- #1

mathmari

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We have the initital value problem $$\begin{cases}y'(t)=1/f(t, y(t)) \\ y(t_0)=y_0\end{cases} \ \ \ \ \ (1)$$ where the function $f:\mathbb{R}^2\rightarrow (0,\infty)$ is continuous in $\mathbb{R}^2$ and continuously differentiable as for $y$ in a domain that contains the point $(t_0, y_0)$.

Show that there exists $h>0$ such that the following two conditions are satisfied:

- The problem (1) has a solution $\phi=\phi(t)$ that is defined at least for each $t\in (t_0-h, t_0+h)$.

- In the interval $(t_0-h, t_0+h)$ there is no other solution of the problem (1). (i.e. if a function $\psi$ is a solution of the problem (1), then $\psi (t)=\phi (t)$, if $t\in (t_0-h, t_0+h)$)

We have that $f$ is continuous, then $\frac{1}{f}$ is also continuous, since it doesn't get the value $0$. Is this correct?

We consider a region $R=\left \{(t,y) : |t-t_0|\leq a, \ |y-y_0|\leq b\right \}$ with $a,b>0$.

Do we have to show that $\frac{1}{f}$ is bounded in $R$ ?

Because then the IVP (1) would have at least one solution $\phi = \phi (t)$ defined in the interval $|t − t_0| \leq h$ where $h=\min \left \{a, \frac{b}{K}\right \}$, where $K$ is the maximum value of $\frac{1}{f}$ in $R$, or not?

(Wondering) For the second time we use the uniqueness theorem, or not?

Do we have to use for that the Lipschitz condition?

We have that $$\left |\frac{\frac{1}{f(t_1)}-\frac{1}{f(t_2)}}{t_1-t_2}\right |=\left |\frac{\frac{f(t_2)-f(t_1)}{f(t_1)f(t_2)}}{t_1-t_2}\right |=\left |\frac{f(t_2)-f(t_1)}{(t_1-t_2)f(t_1)f(t_2)}\right |=\frac{|f(t_2)-f(t_1)|}{|t_1-t_2||f(t_1)||f(t_2)|}$$ Since $f$ is continuous we get that $|f(t_2)-f(t_1)|\leq C|t_2-t_1|\Rightarrow \frac{|f(t_2)-f(t_1)|}{|t_2-t_1|}\leq C$. So we get $$\left |\frac{\frac{1}{f(t_1)}-\frac{1}{f(t_2)}}{t_1-t_2}\right |\leq C\cdot \frac{1}{|f(t_1)||f(t_2)|}$$ Is everything correct so far? How could we continue?

(Wondering)