Find a function $f(x)$ such that$\int_0^txf(x)f(\sqrt{t^2-x^2})dx=\sin(t^2)$

Question

I am looking for a function that satisfies the integral equation $\int_0^txf(x)f(\sqrt{t^2-x^2})dx=\sin(t^2)$. It's some sort of convolution.

If we denote the above equation by $ f(t)*f(t)=\sin(t^2)$ I was also able to show that $f(t)*(2\cos(t^2))=t^2f(t).$

The definition of $ f(t)*g(t)=\int_0^txf(x)g(\sqrt{t^2-x^2})dx$.

I also got $f(t)*\dfrac{f'(t)}{t}+f(0)f(t)=2\cos(t^2)$ through differentiation ($t>0$). That's all.

Answer 1

$$\int_0^t f(x)f(\sqrt{t^2-x^2})xdx=\sin(t^2)$$ HINT :

Let $\quad f(x)=g(x^2)$ $$\int_0^t g(x^2)g(t^2-x^2)\frac{d(x^2)}{2}=\sin(t^2)$$ Let $\quad\begin{cases} x^2=X \\t^2=T\end{cases}$ $$\int_0^{\sqrt{T}} g(X)g(T-X)dX=2\sin(T)$$

Answer 2

Using the hint

$\int_0^t g(X)g(t^2-X)dX=2\sin(t^2)$ we can write $\mathcal{L} \left \{ g(t^2) *g(t^2) \right\}= \mathcal{L} \left \{ 2\sin(t^2 ) \right \}$.

Let $T=t^2$ and $ G(s)=\mathcal{L}\{g(T)\}$. Then $ G(s)G(s)=\mathcal{L} \left \{ 2\sin(T ) \right \}=\dfrac{2}{s^2+1}$.

Therefore $ G(s) =\dfrac{ \sqrt{2}}{\sqrt{s^2+1}} $ and $g(T)=\sqrt{2}\mathcal{L}^{-1} \left \{ \dfrac{1}{\sqrt{s^2+1}} \right \}=\sqrt{2}J_0(T) $.

Thus the answer is $f(t)=g(t^2)=\sqrt{2}J_0(t^2).$