Let's take the following three SDEs:
$$dr=u(r,t)dt + w(r,t)dX$$ $$u(r,t)=a(t)-br$$ $$w(r,t)=c$$
where $b$ and $c$ are constants and $a(t)$ an arbitrary function of time $t$.
If Zero Coupon Bond $Z(r,T,T)=1$ for this model has the form
$$Z(r,t,T)=e^{(A(t,T)-B(t,T)r)}$$
How do you find $A$ and $B$?
I have derived the PDE for this model using no arbitrage condition. Substituting this in the PDE is not giving the right answers.
Here is a solution without using the PDE technique, which is preferred as we do not need to assume the affine form of a zero-coupon price from the start.
we assume that, under the risk-neutral measure, \begin{align*} dr_t = (\theta(t)-a r_t) dt + \sigma dW_t, \end{align*} where $a$ and $\sigma$ are constants, $a(t)$ is a deterministic function, and $W_t$ is a standard Brownian motion. We seek to compute the zero-coupon bond price defined by \begin{align*} P(t, T) &= E\left(e^{-\int_t^T r_s ds} \mid \mathcal{F}_t \right), \end{align*} where $\mathcal{F}_t$ is the information set up to time $t$. Note that \begin{align*} d\left(e^{at} r_t\right) &= be^{at}r_t dt + e^{at} dr_t\\ &=\theta(t)e^{at} dt + \sigma e^{at} dW_t. \end{align*} Then, for $s \geq t \geq 0$, \begin{align*} e^{as} r_s = e^{at} r_t + \int_t^s \theta(u)e^{au} du + \int_t^s \sigma e^{au} dW_u. \end{align*} That is, \begin{align*} r_s = e^{-a(s-t)} r_t + \int_t^s \theta(u)e^{-a(s-u)} du + \int_t^s \sigma e^{-a(s-u)} dW_u. \end{align*} We then have the integral \begin{align*} &\ \int_t^T r_s ds \\ =&\ r_t \int_t^T e^{-a(s-t)} ds + \int_t^T\!\!\!\!\int_t^s \theta(u)e^{-a(s-u)} du ds + \int_t^T\!\!\!\!\int_t^s\sigma e^{-a(s-u)} dW_u ds\\ =&\ \frac{1}{a}\Big(1-e^{-a(T-t)} \Big) r_t + \int_t^T\!\!\!\!\int_u^T \theta(u)e^{-a(s-u)} ds du + \int_t^T\!\!\!\!\int_u^T \sigma e^{-a(s-u)} ds dW_u\\ =&\ \frac{1}{a}\Big(1-e^{-a(T-t)} \Big) r_t + \int_t^T\!\! \frac{\theta(u)}{a}\Big(1-e^{-a(T-u)} \Big)du + \int_t^T \!\!\frac{\sigma}{a}\Big(1-e^{-a(T-u)} \Big)dW_u. \end{align*} Let \begin{align*} B(t, T) = \frac{1}{a}\Big(1-e^{-a(T-t)} \Big). \end{align*} Then, \begin{align*} \int_t^T r_s ds &= B(t, T) r_t + \int_t^T \theta(u) B(u, T) du + \int_t^T \sigma B(u, T) dW_u. \end{align*} Moreover, the zero-coupon bond price is then given by \begin{align*} P(t, T) &= E\left(e^{-\int_t^T r_s ds} \mid \mathcal{F}_t \right)\\ &=\exp\left(-B(t, T) r_t - \int_t^T \theta(u) B(u, T) du + \frac{1}{2}\int_t^T \sigma^2 B(u, T)^2 du\right). \end{align*} Note that \begin{align*} \int_t^T \sigma^2 B(u, T)^2 du &= \frac{\sigma^2}{a^2}\int_t^T \left(1 - 2e^{-a(T-u)} + e^{-2a(T-u)}\right) du\\ &=\frac{\sigma^2}{a^2}\left(T-t-\frac{2}{a}\Big(1-e^{-a(T-t)}\Big) +\frac{1}{2a} \Big(1-e^{-2a(T-t)}\Big) \right)\\ &= \frac{\sigma^2}{a^2}\left(T-t -\frac{1}{2a}\Big(1-e^{-a(T-t)}\Big)^2-\frac{1}{a}\Big(1-e^{-a(T-t)}\Big)\right)\\ &= -\frac{\sigma^2}{a^2}\big(B(t, T) -T+t\big)-\frac{\sigma^2}{2a}B(t, T)^2. \end{align*} Then \begin{align*} P(t, T) &= A(t, T) e^{-B(t, T) r_t}, \end{align*} where \begin{align*} A(t, T) &= \exp\left(- \int_t^T \theta(u) B(u, T) du -\frac{\sigma^2}{2a^2}\big(B(t, T) -T+t\big)-\frac{\sigma^2}{4a}B(t, T)^2\right). \end{align*}
See http://www.math.nyu.edu/~benartzi/Slides10.3.pdf for another derivation using the PDE approach.
