Let us suppose we want to solve a general SDE given as

[$]dX(t)=\mu(X) dt + \sigma(X) dz(t)[$] Equation (1)

The fokker planck equation of this SDE is given as

[$]\frac{\partial p(X,t)}{\partial t} = -\frac{\partial [\mu(X) p(X,t)]}{\partial X} + .5 \frac{\partial^2 [{\sigma(X)}^2 p(X,t)]}{\partial X^2}[$] Equation(2)

First a very brief primer on derivatives of normal and derivatives of the SDE variable.

I denote the standard normal density as p(Z)

We want to convert derivatives of the SDE density of X(t) into derivatives of standard normal density. Here are the main equations

[$]p(Z)=\frac{1}{\sqrt{2 \pi}} exp[-.5 Z^2][$] Equation(3)

derivatives of standard normal density are given in terms of hermite polynomials.

[$]\frac{\partial p(Z)}{\partial Z}=-Z p(Z)[$] Equation(4)

[$]\frac{\partial^2 p(Z)}{\partial Z^2}=(Z^2-1) p(Z)[$] Equation (5)

now we want to convert the density of X(t) and its derivatives in terms of density of Z in which we also use above equations (4-5). In the density of X, each value of X(t) is associated with a particular value of Z through constant CDF points on corresponding densities. So we calculate the density and derivatives of X(t) in terms of Z.

[$]p(X)=p(Z) |\frac{\partial Z}{\partial X}|[$] Equation(6)This follows from the standard change of variables in a density.

[$]\frac{\partial p(X)}{\partial X}=\frac{\partial p(Z)}{\partial Z} {(\frac{\partial Z}{\partial X})}^2+p(Z) \frac{\partial^2 Z}{\partial X^2} =Z p(Z){(\frac{\partial Z}{\partial X})}^2+p(Z) \frac{\partial^2 Z}{\partial X^2} [$] Equation(7)

[$]\frac{\partial^2 p(X)}{\partial X^2}=\frac{\partial^2 p(Z)}{\partial Z^2} {(\frac{\partial Z}{\partial X})}^3+3 \frac{\partial p(Z)}{\partial Z} \frac{\partial Z}{\partial X} \frac{\partial^2 Z}{\partial X^2}+p(Z) \frac{\partial^3 Z}{\partial X^3}[$]

[$]=(Z^2-1) p(Z) {(\frac{\partial Z}{\partial X})}^3 -3 Z p(Z) \frac{\partial Z}{\partial X} \frac{\partial^2 Z}{\partial X^2} +p(Z) \frac{\partial^3 Z}{\partial X^3}[$] Equation(8)

[$]\frac{\partial p(X)}{\partial t}=\frac{\partial [p(Z) \frac{\partial Z}{\partial X}]}{\partial t}= \frac{\partial p(Z)}{\partial Z} {(\frac{\partial Z}{\partial X})}^2 \frac{\partial X}{\partial t}+ p(Z) \frac{\partial^2 Z}{\partial X^2} \frac{\partial X}{\partial t}[$]

[$]=Zp(Z) {(\frac{\partial Z}{\partial X})}^2 \frac{\partial X}{\partial t}+ p(Z) \frac{\partial^2 Z}{\partial X^2} \frac{\partial X}{\partial t}[$] Equation(9)

Now I write the fokker planck equation after applying the derivatives as

[$] \frac{\partial p(X,t)}{\partial t} = -\mu(X) \frac{\partial p(X,t)}{\partial X} - \frac{\partial \mu(X)}{\partial X} p(X,t)[$]

[$]+({(\frac{\partial \sigma(X)}{\partial X})}^2+\sigma(X) \frac{\partial^2 \sigma(X)}{\partial X^2}) p(X,t)[$]

[$]+2 \sigma(X) \frac{\partial \sigma(X)}{\partial X} \frac{\partial p(X,t)}{\partial X}+ .5 {\sigma(X)}^2 \frac{\partial^2 p(X,t)}{\partial X^2}[$] Equation(10)

