# Integral-differential equation for forward rates

I am struggling in this question:

Let $P(t,T)$ denote the price of a zero-coupon bond (with marturity at time $T$) at time $t \in [0,T]$.

As usual, at time $t$ for maturity $T$, the forward rate is defined by

$$f(t,T)= - \frac{\partial}{\partial T} \log P(t,T)$$.

Consider a short interest rate process $(r_t)$ satisfying the following dynamics: $$dr_t = a(r_t) \,dt + b(r_t) \, dW_t$$ for two smooth functions $a$ and $b$.

Let the function $G: [0,T] \times \mathbb{R} \rightarrow \mathbb{R}$ satisfy the following integral-differential equation: $$\frac{\partial G}{\partial t} (t,r) = a(r) \frac{\partial G}{\partial r} (t,r) + \frac{ b(r)^2}{2} \frac{{\partial}^2 G}{\partial r^2} (t,r) - b(r)^2 \frac{\partial G}{\partial r} (t,r) \int_0^t \frac{\partial G}{\partial r} (s,r) \,ds,$$ with initial condition $G(0,r)=r$.

We want to show that there is no arbitrage if the forward rate function is defined by $f(t,T) = G(T-t, r_t)$.

The main problem I encounter is the fact that $\frac{1}{G(t,r)} b(r)^2 \frac{\partial G}{\partial r} (t,r) \int_0^t \frac{\partial G}{\partial r} (s,r)\,ds$ isn't a function of $r$ only. Therefore, I don't know how to apply Feynman-Kac in this situation.

Any suggestions on how to transform this to Feynman-Kac?

Integrate the integral-differential equation from 0 to $T-t$, we obtain that \begin{align*} G(T-t, r) - r &= a(r) \int_0^{T-t}\frac{\partial G}{\partial r} (s,r)ds + \frac{ b(r)^2}{2}\int_0^{T-t} \frac{{\partial}^2 G}{\partial r^2} (s,r)ds\\ &\qquad - b(r)^2 \int_0^{T-t}\frac{\partial G}{\partial r} (u,r) \int_0^u\frac{\partial G}{\partial r} (s,r) \,ds du\\ &=a(r) \int_0^{T-t}\frac{\partial G}{\partial r} (s,r)ds + \frac{ b(r)^2}{2}\int_0^{T-t} \frac{{\partial}^2 G}{\partial r^2} (s,r)ds\\ &\qquad - \frac{ b(r)^2}{2}\left(\int_0^{T-t}\frac{\partial G}{\partial r} (s,r) \,ds\right)^2.\tag{1} \end{align*}
Let $$\widehat{P}(t,T) = e^{-\int_0^{T-t} G(s, r_t)ds}.$$ Then \begin{align*} \frac{\partial \widehat{P}}{\partial t} &= \widehat{P}\, G(T-t, r_t),\\ \frac{\partial \widehat{P}}{\partial r} &=-\widehat{P}\int_0^{T-t}\frac{\partial G(s, r)}{\partial r}ds,\\ \frac{\partial^2 \widehat{P}}{\partial r^2}&=\widehat{P}\left(\int_0^{T-t}\frac{\partial G(s, r)}{\partial r}ds \right)^2 - \widehat{P}\int_0^{T-t}\frac{\partial^2 G(s, r)}{\partial r^2}ds. \end{align*} Moreover, from $(1)$, \begin{align*} &\ \frac{b(r)^2}{2}\frac{\partial^2\widehat{P}}{\partial r^2}+a(r)\frac{\partial \widehat{P}}{\partial r}+\frac{\partial \widehat{P}}{\partial t}-r\widehat{P}\\ =&\ \widehat{P}\Bigg[\frac{b(r)^2}{2}\left(\int_0^{T-t}\frac{\partial G(s, r)}{\partial r}ds \right)^2 - \frac{b(r)^2}{2}\int_0^{T-t}\frac{\partial^2 G(s, r)}{\partial r^2}ds \\ &\qquad - a(r)\int_0^{T-t}\frac{\partial G(s, r)}{\partial r}ds + G(T-t, r_t) -r\Bigg]. \end{align*} That is, $$\frac{b(r)^2}{2}\frac{\partial^2\widehat{P}}{\partial r^2}+a(r)\frac{\partial \widehat{P}}{\partial r}+\frac{\partial \widehat{P}}{\partial t}-r\widehat{P}=0.$$ Now, you can apply the Feynman-Kac formula to obtain that \begin{align*} \widehat{P}(t, T) = E\left(e^{-\int_t^T r_s ds} \mid \mathcal{F}_t \right). \end{align*} Therefore, $\widehat{P}(t, T) = P(t, T)$ is the zero-coupon bond price. Consequently, \begin{align*} f(t, T) &= - \frac{\partial}{\partial T} \ln \widehat{P}(t,T)\\ &=G(T-t, r_t). \end{align*}