Skip to main content
MathOverflow

Return to Answer

deleted 2 characters in body
Source Link
Giorgio Metafune
Giorgio Metafune
  • 6.3k
  • 2
  • 11
  • 24

The answer is yes and the proof splits in some steps. In what follows $\sigma$ is just positive, decreasing and in $L^1(0, \infty)$.

  1. If $y$ is a bounded solution of the ODE (forgetting the initial condition), then $$ y(t)=\gamma e^{\alpha t}\int_t^\infty \sigma (s)(1-y^2(s))e^{-\alpha s}\, ds $$ and $y(t) \to 0$ at infinity.

This follows by writing $(e^{-\alpha t} y)'=-\gamma \sigma (t)(1-y^2)e^{-\alpha t}$ and integrating from $t$ to $\infty$. If $|y| \leq M$, the formula above yileds $|y(t)| \leq \gamma \alpha^{-1}(1+M^2) \sigma (t)$.

  1. If $t_\gamma $ is choosen so that $\int_{t_\gamma}^\infty \gamma \sigma (s) \, ds \leq \frac 14$, then there exists a unique solution $y_\gamma$ defined for $t \geq t_\gamma$ satisfying $0 \leq y_\gamma \leq 1$. This follows from a contraction argument considering the map $$ Ty(t)=\gamma e^{\alpha t}\int_t^\infty \sigma (s)(1-y^2(s))e^{-\alpha s}\, ds $$ from $C=\{y:[t_\gamma, \infty[\mapsto [0,1]\}$ into itself which is a contraction for the sup-norm, by elementary computations as above.

  2. Now I consider the profiles where $y'=0$, given by the quadratic expression $\alpha y-\gamma \sigma (1-y^2)=0$. The upper profile is $$\frac{-\alpha +\sqrt{\alpha^2 +4\gamma^2 \sigma^2}}{2\gamma \sigma}= \frac{s}{1+\sqrt{1+s^2}}, \quad s=2\gamma \alpha^{-1} \sigma. $$ Note that $0 \leq \phi \leq 1$ and $\phi$ is decreasing in $t$ (since it is increasing in $s$ and $\sigma$ is decreasing). In particular, $\phi$ is a supersolutionsubsolution of the ODE.

The lower profile is negative. A solution is decreasing when it is between the two profiles, increasing outside.

  1. The solution $y_\gamma$ constructed in 2. for $t \geq t_\gamma$ can be continued to $[0, t_\gamma]$ and satisfies $0 \leq y_\gamma \leq \phi \leq 1$ for every $t \geq 0$.

In fact, if $y_\gamma (t_0) \geq \phi (t_0)$, then $y_\gamma \geq \phi$ after $t_0$, by comparison and then $y_\gamma$ would be increasing for $t \geq t_0$ and could not vanish at infinity.

  1. The solution $y_\gamma$ depends continuosly on $\gamma$. In fact if $y_i=y_{\gamma_i}$ \begin{align*} y_2-y_1=&\gamma_2 e^{\alpha t}\int_t^\infty \sigma (s)(y_1^2-y_2^2) e^{-\alpha s}ds +(\gamma_2-\gamma_1)e^{\alpha t}\int_t^\infty \sigma (s)e^{-\alpha s}ds\\ &-(\gamma_2-\gamma_1)e^{\alpha t}\int_y^\infty y_1^2\sigma (s)e^{-\alpha s}ds \end{align*} and then $$|y_2(t)-y_1(t)| \leq C|\gamma_2-\gamma_1|+2\gamma_2 \int_t^\infty \sigma (s)|y_2(s)-y_1(s)|\, ds. $$ Gronwall's lemma (with initial point $\infty$, same proof as in the usual case) gives $|y_2(t)-y_1(t)| \leq C|\gamma_2-\gamma_1| \exp{\int_t^\infty 2 \gamma_2 \sigma (s)ds} \leq K|\gamma_2-\gamma_1|$.

  2. Since $y_0(0)=0$ to show that there exists $\gamma$ such that $y_\gamma (0)= \frac 23$ it suffices to show that $\limsup_{\gamma \to \infty} y_\gamma (0)=1$.

If not, $y_\gamma (0)^2 \leq 1-\delta$ for $\gamma$ large and, since $y_\gamma $ is decreasing, the integral equation yields $$ y_\gamma (t)\geq \delta \gamma e^{\alpha t} \int_t^\infty \sigma (s) e^{-\alpha s}\, ds \to \infty $$ as $\gamma \to \infty$, in contrast with $0 \leq y_\gamma \leq 1$ for all $\gamma$.

