Question about the basic existence theorem for linear ordinary differential equation!

Question

Basic existence theorem states that

Suppose we have an $n$-th order ordinary differential equation

$$a_0(x)\frac{d^ny}{dx^n}+a_1(x)\frac{d^{n-1}y}{dx^{n-1}}+...+a_{n-1}(x)\frac{dy}{dx}+a_n(x)y=F(x)$$

The differential equation is said to be homogeneous if the $F(x)=0$

For this basic existence theorem to hold we have the following two hypothesis

1. The coefficients such as $a_0,a_1,a_2...a_n$ & the non homogeneous term $F$ if it exists must be continuous over certain real interval defined as $a\leq x\leq b$. $a_0 \ne 0$

2. Let $x_0$ be any point over $a \leq x \leq b$ and $c_0,c_1,...c_{n-1}$ be arbitrary real constant!

Conclusion there exist a unique solution such that

$f(x_0)=c_0,\ f'(x_0)=c_1,\ ...,\ f^{(n-1)}(x_0)=c_{n-1}$. Hence this solution is defined over $[a,b]$.

My question is

1. why must the coefficients and the non-homogeneous term posses continuity? What will happen if it does not possess continuity? Secondly, does that mean we can ignore differentiability?

2. Why there is a unique solution when the first assumption holds true?

3. I notice that the second part happens until $f_{(n-1)}(x_0)=c_{n-1}$. How about $f^{(n)}$. Does $n$-th derivative has no effect on the theorem at all?

Can someone shed some light into this matter?

Answer 1

1. If there is no continuity, then there is no existence theorem. The existence theorem of Peano requires continuity in $x$ and $y$.

2. Because linear functions are always automatically Lipschitz continuous. If the ODE is linear in $y$ (and its derivatives) then it satisfies the Lipschitz condition of the Picard-Lindelöf theorem.

3. The higher derivatives are determined by the differential equation, if they exists. For the $n$th derivative you get directly $$ y^{(n)}(x)=\frac1{a_0(x)}(F(x)-a_1(x)y^{(n-1)}(x)-…-a_n(x)y(x)) $$