Euler-Lagrange Equation has no solution? I've been asked to compute the Euler-Lagrange equation and second variation of the functional $$I[y]=\int_{a}^{b}(y'^2+y^4)dx$$
with boundary conditions $y(a)=\alpha$, $y(b)=\beta$. It's easy to see that $$I[y+\delta y]=I[y]+\int_{a}^{b}\delta y(4y^{3}-2y'') dx+\int_{a}^{b}(6y^{2}\delta y^{2}+\delta y'^{2})dx$$
So the Euler-Lagrange equation integrates to give $y^{4}-y'^{2}=k$, where $k$ is a constant of integration. We're then asked to solve this equation when $\alpha=\beta=0$. The equation is separable, but to my shame I can't do the integration (I think it involves special functions), so I looked for a different way. Completing the square on $I$ gives $$\int_{a}^{b}(y'^2+y^4)dx=\int_{a}^{b}(y'+y^2)^{2}dx-\int_{a}^{b}2y^{2}y'dx$$
But the final term is just $\left[\frac{2}{3}y^{3}\right]^{y=0}_{y=0}=0$, and the other two integrals are non-negative. The only way to extremise the RHS (I think) is to minimise it, and besides the second variation is non-negative, so we make the RHS zero by allowing $y'=-y^{2}$. But then the LHS forces both $y'=0$ and $y=0$ on all of $[a,b]$. Does this mean that the only solution is the zero function? Then, since $\delta^{2}I=\int_{a}^{b}\delta y'^{2}dx$, and this is positive unless $\delta y$ is constant (and hence $0$, by the boundary condition), do we have the zero function as the actual solution?  
Sorry if this sounds a little incoherent, when I started writing the answer I forgot to consider $y(x)=0$ and found the other solutions to $y'=-y^{2}$, which can't possibly satisfy the boundary conditions. In fact, looking at it now, I'm starting to think that even completing the square was unnecessary.
 A: OP has essentially already proven that there is only a trivial solution $y\equiv 0$. See also the answer by John Ma. Now OP is pondering Why? Perhaps he would appreciate a bit of physics intuition: The Lagrangian $L=T-V$ describes a non-relativistic point particle in 1D in an inverted (=unstable) quartic potential $V \propto -y^4$, where $x$ and $y$ play the roles of time and position, respectively. The particle is initially and finally at the unstable equilibrium $y=0$. Intuitively (and/or by examining the equation of motion), if the particle leaves/crosses the unstable equilibrium $y=0$, then it would never return. Hence it never left in the first place. In other words, it sits still at the unstable equilibrium $y\equiv0$. 
A: The Euler-Lagrange equation 
$$y^4 - (y')^2 = k$$
is actually not too hard to solve once you know $y(a) = y(b) = 0$. Note that $y(a) = 0$ imply $k \le 0$. If $k=0$, then $(y')^2 = y^4$ and you can check that the only solution (using $y(a) =0$ and $y'' - 4y^3 = 0$) is $y(t) = 0$. 
If $k<0$, then 
$$(y')^2 = \sqrt{y^4 - k} \ge \sqrt{-k} >0 . $$
Then $y' \ge (-k)^{\frac 14} $ or $y' \le -(-k)^{\frac 14}$. In both cases, $y(b) = 0$ is not possible. Thus $y(t) = 0$ is the only solution to the Euler-Lagrange equation. 
