Suppose we have the following PDE:
$$u(x, t) = F(4 x - t) + G(2 x - t),$$
where $F$ and $G$ are arbitrary twice differentiable functions. Let $x \in (- \infty, + \infty)$ and $t > 0$ with ICs
$$u(x, 0) = 0, \quad u_t(x, 0) = f(x).$$
Solve for $u(x, t)$.
My attempt: From the first IC we get $F(4 x) = - G(2 x)$ and $F'(4 x) = - G'(2 x)$, where $\frac{d}{d x} F(4 x) = 4 F'(4 x)$ by the chain rule. Then, the second IC gives
$$u_t(x, 0) = - F'(4 x) - G'(2 x) = f(x).$$
Integrate the above equation from $0$ to $x$ yields:
$$\int_0 ^ x f(\tau) \; d \tau = \int_0 ^ x - \frac{d}{d (4 x)} F(4 x) - \frac{d}{d (4 x)} G(2 x) \; d x$$
in which we have $\int_0 ^ x \frac{d}{d (4 x)} F \; d x = \frac{1}{4} F$ and so,
$$\int_0 ^ x f(\tau) \; d \tau = \int_0 ^ x - \frac{1}{4} F(4 x) - \frac{1}{2} G(2 x) \; d x + C,$$
where $C = \frac{1}{4} F(0) + \frac{1}{2} G(0)$.
Thus, we have
\begin{align} F(4 x) & = 4 \int_0 ^ x f(\tau) \; d \tau - 4 C, \\ G(2 x) & = - 4 \int_0 ^ x f(\tau) \; d \tau + 4 C, \\ \end{align}
Therefore,
$$u(x, 0) = 4 \int_{- x} ^ x f(\tau) \; d \tau.$$
Here I'm stuck. I'm not sure how to figure out the bounds for $t$. I tried the bounds $x - t$ and $- x - t$ but I couldn't verify the solution (I used simply $f(x) = x$).
Would appreciate any help or hints. Thank you.