Write down the Euler-Lagrange equation for the following problem: $$ \min { \int_0^1 f(u(x),u'(x))dx; \quad u(0)=0, u(1)=1 } $$ Solve this Euler-Lagrange equation when $f(u,z) = z^2+u^2$
I have gotten the Euler-Lagrange into the form: $df/du=d/dx(df/du')$ And differentiated $f(u,z) = z^2+u^2$, But cannot proceed much further.
The Euler-Lagrange equation for this is exactly as you pointed out $$ \frac{d f}{d u} - \frac{d}{dx} \frac{d f}{d u'} = 0 $$ Now note that $$ \frac{d f}{d u} = 2 u \\ \frac{d f}{d u'} = 2 u' \\ \frac{d}{dx} \frac{d f}{d u'} = 2 u'' $$ So we conclude that our system of equations is $$ \left\{ \begin{array}{l} 2 u - 2u'' = 0 \\ u(0) = 0, u(1) = 1 \end{array} \right. $$ And of course we can simplify this to solving (since a solution to the above system is a solution to the bottom and vice versa). $$ \left\{ \begin{array}{l} u - u'' = 0 \\ u(0) = 0, u(1) = 1 \end{array} \right. $$
Note: An important note I want to put here that you might be struggling with (I know I did) is that $\frac{d f}{d u}$ means to take the derivative of $f$ with respect to the first parameter - to me it was misleading that we were taking the derivative of a function with respect to another function. Likewise $\frac{d f}{d u'}$ means taking the derivative with respect to the second parameter.
Now as for solving this equation consider a solution of the form $$ u(x) = ae^{b x} $$ So that we get the constraint that $$ a e^{b x} - ab^2 e^{b x} = 0 \implies 1 - b^2 = 0 \implies b^2 - 1 = 0 $$ Thus we must have $$ b = \frac{\pm \sqrt{4}}{2} = \pm 1 $$ Since this is a homogenous equation we can add together these solutions and it will still be a solution so that we get a general equation of $$ u_g(x) = a_1 e^{x} + a_2 e^{-x} $$ Finally we incorporate boundary conditions to get that $$ u_g(0) = a_1 + a_2 = 0 \\ u_g(1) = a_1 e + a_2 e^{-1} = 1 $$ Which we can solve by seeing $$ -a_1 = a_2 \implies a_1 e - a_1 e^{-1} = 1 \implies a_1 = \frac{1}{e - e^{-1}} = \frac{1}{2} \mathrm{csch}(1) $$ So finally we get $$ u_s = \frac{1}{2} \mathrm{csch}(1) e^{x} - \frac{1}{2} \mathrm{csch}(1) e^{-x} $$