The answer is yes and the proof splits in some steps. In what follows $\sigma$ is just positive, decreasing and in $L^1(0, \infty)$.

  1. If $y$ is a bounded solution of the ODE (forgetting the initial condition), then $$ y(t)=\gamma e^{\alpha t}\int_t^\infty \sigma (s)(1-y^2(s))e^{-\alpha s}\, ds $$ and $y(t) \to 0$ at infinity.

This follows by writing $(e^{-\alpha t} y)'=-\gamma \sigma (t)(1-y^2)e^{-\alpha t}$ and integrating from $t$ to $\infty$. If $|y| \leq M$, the formula above yileds $|y(t)| \leq \gamma \alpha^{-1}(1+M^2) \sigma (t)$.

  1. If $t_\gamma $ is choosen so that $\int_{t_\gamma}^\infty \gamma \sigma (s) \, ds \leq \frac 14$, then there exists a unique solution $y_\gamma$ defined for $t \geq t_\gamma$ satisfying $0 \leq y_\gamma \leq 1$. This follows from a contraction argument considering the map $$ Ty(t)=\gamma e^{\alpha t}\int_t^\infty \sigma (s)(1-y^2(s))e^{-\alpha s}\, ds $$ from $C=\{y:[t_\gamma, \infty[\mapsto [0,1]\}$ into itself which is a contraction for the sup-norm, by elementary computations as above.

  2. Now I consider the profiles where $y'=0$, given by the quadratic expression $\alpha y-\gamma \sigma (1-y^2)=0$. The upper profile is $$\frac{-\alpha +\sqrt{\alpha^2 +4\gamma^2 \sigma^2}}{2\gamma \sigma}= \frac{s}{1+\sqrt{1+s^2}}, \quad s=2\gamma \alpha^{-1} \sigma. $$ Note that $0 \leq \phi \leq 1$ and $\phi$ is decreasing in $t$ (since it is increasing in $s$ and $\sigma$ is decreasing). In particular, $\phi$ is a supersolution of the ODE.

The lower profile is negative. A solution is decreasing when it is between the two profiles, increasing outside.

  1. The solution $y_\gamma$ constructed in 2. for $t \geq t_\gamma$ can be continued to $[0, t_\gamma]$ and satisfies $0 \leq y_\gamma \leq \phi \leq 1$ for every $t \geq 0$.

In fact, if $y_\gamma (t_0) \geq \phi (t_0)$, then $y_\gamma \geq \phi$ after $t_0$, by comparison and then $y_\gamma$ would be increasing for $t \geq t_0$ and could not vanish at infinity.

  1. The solution $y_\gamma$ depends continuosly on $\gamma$. In fact if $y_i=y_{\gamma_i}$ \begin{align*} y_2-y_1=&\gamma_2 e^{\alpha t}\int_t^\infty \sigma (s)(y_1^2-y_2^2) e^{-\alpha s}ds +(\gamma_2-\gamma_1)e^{\alpha t}\int_t^\infty \sigma (s)e^{-\alpha s}ds\\ &-(\gamma_2-\gamma_1)e^{\alpha t}\int_y^\infty y_1^2\sigma (s)e^{-\alpha s}ds \end{align*} and then $$|y_2(t)-y_1(t)| \leq C|\gamma_2-\gamma_1|+2\gamma_2 \int_t^\infty \sigma (s)|y_2(s)-y_1(s)|\, ds. $$ Gronwall's lemma (with initial point $\infty$, same proof as in the usual case) gives $|y_2(t)-y_1(t)| \leq C|\gamma_2-\gamma_1| \exp{\int_t^\infty 2 \gamma_2 \sigma (s)ds} \leq K|\gamma_2-\gamma_1|$.

  2. Since $y_0(0)=0$ to show that there exists $\gamma$ such that $y_\gamma (0)= \frac 23$ it suffices to show that $\limsup_{\gamma \to \infty} y_\gamma (0)=1$.

If not, $y_\gamma (0)^2 \leq 1-\delta$ for $\gamma$ large and, since $y_\gamma $ is decreasing, the integral equation yields $$ y_\gamma (t)\geq \delta \gamma e^{\alpha t} \int_t^\infty \sigma (s) e^{-\alpha s}\, ds \to \infty $$ as $\gamma \to \infty$, in contrast with $0 \leq y_\gamma \leq 1$ for all $\gamma$.

