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I was wondering if there was knowledge to the following differential equation:

$$y' = a(y) \cdot y$$ with $a$ is defined as being $a_0$ when $y < b$ and $a_1$ elsewhere with $a_0 , a_1, b$ reals.

The family of functions is obviously $Span(\exp{(a_{0/1} t)})$ on each individual interval but how do you link them.

With an initial condition, the problem is just a "glueing" solution problem. However, without, I didn't managed to get more about it.

What is the family of functions satisfying the differential equations

PackSciences
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1 Answers1

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You would want the solution to be continuous at $y = b$. If we impose the single initial condition $y(0) = y_0$, then the solution for $y<b$ is given by $$y_1(t) = y_0e^{a_0t}.$$ Treating the solution for $y \geq b$ ($y_2$) independently, we find that $$y_2(t) = y_{02}e^{a_1t},$$ where $y_{02}$ is the initial condition for $y_2$. For continuity, we impose $$y_1(t_b) = y_2(t_b),$$ where $y = b$ at $t = t_b$. This leads to the relationship $$y_{02} = y_{0}e^{a_0/a_1}.$$ Thus, the continuous solution is $$y(t) = y_0e^{a_0t},y<b,$$ $$y(t) = y_0e^{a_0/a_1}e^{a_1t},y \geq b.$$

Of course, as you pointed out there is an infinite family of solutions, and in this set we can certainly include those solutions that are discontinuous at $t_b$.

One other thing to keep in mind is that you need to be careful with the signs of $a_0$ and $a_1$. For the above continuous solution to work, you would need the two constants to have opposite signs, so that the solution decreases from $b$ in one direction, and increases from $b$ in the other.

D.B.
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