Proving that a differential equation has unique solution. Let $P,Q,f:[-1,1]\to\mathbb{R}$ continuous and $a,b\in\mathbb{R}$. Then I want to prove that the IVP
$$\begin{cases} u''(x)+P(x)u'(x)+Q(x)u(x)=f(x)\\u(0)=a,\qquad u'(0)=b\end{cases}$$
has a unique solution in a neighborhood of x=0.
To this end I have to show that the function $F(x,y,t)$ is lipschitz in the first variables and the constant independent of the parameter $t$.
(This comes when this problem is converted to solve the integral equation and invoke the condition that the operator $\Phi(x)(t)=x_0 + \int_0^tF(x(s),s)ds$ has only one fixed point and the use the Banach fixed point theorem)
So I chose $F(x,y,t)=f(x)-P(x)-Q(x)$ then:
$$|F(x,y,t)-F(u,v,t)|=|(f(x)-f(u))+(P(u)-P(x))+(Q(u)-Q(x))|$$
Now, by continuity of $f,P,Q$ in a compact set we have that they are uniformly continuous so we pick $\delta = \{ \delta_1,\delta_2, \delta_3 \}$ so if $|x-y|<\delta$ we get 
$$|F(x,y,t)-F(u,v,t)|< \epsilon<\epsilon |x-y|$$
but the thing is that I don't know how to extend this property to all $x,y \in [-1,1]$
Can someone help me to fix this problem or provide another way to prove this Lipschitz condition ?  because I don't know other way to proceed.
Thanks a lot in advance.
 A: Expanding the comment about transforming second order equation to the system:
$$ \frac{dx}{dt} = y, \; \frac{dy}{dt} = f(t) - P(t)y- Q(t)x . $$
RHS is a vector now. Its first component is smooth w.r.t. all variables. 
The second component $F(x, y, t) = f(t) - P(t)y- Q(t)x$ is continuous w.r.t. $t$ and continuously differentiable in variables $x$ and $y$, because   $ \frac{\partial F}{\partial x} = -Q(t), \frac{\partial F}{\partial y} = -P(t)  $ and both are continuous functions. Hence $F(x, y, t)$ is Lipschitz in variables $x$ and $y$. Therefore, we can apply existence and uniqueness theorem in this case.

If function $F(x)$ is continuously differentiable w.r.t. $x$, then it's Lipschitz continuous w.r.t. to them

Since $F(x)$ is continuously differentiable with, we can write $F(x_1) - F(x_2)$ 
as 
$(x_1 - x_2) \cdot F'_{x} (x_1 + \theta (x_2 - x_1))$ 
where 
$\theta \in \lbrack 0, 1 \rbrack$. 
From here we have an estimation: 
$$\vert F(x_1) - F(x_2)\vert = \vert (x_1 - x_2) \cdot F'_{x} (x_1 + \theta (x_2 - x_1)) \vert = \vert x_1 - x_2 \vert \cdot \vert F'_{x} (x_1 + \theta (x_2 - x_1)) \vert \leqslant C \cdot \vert x_1 - x_2 \vert,$$
where $C$ is the supremum of $F'_x (x)$ in some neighbourhood of point $x_1$ (it exists for compact neighbourhoods because $F'_x$ is continuous).