Now we write the density of X(t) in terms of density of standard normal Z by substituting from above equations (6-9) in Fokker-Planck Equation (10)

[$]Zp(Z) {(\frac{\partial Z}{\partial X})}^2 \frac{\partial X}{\partial t}+ p(Z) \frac{\partial^2 Z}{\partial X^2} \frac{\partial X}{\partial t}[$]

[$]= -\mu(X) (Z p(Z){(\frac{\partial Z}{\partial X})}^2+p(Z) \frac{\partial^2 Z}{\partial X^2}) - \frac{\partial \mu(X)}{\partial X} p(Z) |\frac{\partial Z}{\partial X}|[$]

[$]+({(\frac{\partial \sigma(X)}{\partial X})}^2+\sigma(X) \frac{\partial^2 \sigma(X)}{\partial X^2}) p(Z) |\frac{\partial Z}{\partial X}|[$]

[$]+2 \sigma(X) \frac{\partial \sigma(X)}{\partial X} (Z p(Z){(\frac{\partial Z}{\partial X})}^2+p(Z) \frac{\partial^2 Z}{\partial X^2})[$]

[$]+ .5 {\sigma(X)}^2 ((Z^2-1) p(Z) {(\frac{\partial Z}{\partial X})}^3 -3 Z p(Z) \frac{\partial Z}{\partial X} \frac{\partial^2 Z}{\partial X^2} +p(Z) \frac{\partial^3 Z}{\partial X^3}) [$] Equation(11)

We see that [$]p(Z)[$] which is a static density is common throughout the equations. We eliminate it and write the new equation as

[$]Z{(\frac{\partial Z}{\partial X})}^2 \frac{\partial X}{\partial t}+ \frac{\partial^2 Z}{\partial X^2} \frac{\partial X}{\partial t}[$]

[$]=- \mu(X) (Z {(\frac{\partial Z}{\partial X})}^2+ \frac{\partial^2 Z}{\partial X^2}) - \frac{\partial \mu(X)}{\partial X} |\frac{\partial Z}{\partial X}|[$]

[$]+({(\frac{\partial \sigma(X)}{\partial X})}^2+\sigma(X) \frac{\partial^2 \sigma(X)}{\partial X^2}) |\frac{\partial Z}{\partial X}|[$]

[$]+2 \sigma(X) \frac{\partial \sigma(X)}{\partial X} (Z {(\frac{\partial Z}{\partial X})}^2+ \frac{\partial^2 Z}{\partial X^2})[$]

[$]+ .5 {\sigma(X)}^2 ((Z^2-1) {(\frac{\partial Z}{\partial X})}^3 -3 Z \frac{\partial Z}{\partial X} \frac{\partial^2 Z}{\partial X^2} + \frac{\partial^3 Z}{\partial X^3})[$] Equation(12)

We multiply both sides of equations with dt and rearrange the above equation(12) to write

[$]dX(Z{(\frac{\partial Z}{\partial X})}^2 + \frac{\partial^2 Z}{\partial X^2} )[$]

[$]= -\mu(X) (Z {(\frac{\partial Z}{\partial X})}^2+ \frac{\partial^2 Z}{\partial X^2}) dt[$]

[$]+2 \sigma(X) \frac{\partial \sigma(X)}{\partial X} (Z {(\frac{\partial Z}{\partial X})}^2+ \frac{\partial^2 Z}{\partial X^2}) dt[$]

[$]+ .5 {\sigma(X)}^2 ((Z^2-1) {(\frac{\partial Z}{\partial X})}^3 -3 Z \frac{\partial Z}{\partial X} \frac{\partial^2 Z}{\partial X^2} + \frac{\partial^3 Z}{\partial X^3}) dt[$]

[$] - \frac{\partial \mu(X)}{\partial X} |\frac{\partial Z}{\partial X}| dt[$]

[$]+({(\frac{\partial \sigma(X)}{\partial X})}^2+\sigma(X) \frac{\partial^2 \sigma(X)}{\partial X^2}) |\frac{\partial Z}{\partial X}| dt[$] Equation(13)

