For which initial values does this IVP have a solution or a unique solution?

Question

I am extremely lost on this question. It asks, for which values of $t_0, a_1, a_2, a_3$ does the IVP

$$y''' = \sin(t) e^y + \frac{(y'')^{5/3}}{\cos(t)} + (y')^{1/3}, \ y(t_0) = a_1, y'(t_0) = a_2, y''(t_0) = a_3$$

have a solution and for which values does it have a unique solution?

I have a theorem that says that for it to have a solution, $y''' = g(t,y,y',y'')$ has to be continuous, so I said it has no solution for $t = (2n+1)\pi/2$ since that would make $\cos(t) = 0$ and the function would be undefined.

It also says that is $g$ is Lipschitz in $y = (y,y',y'')$ then the solution is unique. I have no idea how to use this, or how the initial conditions come into play here, as it seems like these requirements don't utilize them at all?

Any insight helps.

Edit: So I have shown that the partial derivatives are continuous for $y' \neq 0$. So I think that I have a unique solution in a neighborhood of $(t_0,x_0)$ if $t_0 \neq (2n+1)\pi/2$ and $a_2 \neq 0$. Does this sound correct?

Answer 1

You are on the right direction. Remember by the following lemma: Show that g is continuously differentiable with respect to y on some closed "rectangle" R. Then g is Lipschitz with respect to y on R.

Hence, after evaluating function g such that: (∂g)/(∂ y[1])=sin(t)e^(y[1])

(∂g)/(∂ y[2])=1/(3)y[2]^(-2/(3))

(∂g)/(∂ y[3])=((5)/(3))y[3]^(2/(3))cos^(-1)(t)

Since $g(t,y,y′,y′′)$ continuous and differentiable, then is locally Lipschitz iff $t_{{0}}$ = kπ + π/2, $y_{{i}}=0$.

Thus, the solution of the IVP exists and is unique in $y_{{i}}$, for i = 1,2,3, when $t_{{0}}$ = kπ + π/2 and and $a_{{2}}$ = 0, for all k values in the set of integers.