Solving $x''(t)=2x(t)^3$ with $x(0)=x(1)=0$.

Question

I have to find the minimum of the functional

$$ J(x) = \frac{1}{2} \int_0^1 \left(x'(t) +x^2(t) \right)^2\, dt $$

I have calculated the Euler-Lagrange equation, but I can't solve the corresponding differential equation: $$x''(t)=2x(t)^3$$ with the conditions $x(0)=x(1)=0$.

Can someone explain me please how to solve it?

Answer 1

The integrand function is $$f(x,x',t)=\left(x'(t) +x^2(t) \right)^2$$ Euler-Lagrange equation $$\frac {\partial f}{\partial x}-{\frac {\operatorname {d} }{\operatorname {d} t}}\left({\frac {\partial f}{\partial {x'}}}\right)=0$$ leads to the ODE $$4 x(t) \left(x'(t)+x(t)^2\right)-2 \left(x''(t)+2 x(t) x'(t)\right)=0$$ expanding and simplifying $$2 x(t)^3-x''(t)=0;\;x(0)=x(1)=0$$

the unique function which satisfies the DE and the initial conditions is $x(t)=0$.

Answer 2

$$\frac{d^2x}{dt^2}=2x^3 \implies 2\frac{dx}{dt}\frac{d^2x}{dt^2}=4x^3 \frac{dx}{dx}$$ $$\frac{d}{dt}\left ( \frac{dx}{dt}\right)^2 =2x^3 \frac{dx}{dt}.$$ Integrating w.r.t. $t$, we get $$\int \frac{d}{dt}\left ( \frac{dx}{dt}\right)^2 dt =\int 4 x^3 dx \implies \left(\frac{dx}{dt}\right)^2 = x^4+A \implies \frac{dx}{dt}=\sqrt{x^4+A}$$ $$\implies \int \frac{dx}{\sqrt{x^4+A}}=\int dt +B=t+B.$$