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$\def\R{\mathbb R}$$\def\aha{{1/2}}$$\def\maha{{1/2}}$ Hi everyone,

I am interested in the following problem:

Let consider the heat equation problem:

$$\forall (t,x) \in \mathbb{R}_+\times\mathbb{R}, ~\partial_t u(t,x) = \partial_{xx}u(t,x)$$

with the initial condition:

$$\forall x \in \mathbb{R},~u(0,x) = f(x)$$

where $f$ is a smooth and bounded function ($\|f\|_\infty\leq 1$), negative before $0$ and positive after. If we denote $x_t$ the zero of the function $u(t,\cdot)$ (it is unique, I can add details), can we found a constant $K>0$ (dependent of $f$) such that:

$$\forall t\geq 0, |x_t|\leq Kt \text{ ?}$$

What has been found:

For $t\geq 1$, there exist a such constant, the proof is in this post.

For $t\to 0$ it can be proved that $x_t = O(t)$, the proof is in this post.

The problem

I would like to find an explicit constant to obtain the same result on $[0,1]$.

My attempt

My attempt follows the advice of @Lorenzo Pompili in the first post. The objective is to prove that $u(t,x)>0$ when $x>Kt$. By symetry, we would have that $|x_t|<Kt$.

We use the fact that:

\begin{equation} \forall x\leq 0, f(x) \geq x(1+\sup_{y\in[-1,0]}|f'(y)|) := Cx. \end{equation}

and we assume that it exists $\alpha>0$ such that:

$$\forall x \in [0,1],f(x)\geq \alpha x.$$ As done in the first post, we have that:

$$\forall (t,x)\in\mathbb{R}\times\mathbb{R}, u(t,x)=\frac{1}{\sqrt{4\pi}}\int_\mathbb{R}f(t^\maha z)e^{-\frac{(xt^{-\maha}-z)^2}{4}}dz.$$

we have :

$$u(t,x)\geq \frac{1}{\sqrt{4\pi}}\int_{-\infty}^0 f(t^\maha z)e^{-\frac{(t^{-\maha} x-z)^2}{4}} dz + \alpha t^\maha\frac{1}{\sqrt{4\pi}} \int_{0}^{t^{-\maha}} ze^{-\frac{(t^{-\maha} x-z)^2}{4}} dz := A+B.$$

By the bound mentionned above, we have:

$$A\geq Ct^\maha\frac{1}{\sqrt{4\pi}}\int_{-\infty}^0 ze^{-\frac{(t^\maha x-z)^2}{4}} dz = -2Ct^\maha \frac{e^{-\frac{x^2}{4t}}}{\sqrt{4\pi}}+Cx\Phi(-t^{-\maha}x) \text{ (change of variable)}$$

$$B = 2\alpha\frac{1}{\sqrt{4\pi}}\left(e^{-\frac{x^2}{4t}}-e^{-\frac{(x-1)^2}{4t}}\right)+\alpha x (\Phi(-t^{-\maha}(x-1))-\Phi(-t^{-\maha}x))$$

where $\Phi(x):=\int_{-\infty}^x\frac{1}{\sqrt{4\pi}}e^{-z^2/4}dz$.

If we denote $H$ the function:

$$H(t) = 2\alpha\frac{1}{\sqrt{4\pi}}\left(e^{-\frac{x^2}{4t}}-e^{-\frac{(x-1)^2}{4t}}\right)+\alpha x (\Phi(-t^{-\maha}(x-1))-\Phi(-t^{-\maha}x))+-2Ct^\maha \frac{e^{-\frac{x^2}{4t}}}{\sqrt{4\pi}}+Cx\Phi(-t^{-\maha}x)$$

we have $\lim_{t\to0}H(t) = \alpha x $ if $0<x<1$ and $\lim_{t\to0}H(t) = 0$, if $x>1$.

I am now blocked because I can't prove that this quantity is positive if $x>Kt$ where $K$ is a constant that should be found.

Does anyone have an idea on that ?

Thank you very much!

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    $\begingroup$ Hi again! It seems to me that the bound for positive $x$ is too weak, because for small $t$ the main contribution to $u(t,x)$ comes from values of $f$ close to $x=0$. I think the bound you want to use for positive $x$ is: $f(x)\geq \varepsilon x$ on (say) the interval $[0,1]$, with suitable $\varepsilon>0$. To have $\varepsilon>0$, of course you need to assume $f'(0)\neq 0$, and honestly I don't know what happens in the case $f'(0)=0$... $\endgroup$ Commented Oct 22, 2023 at 12:10
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    $\begingroup$ Admittedly, I have not much intuition on what happens when $f'(0)=0$, but I feel that the more derivatives vanish at $x=0$, the worse can be the behaviour of $x_t$ for small $t$. So it could be that you have to assume $f'(0)>0$ in the main statement to have a constant $K$ that works for all $t>0$. $\endgroup$ Commented Oct 22, 2023 at 12:35
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    $\begingroup$ Thank you for your answer, you are absolutely right for the $x$ multiplication. I will take into account your comments and will update the post today. Thank you again! $\endgroup$
    – NancyBoy
    Commented Oct 22, 2023 at 14:54
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    $\begingroup$ (i) the point is that $x_t$ satisfies a differential equation, $\frac{d}{dt}x_t=\frac{u_{xx}(t,x)}{u_x(t,x)}$, so if the derivative vanishes at $t=0$, it seems that the derivative of $t\mapsto x_t$ at $t=0$ could be unbounded, or not defined. (ii) I guess it depends on whether $f'(0)>0$ or not. If yes, then it should be immediate because $\partial_x u(t,x)$ also satisfies the heat equation and it is smooth at $t=0$, so $\partial_xu(t,x)$ must be continuous, hence the derivative can not be zero in a neighbourhood of $x=0$, $t=0$. If $f'(0)=0$ I don't know $\endgroup$ Commented Oct 22, 2023 at 17:14
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    $\begingroup$ Your estimate looks better now, it should suffice. To continue, note that the leading parts are the ones with $\Phi$, because the other terms go to zero for small $t$. Try to compare the terms with $\Phi$ first, and then try to add the other terms and see how to deal with them. It could be messy but probably not impossible, you should try and do some computation and estimates. There is no unique nice way of proving such statements, try and see what you get, and then if it is not enough come back and try to use better bounds $\endgroup$ Commented Oct 22, 2023 at 17:15

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