We multiply both sides of the above equation 13 with [$]{(\frac{\partial X}{\partial Z})}^3[$] to get

[$]dX(Z (\frac{\partial X}{\partial Z}) + {(\frac{\partial X}{\partial Z})}^3 \frac{\partial^2 Z}{\partial X^2} )[$]

[$]= -\mu(X) (Z (\frac{\partial X}{\partial Z})+ {(\frac{\partial X}{\partial Z})}^3 \frac{\partial^2 Z}{\partial X^2}) dt[$]

[$]+2 \sigma(X) \frac{\partial \sigma(X)}{\partial X} (Z (\frac{\partial X}{\partial Z})+ {(\frac{\partial X}{\partial Z})}^3 \frac{\partial^2 Z}{\partial X^2}) dt[$]

[$]- .5 {\sigma(X)}^2 dt[$]

[$]+ .5 {\sigma(X)}^2 (Z^2 -3 Z {(\frac{\partial X}{\partial Z})}^2 \frac{\partial^2 Z}{\partial X^2} +{(\frac{\partial X}{\partial Z})}^3 \frac{\partial^3 Z}{\partial X^3}) dt[$]

[$] - \frac{\partial \mu(X)}{\partial X} {(\frac{\partial X}{\partial Z})}^2 dt[$]

[$]+({(\frac{\partial \sigma(X)}{\partial X})}^2+\sigma(X) \frac{\partial^2 \sigma(X)}{\partial X^2}) {(\frac{\partial X}{\partial Z})}^2 dt[$] Equation(14)

We use the relationship [$]{(\frac{\partial X}{\partial Z})}^3 \frac{\partial^2 Z}{\partial X^2} =- \frac{\partial^2 X}{\partial Z^2}[$] in above equation (14) to get

the following equation

[$]dX(Z (\frac{\partial X}{\partial Z}) - \frac{\partial^2 X}{\partial Z^2} )[$]

[$]=- \mu(X) (Z (\frac{\partial X}{\partial Z})- \frac{\partial^2 X}{\partial Z^2}) dt[$]

[$]+2 \sigma(X) \frac{\partial \sigma(X)}{\partial X} (Z (\frac{\partial X}{\partial Z})- \frac{\partial^2 X}{\partial Z^2}) dt[$]

[$]- .5 {\sigma(X)}^2 dt[$]

[$]+ .5 {\sigma(X)}^2 (Z^2 +3 Z {(\frac{\partial Z}{\partial X})} \frac{\partial^2 X}{\partial Z^2} +{(\frac{\partial X}{\partial Z})}^3 \frac{\partial^3 Z}{\partial X^3}) dt[$]

[$] - \frac{\partial \mu(X)}{\partial X} {(\frac{\partial X}{\partial Z})}^2 dt[$]

[$]+({(\frac{\partial \sigma(X)}{\partial X})}^2+\sigma(X) \frac{\partial^2 \sigma(X)}{\partial X^2}) {(\frac{\partial X}{\partial Z})}^2 dt[$] Equation(15)

The first four terms of the equation can be solved to give a differential of following squared form

[$]dX(Z (\frac{\partial X}{\partial Z}) - \frac{\partial^2 X}{\partial Z^2} )[$]

[$]= -\mu(X) (Z (\frac{\partial X}{\partial Z})- \frac{\partial^2 X}{\partial Z^2}) dt[$]

[$]+2 \sigma(X) \frac{\partial \sigma(X)}{\partial X} (Z (\frac{\partial X}{\partial Z})- \frac{\partial^2 X}{\partial Z^2}) dt[$]

[$]- .5 {\sigma(X)}^2 dt[$]

[$]=.5 d\Big[\int_{t_0}^{t_1} dw - \int_{t_0}^{t_1} \mu(X) ds - \int_{t_0}^{t_1} 2 \sigma(X) \frac{\partial \sigma(X)}{\partial X} ds [$]

