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Consider the following equation of $\beta$ : $\beta(0)=2$ and

$$\dot\beta(t)\beta(t)^2 = \frac{1}{2}\int_0^t\frac{\dot\beta(s)(t-s)}{A_{\beta}(s,t)^{3/2}}f\left(\frac{t-s}{\sqrt{A_{\beta}(s,t)}}\right)ds,\quad \forall t>0,$$

where $\dot \beta$ denotes the derivative of $\beta$,

$$A_{\beta}(s,t) := \int_s^t\frac{du}{\beta(u)^2},\quad \forall 0\le s\le t$$

and

$$f(x):=\frac{1}{\sqrt{2\pi}}e^{-x^2/2},\quad \forall x\in\mathbb R.$$

Does this equation have a (unique) solution? Any answer, comments or references are highly appreciated.

Consider the following equation of $\beta$ : $\beta(0)=2$ and

$$\dot\beta(t)\beta(t)^2 = \frac{1}{2}\int_0^t\frac{\dot\beta(s)(t-s)}{A_{\beta}(s,t)^{3/2}}f\left(\frac{t-s}{\sqrt{A_{\beta}(s,t)}}\right)ds,\quad \forall t>0,$$

where $\dot \beta$ denotes the derivative of $\beta$,

$$A_{\beta}(s,t) := \int_s^t\frac{du}{\beta(u)^2},\quad \forall 0\le s\le t$$

and

$$f(x):=\frac{1}{\sqrt{2\pi}}e^{-x^2/2},\quad \forall x\in\mathbb R.$$

Does this equation have a solution that is strictly decreasing? Any answer, comments or references are highly appreciated.

Consider the following equation of $\beta$ : $\beta(0)=2$ and

$$\dot\beta(t)\beta(t)^2 = \frac{1}{2}\int_0^t\frac{\dot\beta(s)(t-s)}{A_{\beta}(s,t)}f\left(\frac{t-s}{\sqrt{A_{\beta}(s,t)}}\right)ds,\quad \forall t>0,$$

where $\dot \beta$ denotes the derivative of $\beta$,

$$A_{\beta}(s,t) := \int_s^t\frac{du}{\beta(u)^2},\quad \forall 0\le s\le t$$

and

$$f(x):=\frac{1}{\sqrt{2\pi}}e^{-x^2/2},\quad \forall x\in\mathbb R.$$

Does this equation have a (unique) solution? Any answer, comments or references are highly appreciated.

Consider the following equation of $\beta$ : $\beta(0)=2$ and

$$\dot\beta(t)\beta(t)^2 = \frac{1}{2}\int_0^t\frac{\dot\beta(s)(t-s)}{A_{\beta}(s,t)^{3/2}}f\left(\frac{t-s}{\sqrt{A_{\beta}(s,t)}}\right)ds,\quad \forall t>0,$$

where $\dot \beta$ denotes the derivative of $\beta$,

$$A_{\beta}(s,t) := \int_s^t\frac{du}{\beta(u)^2},\quad \forall 0\le s\le t$$

and

$$f(x):=\frac{1}{\sqrt{2\pi}}e^{-x^2/2},\quad \forall x\in\mathbb R.$$

Does this equation have a (unique) solution? Any answer, comments or references are highly appreciated.

Consider the following equation of $\beta$ : $\beta(0)=2$ and

$$\dot\beta(t)\beta(t)^2 = \frac{1}{2}\int_0^t\frac{\dot\beta(s)(t-s)}{A_{\beta}(s,t)^{3/2}}f\left(\frac{t-s}{\sqrt{A_{\beta}(s,t)}}\right)ds,\quad \forall t>0,$$

where $\dot \beta$ denotes the derivative of $\beta$,

$$A_{\beta}(s,t) := \int_s^t\frac{du}{\beta(u)^2},\quad \forall 0\le s\le t$$

and

$$f(x):=\frac{1}{\sqrt{2\pi}}e^{-x^2/2},\quad \forall x\in\mathbb R.$$

Does this equation have a (unique) solution? Any answer, comments or references are highly appreciated.

Consider the following equation of $\beta$ : $\beta(0)=2$ and

$$\dot\beta(t)\beta(t)^2 = \frac{1}{2}\int_0^t\frac{\dot\beta(s)(t-s)}{A_{\beta}(s,t)}f\left(\frac{t-s}{\sqrt{A_{\beta}(s,t)}}\right)ds,\quad \forall t>0,$$

where $\dot \beta$ denotes the derivative of $\beta$,

$$A_{\beta}(s,t) := \int_s^t\frac{du}{\beta(u)^2},\quad \forall 0\le s\le t$$

and

$$f(x):=\frac{1}{\sqrt{2\pi}}e^{-x^2/2},\quad \forall x\in\mathbb R.$$

Does this equation have a (unique) solution? Any answer, comments or references are highly appreciated.