Return to Answer

edited body

Source Link

edited Sep 11 at 6:47

I have found a solution to my question. I have posted the solution below. Thanks to everyone for insightful discussions.

Define a new variable $U_t := X_t/\sigma_t$. Then, using the ODE , we get the following new ODE:

\begin{equation} \label{eq: ODE U} \frac{dU_t}{dt} = \frac{y (1-t)}{\sigma_t^3} \tanh \left( \frac{a t U_t}{\sigma_t}\right), \quad U_0>0. \end{equation} First, note that to prove our claim, it is enough to show that $U_t>0$ for all $t>0$. We will now proceed via a contra-positive argument. Let there exists $t\in [0,1]$ such that $U_t \le 0$. To be more precise, let $$t_0 = \inf\{t \in [0,1] : U_t \le 0 \}.$$ Note that $t_0>0$ (as $U_0>0$) and $U_{t_0} \le 0$ as $t_0$ is infimum of a closed set by inverse map theorem. Now, we have $$ \int_0^{t_0} \dot{U}_t\; dt = U_{t_0} - U_0 <0. $$ Due to mean value theorem there exists $\tau \in (0,t_0)$ such that $\dot{U}_\tau <0$ , which implies $$ \frac{y (1-\tau)}{\sigma_\tau^3} \tanh \left( \frac{a \tau U_\tau}{\sigma_\tau}\right) <0 \Leftrightarrow U_\tau <0. $$$$ \frac{a (1-\tau)}{\sigma_\tau^3} \tanh \left( \frac{a \tau U_\tau}{\sigma_\tau}\right) <0 \Leftrightarrow U_\tau <0. $$ This is a direct contradiction to the definition of $t_0$. Therefore, we have $U_t>0$ for all $t>0$, which evidently yields that $X_t>0$ for all $t \in [0,1]$. This further shows that if $U_t$ is a non-decreasing (maybe strictly increasing) of $t$ if $U_0 = X_0>0$.

I have found a solution to my question. I have posted the solution below. Thanks to everyone for insightful discussions.

Define a new variable $U_t := X_t/\sigma_t$. Then, using the ODE , we get the following new ODE:

\begin{equation} \label{eq: ODE U} \frac{dU_t}{dt} = \frac{y (1-t)}{\sigma_t^3} \tanh \left( \frac{a t U_t}{\sigma_t}\right), \quad U_0>0. \end{equation} First, note that to prove our claim, it is enough to show that $U_t>0$ for all $t>0$. We will now proceed via a contra-positive argument. Let there exists $t\in [0,1]$ such that $U_t \le 0$. To be more precise, let $$t_0 = \inf\{t \in [0,1] : U_t \le 0 \}.$$ Note that $t_0>0$ (as $U_0>0$) and $U_{t_0} \le 0$ as $t_0$ is infimum of a closed set by inverse map theorem. Now, we have $$ \int_0^{t_0} \dot{U}_t\; dt = U_{t_0} - U_0 <0. $$ Due to mean value theorem there exists $\tau \in (0,t_0)$ such that $\dot{U}_\tau <0$ , which implies $$ \frac{y (1-\tau)}{\sigma_\tau^3} \tanh \left( \frac{a \tau U_\tau}{\sigma_\tau}\right) <0 \Leftrightarrow U_\tau <0. $$ This is a direct contradiction to the definition of $t_0$. Therefore, we have $U_t>0$ for all $t>0$, which evidently yields that $X_t>0$ for all $t \in [0,1]$. This further shows that if $U_t$ is a non-decreasing (maybe strictly increasing) of $t$ if $U_0 = X_0>0$.

I have found a solution to my question. I have posted the solution below. Thanks to everyone for insightful discussions.

Define a new variable $U_t := X_t/\sigma_t$. Then, using the ODE , we get the following new ODE:

\begin{equation} \label{eq: ODE U} \frac{dU_t}{dt} = \frac{y (1-t)}{\sigma_t^3} \tanh \left( \frac{a t U_t}{\sigma_t}\right), \quad U_0>0. \end{equation} First, note that to prove our claim, it is enough to show that $U_t>0$ for all $t>0$. We will now proceed via a contra-positive argument. Let there exists $t\in [0,1]$ such that $U_t \le 0$. To be more precise, let $$t_0 = \inf\{t \in [0,1] : U_t \le 0 \}.$$ Note that $t_0>0$ (as $U_0>0$) and $U_{t_0} \le 0$ as $t_0$ is infimum of a closed set by inverse map theorem. Now, we have $$ \int_0^{t_0} \dot{U}_t\; dt = U_{t_0} - U_0 <0. $$ Due to mean value theorem there exists $\tau \in (0,t_0)$ such that $\dot{U}_\tau <0$ , which implies $$ \frac{a (1-\tau)}{\sigma_\tau^3} \tanh \left( \frac{a \tau U_\tau}{\sigma_\tau}\right) <0 \Leftrightarrow U_\tau <0. $$ This is a direct contradiction to the definition of $t_0$. Therefore, we have $U_t>0$ for all $t>0$, which evidently yields that $X_t>0$ for all $t \in [0,1]$. This further shows that if $U_t$ is a non-decreasing (maybe strictly increasing) of $t$ if $U_0 = X_0>0$.

Source Link

created Sep 11 at 6:38

De vinci

I have found a solution to my question. I have posted the solution below. Thanks to everyone for insightful discussions.

Define a new variable $U_t := X_t/\sigma_t$. Then, using the ODE , we get the following new ODE:

\begin{equation} \label{eq: ODE U} \frac{dU_t}{dt} = \frac{y (1-t)}{\sigma_t^3} \tanh \left( \frac{a t U_t}{\sigma_t}\right), \quad U_0>0. \end{equation} First, note that to prove our claim, it is enough to show that $U_t>0$ for all $t>0$. We will now proceed via a contra-positive argument. Let there exists $t\in [0,1]$ such that $U_t \le 0$. To be more precise, let $$t_0 = \inf\{t \in [0,1] : U_t \le 0 \}.$$ Note that $t_0>0$ (as $U_0>0$) and $U_{t_0} \le 0$ as $t_0$ is infimum of a closed set by inverse map theorem. Now, we have $$ \int_0^{t_0} \dot{U}_t\; dt = U_{t_0} - U_0 <0. $$ Due to mean value theorem there exists $\tau \in (0,t_0)$ such that $\dot{U}_\tau <0$ , which implies $$ \frac{y (1-\tau)}{\sigma_\tau^3} \tanh \left( \frac{a \tau U_\tau}{\sigma_\tau}\right) <0 \Leftrightarrow U_\tau <0. $$ This is a direct contradiction to the definition of $t_0$. Therefore, we have $U_t>0$ for all $t>0$, which evidently yields that $X_t>0$ for all $t \in [0,1]$. This further shows that if $U_t$ is a non-decreasing (maybe strictly increasing) of $t$ if $U_0 = X_0>0$.