[$] + ( Z (\frac{\partial X}{\partial Z}) - \frac{\partial^2 X}{\partial Z^2})\Big]^2 [$] Equation(16)

I want to explain that the above term (equation 16)

[$].5 d\Big[\int_{t_0}^{t_1} dX - \int_{t_0}^{t_1} \mu(X) ds - \int_{t_0}^{t_1} 2 \sigma(X) \frac{\partial \sigma(X)}{\partial X} ds [$]

[$] + ( Z (\frac{\partial X}{\partial Z}) - \frac{\partial^2 X}{\partial Z^2})\Big]^2 [$]

equals

[$] +.5 \Big[( Z (\frac{\partial X}{\partial Z}) - \frac{\partial^2 X}{\partial Z^2})\Big]^2 [$] at time [$]t_0[$] since the first three integral terms vanish at start time.

and equation 16 equals at time [$]t_1[$]

[$].5 \Big[X(t_1)-X(t_0) - \int_{t_0}^{t_1} \mu(X) ds - \int_{t_0}^{t_1} 2 \sigma(X) \frac{\partial \sigma(X)}{\partial X} ds [$]

[$] + ( Z (\frac{\partial X}{\partial Z}) - \frac{\partial^2 X}{\partial Z^2})\Big]^2 [$]

So we can write the above equation 15 after substituting equation 16 in it as

[$].5 d\Big[\int_{t_0}^{t_1} dX - \int_{t_0}^{t_1} \mu(X) ds - \int_{t_0}^{t_1} 2 \sigma(X) \frac{\partial \sigma(X)}{\partial X} ds [$]

[$] + ( Z (\frac{\partial X}{\partial Z}) - \frac{\partial^2 X}{\partial Z^2})\Big]^2 [$] Equation(16)

[$]=+ .5 {\sigma(X)}^2 (Z^2 +3 Z {(\frac{\partial Z}{\partial X})} \frac{\partial^2 X}{\partial Z^2} +{(\frac{\partial X}{\partial Z})}^3 \frac{\partial^3 Z}{\partial X^3}) dt[$]

[$] - \frac{\partial \mu(X)}{\partial X} {(\frac{\partial X}{\partial Z})}^2 dt[$]

[$]+({(\frac{\partial \sigma(X)}{\partial X})}^2+\sigma(X) \frac{\partial^2 \sigma(X)}{\partial X^2}) {(\frac{\partial X}{\partial Z})}^2 dt[$] Equation(17)

We can write the above equation after noting that some terms are getting constant multiplier coefficients not immediately known. So we re-write equation (17) as

[$].5 \Big[X(t_1)-X(t_0) - \int_{t_0}^{t_1} \mu(X) ds - \int_{t_0}^{t_1} 2 \sigma(X) \frac{\partial \sigma(X)}{\partial X} ds [$]

[$] + ( Z (\frac{\partial X}{\partial Z}) - \frac{\partial^2 X}{\partial Z^2})\Big]^2 [$]

[$]= +{\Big[ ( Z (\frac{\partial X}{\partial Z}) - \frac{\partial^2 X}{\partial Z^2})\Big]}^2 [$]

[$]+ .5 {\sigma(X)}^2 (Z^2 +3C_2 Z {(\frac{\partial Z}{\partial X})} \frac{\partial^2 X}{\partial Z^2} +C_4 {(\frac{\partial X}{\partial Z})}^3 \frac{\partial^3 Z}{\partial X^3}) dt[$]

[$] - C_3\frac{\partial \mu(X)}{\partial X} {(\frac{\partial X}{\partial Z})}^2 dt[$]

[$]+C_5({(\frac{\partial \sigma(X)}{\partial X})}^2+\sigma(X) \frac{\partial^2 \sigma(X)}{\partial X^2}) {(\frac{\partial X}{\partial Z})}^2 dt[$] Equation(18)

After taking square root on both sides and re-arranging, we get the evolution equation for X(t) as

[$]X(t_1)=X(t_0) + \int_{t_0}^{t_1} \mu(X) ds + \int_{t_0}^{t_1} 2 \sigma(X) \frac{\partial \sigma(X)}{\partial X} ds [$]

