Initial Value Problem, Lipschitz

Question

Show that the IVP $$y'= t^{-2}(\sin(2t)-2ty) $$

Such that, $y(1)=2$, has a unique solution for $1 \leq t \leq 2 $.

I'm attempting to use Lipschitz Condition and finding the lipschitz Constant but this seems harder then anticipated, if anyone can solve this using Lipschitz it would be much appreciated.

Answer 1

Given $1 \leq t \leq a$ for any $a >1$. Then $\frac{1}{|t|} \leq 1$ and \begin{align} |f(t,y_1) - f(t,y_2)| &= |(\frac{\sin(2t)}{t^2}-\frac{2ty_1}{t^2}) - (\frac{\sin(2t)}{t^2}-\frac{2ty_2}{t^2})|\\ &= |\frac{-2y_1}{t} + \frac{2 y_2}{t}| \\&= 2 \frac{1}{|t|}|y_2 - y_1| \\ &\leq 2 |y_1 - y_2| \end{align}

Answer 2

Let

$F(y, t) = t^{-2}(\sin (2t) - 2ty); \tag{1}$

then our differential equation may be written

$y' = F(y, t); \tag{2}$

note that

$\vert F(y, t) - F(z, t) \vert = \vert t^{-2}(-2ty + 2tz) \vert = \vert 2t^{-1} \vert \vert y - z \vert; \tag{3}$

for $1 \le t \le 2$, we have $1/2 \le t^{-1} \le 1$, hence

$\vert 2t^{-1} \vert \le 2; \tag{4}$

thus, from (3),

$\vert F(y, t) - F(z, t) \vert \le 2\vert y - z \vert \tag{5}$

for $1 \le t \le 2$. (5) shows that $F(y, t)$ is Lipschitz continuous with Lipschitz constant $2$ for all $y, z \in \Bbb R$; furthermore $F(y, t)$ is continuous in $t$ for $1 \le t \le 2$. It follows then from the standard results on existence and uniqueness for ODEs (see for example https://en.m.wikipedia.org/wiki/Picard%E2%80%93Lindel%C3%B6f_theorem) that (1)/(2) has a unique solution on $[1, 2]$ for any $y(1)$; taking $y(1) = 2$ finalizes the particular case at hand.