/2$, as this amounts simply to a linear rescaling of the time variable). $$ u(t,x)=\int_\R f(y)\frac{1}{\sqrt{4\pi t}}e^{-\frac{(x-y)^2}{4t}}dy=\int_\R f(t^\aha z)\frac{1}{\sqrt{4\pi}}e^{-\frac{(t^\maha x-y)^2}{4}}dz. $$ Call $\Phi(x):=\int_{-\infty}^x\frac{1}{4\pi}e^{-z^2/4}dz.$
We split the above integral in two pieces and we estimate both from below: $$ u(t,x)\geq -\int_{-\infty}^0 \frac{1}{\sqrt{4\pi}}e^{-\frac{(t^\maha x-y)^2}{4}} dz + \varepsilon\int_{t^\maha}^{2t^\maha} \frac{1}{\sqrt{4\pi}}e^{-\frac{(t^\maha x-y)^2}{4}} dz=$$ $$ = -\Phi(-t^\maha x) + \varepsilon\left(\Phi(t^\maha(2-x))-\Phi(t^\maha(1-x))\right). $$ Now we use the estimate on $\Phi$ (which can be checked by differentiating both sides) $$ \Phi(z)\leq \frac{1}{\sqrt{4\pi}}\frac{4}{|z|+1}e^{-z^2/4},\quad\forall z\leq 0 $$ and Lagrange formula to show that for $x\geq 0$ and some $g\in[1,2]$ we have $$ u(t,x)\geq \varepsilon\frac{1}{\sqrt{4\pi t}}e^{-\frac{(x-g)^2}{4t}}-\frac{1}{\sqrt{4\pi}}\frac{4}{|x/\sqrt t|+1}e^{-x^2/4t}\geq $$ $$ \geq \varepsilon\frac{1}{\sqrt{4\pi t}}e^{-\frac{(x-1)^2}{4t}}-\frac{1}{\sqrt{4\pi}}\frac{4}{|x/\sqrt t|+1}e^{-x^2/4t}= $$ $$ =\frac{1}{\sqrt{4\pi t}}e^{-\frac{x^2}{4t}}\left(e^{\log(\varepsilon) +\frac{x}{2t}-\frac{1}{4t}}-e^{\log(4)-\log(\frac{x}{t}+\frac{1}{\sqrt t}) }\right).$$ Simplifying, one obtains that for non-negative $x$, $u(t,x)$ is positive whenever $$ \frac{x}{t}>h\left(\frac{1}{2\sqrt t}+\frac{1}{4t}-\log(\varepsilon/4)\right)-t^\maha,$$ where $h=g^{-1}$ is the inverse of the function $g(z)=\log(z)+z/2$. After further estimates on $h$, one can show that $u(t,x)>0$ whenever $t\geq 1$ and $$ x>Ct$$ for some constant $C$ depending on $\varepsilon$ ($C=-\log(\varepsilon/20)$ should work). By symmetry, a similar bound holds from below, which implies that $$ |x_t|\leq Ct $$ for all $t\geq 1$. Combined with the asymptotic formula for small $t$, this should give the desired result.
I am not entirely sure every step of the proof is correct, but in any case this idea should work to establish the bound you want.