Why a DE solution must have certain properties

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I have a DE with an initial value:
$\frac{dy}{dx}=y^{2}+y-2, $ $y(x_{0})=y_{0}$

The question I am asking relates to two answers before it. The first question asks to verify that the existence and uniqueness theorem are satisfied.
Here I took the partial derivative with respect to y. I said that this shows that the function is continuous which is implied by its ability to have a derivative (this answer is incomplete and as a side note, if you could help me with it, I'd be super happy, but that's not my main question).
The next asks to determine all the equilibrium solutions to the DE. I got
$y=-2,y=1$ Lastly, the question is:
Using your answers to the last two questions, explain why any solution $y=y(x)$ to this equation with $-2<y_{0}=y(x_{0})<1$ must have $-2<y(x)<$ for all x and have y be a decreasing function of x.
I don't honestly know.

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You might want to look at the "shape" of the right hand side. Plotting $x^2 + x - 2$ yields a parabola whose negative part is exactly between the two equilibrium points. Now you have to think as if this parabola plots the value of the derivative of $y$ on the vertical axis, taking $y$ on the horizontal axis.

With $y_0$ between the equilibrium points, the derivative is going to be immediately negative: $y$ will decrease, but as we know from the shape of the parabola, even if it does it will stay in the negative region. So it will keep decreasing, until $y$ hits $-2$: but even if it does, the derivative will be zero, so that the function as a whole cannot increase or decrease anymore.

So we must conclude that the function is bounded between the two points, and is everywhere decreasing.

Bonus question: can in fact $y$ reach $-2$ from some initial conditions between $-2$ and $1$? Can it reach $1$? Answering to this will give you the correct bounds (namely, the fact that you can bound the function using $<$ and $>$ rather than $\leq$ or $\geq$).

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Just to answer your first point about the existence and uniqueness theorem; this theorem holds when the function defining your dynamical system is continuously differentiable in a neighborhood around your initial condition $x_0$.

In your system, this function is $f(y)=y^2+y-2$, whose derivative is continuous on the whole real line $\mathbb{R}$. Thus, your system has a unique solution $\forall x_0\in\mathbb{R}$.

Hope this helps as well!