The answer is yes and the proof splits in some steps. In what follows $\sigma$ is just positive, decreasing and in $L^1(0, \infty)$.

  1. If $y$ is a bounded solution of the ODE (forgetting the initial condition), then $$ y(t)=\gamma e^{\alpha t}\int_t^\infty \sigma (s)(1-y^2(s))e^{-\alpha s}\, ds $$ and $y(t) \to 0$ at infinity.

This follows by writing $(e^{-\alpha t} y)'=-\gamma \sigma (t)(1-y^2)e^{-\alpha t}$ and integrating from $t$ to $\infty$. If $|y| \leq M$, the formula above yileds $|y(t)| \leq \gamma \alpha^{-1}(1+M^2) \sigma (t)$.

  1. If $t_\gamma $ is choosen so that $\int_{t_\gamma}^\infty \gamma \sigma (s) \, ds \leq \frac 14$, then there exists a unique solution $y_\gamma$ defined for $t \geq t_\gamma$ satisfying $0 \leq y_\gamma \leq 1$. This follows from a contraction argument considering the map $$ Ty(t)=\gamma e^{\alpha t}\int_t^\infty \sigma (s)(1-y^2(s))e^{-\alpha s}\, ds $$ from $C=\{y:[t_\gamma, \infty[\mapsto [0,1]\}$ into itself which is a contraction for the sup-norm, by elementary computations as above.

  2. Now I consider the profiles where $y'=0$, given by the quadratic expression $\alpha y-\gamma \sigma (1-y^2)=0$. The upper profile is $$\frac{-\alpha +\sqrt{\alpha^2 +4\gamma^2 \sigma^2}}{2\gamma \sigma}= \frac{s}{1+\sqrt{1+s^2}}, \quad s=2\gamma \alpha^{-1} \sigma. $$ Note that $0 \leq \phi \leq 1$ and $\phi$ is decreasing in $t$ (since it is increasing in $s$ and $\sigma$ is decreasing). In particular, $\phi$ is a subsolution of the ODE.

The lower profile is negative. A solution is decreasing when it is between the two profiles, increasing outside.

  1. The solution $y_\gamma$ constructed in 2. for $t \geq t_\gamma$ can be continued to $[0, t_\gamma]$ and satisfies $0 \leq y_\gamma \leq \phi \leq 1$ for every $t \geq 0$.

In fact, if $y_\gamma (t_0) \geq \phi (t_0)$, then $y_\gamma \geq \phi$ after $t_0$, by comparison and then $y_\gamma$ would be increasing for $t \geq t_0$ and could not vanish at infinity.

  1. The solution $y_\gamma$ depends continuosly on $\gamma$. In fact if $y_i=y_{\gamma_i}$ \begin{align*} y_2-y_1=&\gamma_2 e^{\alpha t}\int_t^\infty \sigma (s)(y_1^2-y_2^2) e^{-\alpha s}ds +(\gamma_2-\gamma_1)e^{\alpha t}\int_t^\infty \sigma (s)e^{-\alpha s}ds\\ &-(\gamma_2-\gamma_1)e^{\alpha t}\int_y^\infty y_1^2\sigma (s)e^{-\alpha s}ds \end{align*} and then $$|y_2(t)-y_1(t)| \leq C|\gamma_2-\gamma_1|+2\gamma_2 \int_t^\infty \sigma (s)|y_2(s)-y_1(s)|\, ds. $$ Gronwall's lemma (with initial point $\infty$, same proof as in the usual case) gives $|y_2(t)-y_1(t)| \leq C|\gamma_2-\gamma_1| \exp{\int_t^\infty 2 \gamma_2 \sigma (s)ds} \leq K|\gamma_2-\gamma_1|$.

  2. Since $y_0(0)=0$ to show that there exists $\gamma$ such that $y_\gamma (0)= \frac 23$ it suffices to show that $\limsup_{\gamma \to \infty} y_\gamma (0)=1$.

If not, $y_\gamma (0)^2 \leq 1-\delta$ for $\gamma$ large and, since $y_\gamma $ is decreasing, the integral equation yields $$ y_\gamma (t)\geq \delta \gamma e^{\alpha t} \int_t^\infty \sigma (s) e^{-\alpha s}\, ds \to \infty $$ as $\gamma \to \infty$, in contrast with $0 \leq y_\gamma \leq 1$ for all $\gamma$.

