Show that the integral equation
$$
\phi(x, y)+\int_x^y \mathrm{~d} t \int_x^t \phi(s, t) \mathrm{d} s=\frac{1}{2}(f(x)+f(y)) \quad(x \in \mathbb{R}, x \leqslant y)
$$
is equivalent to the initial value problem
$$
\begin{gathered}
\phi_{x y}(x, y)-\phi(x, y)=0 \quad(x \in \mathbb{R}, x \leqslant y) \\
\phi(x, x)=f(x), \quad \phi_x(x, x)-\phi_y(x, x)=0 \quad(x \in \mathbb{R}) .
\end{gathered}
$$
Let $(K \phi)(x, y)=-\int_x^y \mathrm{~d} t \int_x^t \phi(s, t) \mathrm{d} s(x \in \mathbb{R}, x \leqslant y \leqslant x+a)$ and show that
$$
\left(K^n \phi\right)(x, y)=(-1)^n \int_x^y \mathrm{~d} t \int_x^t \frac{(y-t)^{n-1}(s-x)^{n-1}}{((n-1) !)^2} \phi(s, t) \mathrm{d} s .
$$
By showing that, as a map from $L_2(\mathbb{R} \times[0, a])$ to itself, $\left\|K^n\right\| \leqslant a^n / n$ ! deduce that (3.29) has a unique solution $\phi$. Deduce that the solution of (3.29) is
$$
\phi(x, y)=\frac{1}{2}(f(x)+f(y))-\frac{1}{2} \int_x^y \mathrm{~d} t \int_x^t J_0\left(2(y-t)^{\frac{1}{2}}(s-x)^{\frac{1}{2}}\right)(f(s)+f(t)) \mathrm{d} s
$$
where $J_0(z)=\sum_{n=0}^{\infty}(-1)^n \frac{z^{2 n}}{4^n(n !)^2}$ (the Bessel function of the first kind of order zero).