If $\int^{x}_{0}2x(f(t))^2dt = \bigg(\int^{x}_{0}2f(x-t)dt\bigg)^2$ and $f(1) = 1,$ Then $f(x)$ is

Question

If $\displaystyle \int^{x}_{0}2x(f(t))^2dt = \bigg(\int^{x}_{0}2f(x-t)dt\bigg)^2$ and $f(1) = 1,$ Then $f(x)$ is

Try: Using $\displaystyle \int^{a}_{0}f(x)dx = \int^{a}_{0}f(a-x)dx$

We can write it as $$\displaystyle x\int^{x}_{0}f^2(t)dt = \bigg(\int^{x}_{0}2f(t)dt\bigg)^2 = 4 \bigg(\int^{x}_{0}f(t)dt\bigg)^2\;\;\;(*)$$

Using Leibnitz Rule of Differentiation

$$xf^2(x)+\int^{x}_{0}f^2(t)dt = 8\int^{x}_{0}f(t)dt\cdot f(x)$$

Again Differentiate w r to $x$

$$x\cdot 2f(x)\cdot f'(x)+f^2(x)+f^2(x)=8f^2(x)+8\int^{x}_{0}f(t)dt$$

Could some help me how to solve it, Thanks

Answer 1

After Leibniz, it should be:

$$xf^2(x)+\int^{x}_{0}f^2(t)dt = \color{red}{4}\int^{x}_{0}f(t)dt\cdot f(x)\;\;\;\;\;(**)$$

Let $$g(x) = \int^{x}_{0}f(t)dt$$

if we multiply $(**)$ with $x$ and use $(*)$ we get:

$$x^2f^2(x) + 4g^2(x) = 4xg(x)f(x)$$

implies $$xf(x)-2g(x)=0 \implies g(x)=xf(x)/2$$

Does this help?

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So we have $$xf(x) = 2\int^{x}_{0}f(t)dt$$ and after differentiation we get $$f(x)+xf'(x) = 2f(x)\implies {f'(x)\over f(x)} ={1\over x}$$

So $$(\ln f(x))' = (\ln x )' \implies \ln f(x) =\ln x+ c$$

Since $f(1)=1$ we get $c=0$ and so $\boxed{f(x)=x}$.

Plugging this in to staring formula we see that FE doesn't have a solution.

Answer 2

@Angle made a mistake in their answer, but the method works. You have

$$ x\int_0^x f^2 = 2\left(\int_0^x f \right)^2 $$

and upon differentiating

$$ xf^2 + \int_0^x f^2 = 4f\int_0^x f $$

Multiplying through by $x$ and substituting the original relation

$$ x^2f^2 + 2\left(\int_0^x f\right)^2 = 4xf\int_0^x f $$

Let $g = \int_0^x f$, then

$$ x^2(g')^2 + 2g^2 = 4xgg' $$

Divide through by $g^2$

$$ \left(x\frac{g'}{g} \right)^2 - 4x\frac{g'}{g} + 2 = 0 \implies x\frac{g'}{g} = 2 \pm \sqrt{2} $$

Solving this, we find $g = cx^{2\pm\sqrt{2}}$. Differentiating and applying the given initial condition, the solution shows itself

$$ f(x) = x^{1\pm\sqrt{2}} $$

You can check that both solutions satisfy the original integral equation