$X_t$ bessel square process which satisfies $$\mathop{dx_t}= 2(a+1) \mathop{dt} +2 \sqrt{x_t} \mathop{dB_t}$$ and $u$ is a function which satisfies $x^2 u'' +x u' -u(a^2 + b x^{2p+2})= 0$. How can I show the function:
$$u\left(\sqrt {X_t}\right) ( X_t)^{-a/2} \exp\left(-b/2 \cdot\int_0^tX_s^p \mathop{ds}\right)$$ is a local martingale.
I tried with ito formula but finding derivatives of product of three terms comes with many unmanagable terms. Any suggestions!
Let's introduce some notation first:
$$\begin{align*} Y_t &:= \int_0^t X_s^p \, ds \\ Z_t &:= \underbrace{u(\sqrt{X_t}) \cdot X_t^{-a/2}}_{=:f(X_t)} \cdot \underbrace{\exp \left(- \frac{b}{2} Y_t \right)}_{=:g(Y_t)} \end{align*} $$
By the definition of $X_t$, we have $\langle X,Y \rangle_t=0$ which means that the mixed derivatives in the Itô formula will vanish. Moreover, $\langle Y \rangle_t =0$. Hence,
$$\begin{align*} Z_t - Z_0 &= \int_0^t f'(X_s) \cdot g(Y_s) \, dX_s + \int_0^t f(X_s) \cdot g'(Y_s) \, dY_s + \frac{1}{2} \int_0^t f''(X_s) \cdot g(Y_s) \, d\langle X \rangle_s \\ &=: I_1+I_2+I_3 \end{align*}$$
We write
$$I_j = \int_0^t \varphi_j(s) \, dB_s + \int_0^t \psi_j(s) \, ds, \qquad j=1,2,3.$$
Since we want to prove that $(Z_t)_t$ is a local martingale, it suffices to show $$\psi_1(t)+\psi_2(t)+\psi_3(t)=0$$ for any $t \geq 0$. Using that $\langle X \rangle_t = 4X_t \, dt$ we find by some straight-forward calculations
$$\begin{align*}\psi_1(t) &= f'(X_t) g(Y_t) 2(a+1) \\ &= \left(u'(\sqrt{X_t}) X_t^{-a/2-1/2} -a \, u(\sqrt{X_t}) X_t^{-a/2-1} \right) (a+1) \cdot g(Y_t) \\ \psi_2(t) &= f(X_t) g'(Y_t) \, X_t^p \\ &= - \frac{b}{2} f(X_t) g(Y_t) X_t^p \\ \psi_3(t) &= \frac{1}{2} f''(X_t) g(Y_t) 4X_t \\ &= \bigg( \frac{1}{2} u''(\sqrt{X_t}) X_t^{-a/2-1} + \left(- \frac{a}{2}-1 \right) u'(\sqrt{X_t}) X_t^{-a/2-3/2} \\ &\quad - \frac{a}{2} u'(\sqrt{X_t}) X_t^{-a/2-3/2} - a \left( - \frac{a}{2}-1 \right) u(\sqrt{X_t}) X_t^{-a/2-2} \bigg) g(Y_t) X_t \, dt \end{align*}$$
Hence, by the definition of $f$ and $g$,
$$\begin{align*} &\psi_1(t)+\psi_2(t)+\psi_3(t) \\ &= u(\sqrt{X_t}) X_t^{-a/2} g(Y_t) \cdot \left( - a \cdot (a+1) \, \frac{1}{X_t}- a \, \left( - \frac{a}{2}-1 \right) \frac{1}{X_t}- \frac{b}{2} X_t^p \right) \\ &\quad + u'(\sqrt{X_t}) X_t^{-a/2} g(Y_t) \cdot \left( (a+1) \, \frac{1}{\sqrt{X_t}} + \left( - \frac{a}{2} - \frac{1}{2} \right) \frac{1}{\sqrt{X_t}} - \frac{a}{2} \frac{1}{\sqrt{X_t}}) \right) \\ &\quad + u''(\sqrt{X_t}) X_t^{-a/2} g(Y_t) \\ &= \frac{1}{2} g(Y_t) X_t^{-a/2} \left( u''(\sqrt{X_t}) + u'(\sqrt{X_t}) X_t^{-1/2} - u(\sqrt{X_t}) \cdot \left( a^2 X_t^{-1} +b X_t^p \right) \right) \tag{1} \end{align*}$$
By assumption, $u$ satisfies
$$X_t \cdot u''(\sqrt{X_t}) + \sqrt{X_t} u'(\sqrt{X_t}) - u(\sqrt{X_t}) \cdot (a^2+b X_t^{p+1}) = 0 $$
Dividing by $X_t$ yields
$$u''(\sqrt{X_t}) + \frac{1}{\sqrt{X_t}} u'(\sqrt{X_t}) - u(\sqrt{X_t}) \cdot (a^2 X_t^{-1} +b X_t^{p}) = 0 \tag{2}$$
Combining $(1)$ and $(2)$ we find
$$\psi_1(t)+\psi_2(t)+\psi_3(t)=0.$$
This finishes the proof.