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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 $\lim_{t\to0}H(t) = 0$ if $0<x<1$ and $\lim_{t\to0}H(t) = \alpha x$, 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!

$\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!

$\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:

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

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

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

If we denote $H$ the function:

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

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

$\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 $\lim_{t\to0}H(t) = 0$ if $0<x<1$ and $\lim_{t\to0}H(t) = \alpha x$, 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!

$\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:

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

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

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

Using the Lagrange's formula, we have that:

$$B\geq \varepsilon\frac{1}{\sqrt{4\pi t}}e^{-\frac{(x-g)^2}{4t}}, g\in[1,2].$$

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!

$\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:

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

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

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

If we denote $H$ the function:

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

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