# Existence and uniqueness of solution

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 $t\in (t_0-h, t_0+h)$)



For the first point we use the existence theorem, or not?

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?

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?

• Yes. Use the continuous differentiability of $f$ in a neighbrhood of $(t_0,y_0)$ in order to show that $1/f$ is Lipschitz continuous there. – amsmath Sep 13 '18 at 22:06
• 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? @amsmath – Mary Star Sep 13 '18 at 22:19
• Or did you mean something else? @amsmath – Mary Star Sep 14 '18 at 6:29
• Note that you do not have $f(t)$ but $f(t,y)$. So $f$ depends on two variables. The reasoning I meant goes as follows: You know that $f$ is cont. diff. in a neighborhood of $(t_0,y_0)$. Then so is $1/f$. Now, there is a theorem that says that if $g$ is cont. diff. in a neighborhood of a compact set $K$, then $g$ is Lipschitz-continuous on $K$. The Lipschitz constant is the maximal norm of $\nabla f$ on $K$ (which exists since $\nabla f$ is continuous on $K$). – amsmath Sep 14 '18 at 13:31
• So in general it holds that if f is local Lipschitz continuous as for y, then the IVP has unique total solution on $[0,\infty )$ ? Or is this something else? @amsmath – Mary Star Sep 14 '18 at 14:45