Continuity of solution of Riccati equation with negative source term

Let $t_0 > 0$ and consider the scalar Riccati differential equation

$p'(t) + 2 a(t) p(t) - r(t) \, p(t)^2 + q = 0 \; ,$

with initial condition

$p(t_0) = 0 \; ,$

in which $a$ is a function, $r$ is a non-negative function and $q < 0$ is a constant.

Under what conditions on $a$, $r$ and $q$ does this differential equation have a (continuous) solution on the interval $[0, t_0]$?

Note: In a previous post Willie Wong answered this question for the case where $q$ is a non-negative function, see Existence of global solution of Riccati equation.

• In any case, when $q < 0$, first notice that you must have $\tilde{p}' > 0$ where $\tilde{p}$ is defined as in my previous answer; in other words you don't have a barrier at 0 any more. Using that $\tilde{p}' > \tilde{r} \tilde{p}^2$, you see that so long as $\int \tilde{r}$ is unbounded, you will have finite time blow-ups. If this condition is not satisfied, whether global solutions exist will depend somewhat delicately on the form of $a$ and $r$, as well as on initial condition. – Willie Wong Dec 16 '13 at 12:12
• Basically, for sufficiently small initial data, the source term $q$ is expected to dominate the nonlinear term. So if $$e^{\int_0^t 2a \mathrm{d}s}$$ is integrable we can expect small data global existence of solutions, while large data should lead to blow-up. If the integrating factor itself is allowed to grow unboundedly, then even small initial data can lead to large data at intermediate times, around which time the nonlinear term may take over. Whether this happens depends on the relative strengths of the functions $a,r,q$, and there is, as far as I can see, no simple description. – Willie Wong Dec 16 '13 at 12:22
• There will be no easily stated general conditions. Consider the equation satisfied by $\hat{p} = \lambda p$. This is satisfied with $\hat{q} = \lambda q$ and $\hat{r} = \lambda^{-1} r$, so the three coefficients scale differently. Similarly, consider the equation satisfied by $\breve{p}(t) = p(\eta t)$, this is satisfied with $\breve{a}(t) = \eta a(\eta t)$ and similarly $r$ and $q$. From this scaling and local existence of solutions to the ODE we have that as long as $a, r$ and $q$ are all suitably small (compared to $t_0$) then there must be a continuous solution. – Willie Wong Dec 17 '13 at 8:55
• Thank you very much Willie. If you put this as an answer (instead of a comment), then I will accept the answer. – user112645 Feb 28 '14 at 14:37

When $q < 0$, first notice that you must have $\tilde{p}' > 0$ where $\tilde{p}$ is defined as in my previous answer; in other words you don't have a barrier at 0 any more. Using that $\tilde{p}' > \tilde{r} \tilde{p}^2$, you see that so long as $\int \tilde{r}$ is unbounded, you will have finite time blow-ups. If this condition is not satisfied, whether global solutions exist will depend somewhat delicately on the form of $a$ and $r$, as well as on initial condition.
Basically, for sufficiently small initial data, the source term $q$ is expected to dominate the nonlinear term. So if $$\exp {\int_0^t 2a \mathrm{d}s}$$ is integrable we can expect small data global existence of solutions, while large data should lead to blow-up. If the integrating factor itself is allowed to grow unboundedly, then even small initial data can lead to large data at intermediate times, around which time the nonlinear term may take over. Whether this happens depends on the relative strengths of the functions $a,r,q$, and there is, as far as I can see, no simple description.
To see that there will be no easily stated general conditions: Consider the equation satisfied by $\hat{p} = \lambda p$. This is satisfied with $\hat{q} = \lambda q$ and $\hat{r} = \lambda^{-1} r$, so the three coefficients scale differently. Similarly, consider the equation satisfied by $\breve{p}(t) = p(\eta t)$, this is satisfied with $\breve{a}(t) = \eta a(\eta t)$ and similarly $r$ and $q$. From this scaling and local existence of solutions to the ODE we have that as long as $a, r$ and $q$ are all suitably small (compared to $t_0$) then there must be a continuous solution.