Direct method: differential equation with conditions on the first derivative

Question

I would like to solve the following problem \begin{equation} u'' =u^3+\sin(x)^6 \quad u'(3)=6 \quad u'(6)=3. \end{equation} We can think the equation as a Euler-Lagrange equation associated to the Lagrangian \begin{equation} L(x,s,p)=\frac{1}{2}p^2+\frac{1}{4}s^4+s\sin^6(x) \end{equation} and we have the functional \begin{equation} F(u)=\int_3^6 (\frac{1}{2}(u'(x))^2+\frac{1}{4}u^4(x)+u(x)\sin^6(x))dx \end{equation} My idea was to work with in the sobolev space $H^{1,2}(3,6)$, wich I will call $\mathbb{X}$. I have choose as convergence notion the weak $L^2$ convergence on the derivatives and a strong convergence for the functions. I proved that I can use some form of the Weierstrass Theorem to found a minimun(in $\mathbb{X}$), my question is how can I treat the conditions on the first derivative.

Answer 1

I solved by considering a different integral operator. \begin{equation} L(x,s,p)=\frac{1}{2}p^2+(x-9)p+\frac{1}{4}s^4+s(\sin(x)^6+1) \end{equation} The Euler-Lagrage equation is the same, but while integrating by part we obtain the wanted constraint for the derivative.