I have to find $u$ minimizing $\int_0^1 F$ with $F(x,u(x),u'(x)) = (1-(u'(x))^2)^2+(u(x))^2$ with $u(0) = 0$ and $u(1) = 1$.
I'm relatively new to CoV and got told i should try Euler-Lagrange-Equation which gives me $2u(x) + 4u''(x) - 12{(u'(x))}^2 \cdot u''(x) = 0$.
But then I don't have a clue how to solve this ODE. Can you please tell me if this is the right way and I should learn solving (even ugly) ODEs?
Thank you very much in advance, JohnDoe746
You can use approximation methods such direct variational method if there is no exact solution. Let $\hat u(x)$ be the approximate solution $$\hat u(x)=a\sin\bigg(\frac{\pi x}2\bigg)$$ which satisfies the boundary conditions $\hat u(0)=0$ and $\hat u(1)=1$. Then $$I=\int_0^1\bigg((1-\hat u'(x)^2)^2+u(x)^2\bigg)dx=\int_0^1\bigg(\bigg(1- \bigg(\frac{a\pi}2 \cos\frac{x\pi}2 \bigg)^2\bigg)^2+\bigg(a\sin\frac{\pi x}2\bigg)^2\bigg)dx$$
You can use Wolfram Alpha to find the solution of integral such as $$I=\frac{3\pi^4}{128}a^4-\frac{\pi^2-2}4a^2+1$$ To minimize the integral wrt to $a$ $$\frac{d\,I}{d\,a}=\frac{3\pi^4}{32}a^3-\frac{\pi^2-2}2a=0$$ which has the solution $$a_1=0\qquad a_{2,3}=\pm \frac 4{\pi^2 \sqrt{3}}\sqrt{\pi^2-2}$$ If you reevaluate the integral with $$I(0)=1$$ $$I\bigg(\pm \frac 4{\pi^2 \sqrt{3}}\sqrt{\pi^2-2}\bigg)=0.58$$ which means that you can approximate with $$\hat u(x)=\bigg(\pm \frac 4{\pi^2 \sqrt{3}}\sqrt{\pi^2-2}\bigg)\sin\bigg(\frac{\pi x}2\bigg)$$ to minimize the integral.