Is there any way to express $f(x)$ in terms of $x$, or plot a graph of $f(x)$ with respect to $x$ from these equations? The equations available to me are:
$$f(x) = \int_{0}^{x}f(u) · f(2x-u)  du$$
$$ \int_{0}^{∞}f(x) dx = 1$$
$$f(0) = 0$$
I am unfamiliar with differential equations, and therefore did not know where to start for this.
Is it possible to state a relationship between $f(x)$ and $x$ purely in terms of a single variable, $x$? This can include an equation, or a graph.
 A: I do not think there is a solution unless one of the conditions is relaxed.
Let $u=2v$ in the RHS integral, then we have
$$\tag{1}
f(x)=2\int\limits_0^{x/2}dv \ f(2v)f(2(x-v))
$$
Let $f(2x)=g(x)$, then
$$\tag{2}
g(\frac{x}{2})=2\int\limits_0^{x/2}dv \ g(v)g(x-v)=\int\limits_0^xdv \ g(v)g(x-v)
$$
The second equality on the right is because the integrand is symmetric around the point $v=x/2$, which is done so that the integral is a Laplace convolution. Now take the Laplace transform (henceforth LT) of (2). This yields
$$\tag{3}
2G(2s)=G(s)G(s)
$$
Where $G(s)$ is the LT of $g(x)$. By definition of the LT, we have
$$\tag{4}
G(s)=\int\limits_0^\infty dx \ e^{-xs}g(x)\\
\therefore G(0)=\int\limits_0^\infty dx \ g(x) = \int\limits_0^\infty dx \ f(2x)=\frac{1}{2}\int\limits_0^\infty dx \ f(x)
$$
The condition $\int\limits_0^\infty dx \ f(x)=1$ means that we must have $G(0)=1/2$, while (3) demands $G(0)=0$ or $G(0)=2$ so we have reached an inconsistency, and there is no solution. One could argue that the LT does not exist, so this argument is not valid, but given that $\int\limits_0^\infty dx \ f(x)=1<\infty$, the LT converges at least for $\Re(s) \geq 0$.
Suppose that we relax the condition $\int\limits_0^\infty dx \ f(x)=1$. We then face the functional equation (3) for which a general solution is unlikely. Letting $G(s)=e^{cs}H(s)$, we see that $H(s)$ satisfies the same functional equation (3) as $G(s)$, so the solutions are not unique. Using this form, one solution is $G(s)=e^{-as}$, whose inverse LT is $g(x)=\delta(x-a)$, which is Dirac's delta. Finally, one (formal) solution for $f(x)$ that satisfies the integral equation and the condition $f(0)=0$ is
$$
f(x)=2\delta(x-a) \qquad,\qquad a>0
$$
