# Existence and unicity ODE

I've been reading "A text book on ordinary differential equations" of Sahir Ahmad & Ambrossi. On page 37 we have the following excercise:

"Explain why $$x'+\dfrac{\sin(t)}{e^t+1}x=0$$ cannot has solution $$x(t)$$ such that $$x(1)=1$$ and $$x(2)=-1$$."

As far I know, $$f(t,x)$$ is continuos everywhere since $$e^t\ne -1$$ for every $$t$$, and $$\frac{\partial f}{\partial x}$$ doesn't depends on $$x$$, so its constant and therefore continous, then the existence and unicity theorem must apply on neighborhood of those points. Where am I wrong? Can anyone explain me? Im very stuck af this assignament.

• As $\dfrac{\sin t}{e^t+1}$ remains positive, $x'$ and $x$ have opposite signs, and $x$ cannot decrease below zero. – Yves Daoust Mar 2 at 13:19
• @YvesDaoust: Actually the sign of $\frac{\sin t}{e^t+1}$ does not matter. $\tilde x(t) = 0$ is a solution, therefore all other solutions are identically zero or have no zero at all. – Martin R Mar 3 at 18:09

Let $$F$$ be an antiderivative of the function $$\frac{ \sin t}{e^t+1}.$$ Then the general solution of the differential equation is given by

$$x(t)=C e^{-F(t)},$$

where $$C \in \mathbb R.$$

From $$1=x(1)= Ce^{-F(1)}$$ we get $$C>0.$$

But from $$-1=x(2)=Ce^{-F(2)}$$ we get $$C<0.$$

Hint:

$$\left|\frac{\sin(t)}{1+e^t}x-\frac{\sin(t)}{1+e^t}y\right|\leq |x-y|,$$ for all $$x,y\in\mathbb R$$.

Yes, the conclusion follows directly from the uniqueness part of the Picard–Lindelöf theorem for initial value problems.

• The differential equation is of the form $$x'(t) = f(t, x(t))$$ where $$f(t, x) = -\frac{\sin(t)}{e^t+1} x$$ is continuous, and uniformly Lipschitz continuous in $$x$$.
• $$\tilde x(t) = 0$$ is a solution of that differential equation on the interval $$I = [1, 2]$$.

If $$x$$ is any solution of the differential equation on $$I$$ with $$x(t_0) = 0$$ then $$x$$ and $$\tilde x$$ are both solutions of the same initial value problem $$x'(t) = f(t, x(t))$$, $$x(t_0) = 0$$, and therefore $$x = \tilde x$$.

It follows that any solution of the differential equation on $$I$$ is either identically zero, or has no zeros at all.

In particular, there can be no solution which takes both positive and negative values.