Source Link
Giorgio Metafune
Giorgio Metafune
  • 6.3k
  • 2
  • 11
  • 24

The answer is yes and the proof splits in some steps. In what follows $\sigma$ is just positive, decreasing and in $L^1(0, \infty)$.

  1. If $y$ is a bounded solution of the ODE (forgetting the initial condition), then $$ y(t)=\gamma e^{\alpha t}\int_t^\infty \sigma (s)(1-y^2(s))e^{-\alpha s}\, ds $$ and $y(t) \to 0$ at infinity.

This follows by writing $(e^{-\alpha t} y)'=-\gamma \sigma (t)(1-y^2)e^{-\alpha t}$ and integrating from $t$ to $\infty$. If $|y| \leq M$, the formula above yileds $|y(t)| \leq \gamma \alpha^{-1}(1+M^2) \sigma (t)$.

  1. If $t_\gamma $ is choosen so that $\int_{t_\gamma}^\infty \gamma \sigma (s) \, ds \leq \frac 14$, then there exists a unique solution $y_\gamma$ defined for $t \geq t_\gamma$ satisfying $0 \leq y_\gamma \leq 1$. This follows from a contraction argument considering the map $$ Ty(t)=\gamma e^{\alpha t}\int_t^\infty \sigma (s)(1-y^2(s))e^{-\alpha s}\, ds $$ from $C=\{y:[t_\gamma, \infty[\mapsto [0,1]\}$ into itself which is a contraction for the sup-norm, by elementary computations as above.

  2. Now I consider the profiles where $y'=0$, given by the quadratic expression $\alpha y-\gamma \sigma (1-y^2)=0$. The upper profile is $$\frac{-\alpha +\sqrt{\alpha^2 +4\gamma^2 \sigma^2}}{2\gamma \sigma}= \frac{s}{1+\sqrt{1+s^2}}, \quad s=2\gamma \alpha^{-1} \sigma. $$ Note that $0 \leq \phi \leq 1$ and $\phi$ is decreasing in $t$ (since it is increasing in $s$ and $\sigma$ is decreasing). In particular, $\phi$ is a supersolution of the ODE.

The lower profile is negative. A solution is decreasing when it is between the two profiles, increasing outside.

  1. The solution $y_\gamma$ constructed in 2. for $t \geq t_\gamma$ can be continued to $[0, t_\gamma]$ and satisfies $0 \leq y_\gamma \leq \phi \leq 1$ for every $t \geq 0$.

In fact, if $y_\gamma (t_0) \geq \phi (t_0)$, then $y_\gamma \geq \phi$ after $t_0$, by comparison and then $y_\gamma$ would be increasing for $t \geq t_0$ and could not vanish at infinity.

  1. The solution $y_\gamma$ depends continuosly on $\gamma$. In fact if $y_i=y_{\gamma_i}$ \begin{align*} y_2-y_1=&\gamma_2 e^{\alpha t}\int_t^\infty \sigma (s)(y_1^2-y_2^2) e^{-\alpha s}ds +(\gamma_2-\gamma_1)e^{\alpha t}\int_t^\infty \sigma (s)e^{-\alpha s}ds\\ &-(\gamma_2-\gamma_1)e^{\alpha t}\int_y^\infty y_1^2\sigma (s)e^{-\alpha s}ds \end{align*} and then $$|y_2(t)-y_1(t)| \leq C|\gamma_2-\gamma_1|+2\gamma_2 \int_t^\infty \sigma (s)|y_2(s)-y_1(s)|\, ds. $$ Gronwall's lemma (with initial point $\infty$, same proof as in the usual case) gives $|y_2(t)-y_1(t)| \leq C|\gamma_2-\gamma_1| \exp{\int_t^\infty 2 \gamma_2 \sigma (s)ds} \leq K|\gamma_2-\gamma_1|$.

  2. Since $y_0(0)=0$ to show that there exists $\gamma$ such that $y_\gamma (0)= \frac 23$ it suffices to show that $\limsup_{\gamma \to \infty} y_\gamma (0)=1$.

If not, $y_\gamma (0)^2 \leq 1-\delta$ for $\gamma$ large and, since $y_\gamma $ is decreasing, the integral equation yields $$ y_\gamma (t)\geq \delta \gamma e^{\alpha t} \int_t^\infty \sigma (s) e^{-\alpha s}\, ds \to \infty $$ as $\gamma \to \infty$, in contrast with $0 \leq y_\gamma \leq 1$ for all $\gamma$.