Find $g(x)$ from functional equation involving integrals

Question

If $g(x)$ is continuous function in $[0,\infty)$ satisfying $g(1)=1$ and $$\int _{0}^{x}{2xg^{2}(t)\,dt} =\left(\int _{ 0 }^{ x }{ 2g(x-t)\,dt} \right)^2$$Find $g(x)$.

I differentiated both sides w.r.t $x$. But what to do next? I'm stuck!

Answer 1

Here's some progress.

A simple change of variable shows that the original equality may be rewritten as $\displaystyle \forall x\geq 0, x\int _{0}^{x}{g^{2}(t)\,dt} =2\left(\int _{ 0 }^{ x }{g(t)\,dt} \right)^2$

Differentiating both sides of this equality yields $$\forall x\geq 0, x\int_0^x{g^2(t)} dt +(xg(x))^2=4xg(x)\int_0^x{g(t)} dt$$

Combining both equalities we have $$2\left(\int _{ 0 }^{ x }{g(t)\,dt} \right)^2 +(xg(x))^2=4xg(x)\int_0^x{g(t)} dt$$

Setting $F(x) = \int_0^x g(t)dt$, we have $$\forall x\geq 0, 2F^2(x)+x^2(F'(x))^2=4xF(x)F'(x)$$

which luckily factors as $$ (xF'(x)-(2-\sqrt2)F(x))(xF'(x)-(2+\sqrt2)F(x))=0$$

It would be nice to infer that either $\forall x, F'(x)-(2-\sqrt2)F(x)=0$ or $\forall x, F'(x)-(2+\sqrt2)F(x)=0$ which gives Christian Blatter's solution, but that seems difficult.

Answer 2

I don't know how to find all solutions, but it so happens that $$g(x):=x^{1+\sqrt{2}}\qquad(x\geq0)$$ satisfies the conditions.

How I got that? I remarked that the "Ansatz" $g(x):=x^\alpha$ would lead to a homogeneous identity in $x$. It then only remained to determine the "characteristic value" $\alpha$.