Let $f :\mathbb R\rightarrow\mathbb R$ be a continuously differentiable function such that both $f$ and its derivative $f'$ are Lipschitz for a constant $C > 0$.
Show that there exists an open interval $I$ containing $0$ such that the differential equation with boundary conditions $$ \left\{\begin{array}{l} u'' - \lambda f(u) = 0,\, \text{on }(0,1) \\ u(1) = u(0) = 0 \end{array}\right. $$ has a twice continuously differentiable solution $u_\lambda$, defined on $[0,1]$, for every $\lambda\in I$.
Now, I've had the idea of setting $F:(\lambda,x,y)\in\mathbb R^3\longmapsto \begin{bmatrix}y \\ \lambda f(x)\end{bmatrix}$ and considering the differential equation $$ X'(t) = F(\lambda, X(t)). $$
I can use the Cauchy-Lipschitz theorem to demonstrate solutions $X(t) = (u(t),u'(t))$ defined on an open neighbourhood of $0$ with $u(0) = 0$ exist for $\lambda$ in a neighbourhood of $0$, but I haven't been able to show solutions defined on $[0,1]$ with $u(1) = 0$ exist.
Moreover, I haven't used the hypothesis that $f'$ is Lipschitz. Did I have the right idea ? What am I missing here ?