[$] - Z (\frac{\partial X}{\partial Z}) + \frac{\partial^2 X}{\partial Z^2} [$]

[$]+\sqrt{\Big[ +{\big[ ( Z (\frac{\partial X}{\partial Z}) - \frac{\partial^2 X}{\partial Z^2})\big]}^2 + {\sigma(X)}^2 \big[ Z^2 +3C_2 Z {(\frac{\partial Z}{\partial X})} \frac{\partial^2 X}{\partial Z^2} +C_4 {(\frac{\partial X}{\partial Z})}^3 \frac{\partial^3 Z}{\partial X^3} \big] dt - C_3\frac{\partial \mu(X)}{\partial X} {(\frac{\partial X}{\partial Z})}^2 dt+C_5({(\frac{\partial \sigma(X)}{\partial X})}^2+\sigma(X) \frac{\partial^2 \sigma(X)}{\partial X^2}) {(\frac{\partial X}{\partial Z})}^2 dt \Big]}[$]

Equation(19)

Please note that all the terms on RHS appear under square-root. The last matlab program I posted has two terms outside the square-root and are being linearly added and that is an error. I will post a new program with this solution in a day and it works far better closer to zero.

Here I mention that Lamperti/Bessel form SDE is given as

[$]dX(t)=\mu(X) dt + \sigma dz(t)[$] Equation(20)

and it has a solution given as

[$]X(t_1)=X(t_0) + \int_{t_0}^{t_1} \mu(X) ds - Z (\frac{\partial X}{\partial Z}) + \frac{\partial^2 X}{\partial Z^2} [$]

[$]+\sqrt{\Big[ +{\big[ ( Z (\frac{\partial X}{\partial Z}) - \frac{\partial^2 X}{\partial Z^2})\big]}^2+ {\sigma(X)}^2 \big[Z^2 +3C_2 Z {(\frac{\partial Z}{\partial X})} \frac{\partial^2 X}{\partial Z^2} +C_4 {(\frac{\partial X}{\partial Z})}^3 \frac{\partial^3 Z}{\partial X^3}\big] dt- C_3\frac{\partial \mu(X)}{\partial X} {(\frac{\partial X}{\partial Z})}^2) dt \Big]} [$] Equation (21)

Please pardon any error with signs. I quickly wrote the post but I will carefully check it again and fix any errors in a day or two.

Friends, I redid some of the analysis of analytic solution of Fokker-Planck equation. I would further test it on computer and see how the existing algorithm needs to be improved. There might be some minor errors but I am presenting it in the hope that it would be useful. I will come with improvements in FPE solution algorithm code in a few days and post it here. Here is the analytic analysis I have redone.

I am starting with equation(13) of copied post.

[$]dX(Z{(\frac{\partial Z}{\partial X})}^2 + \frac{\partial^2 Z}{\partial X^2} )[$]

[$]= -\mu(X) (Z {(\frac{\partial Z}{\partial X})}^2+ \frac{\partial^2 Z}{\partial X^2}) dt[$]

[$]+2 \sigma(X) \frac{\partial \sigma(X)}{\partial X} (Z {(\frac{\partial Z}{\partial X})}^2+ \frac{\partial^2 Z}{\partial X^2}) dt[$]

[$]+ .5 {\sigma(X)}^2 ((Z^2-1) {(\frac{\partial Z}{\partial X})}^3 -3 Z \frac{\partial Z}{\partial X} \frac{\partial^2 Z}{\partial X^2} + \frac{\partial^3 Z}{\partial X^3}) dt[$]

[$] - \frac{\partial \mu(X)}{\partial X} |\frac{\partial Z}{\partial X}| dt[$]

[$]+({(\frac{\partial \sigma(X)}{\partial X})}^2+\sigma(X) \frac{\partial^2 \sigma(X)}{\partial X^2}) |\frac{\partial Z}{\partial X}| dt[$] Equation(13)

For simplicity, I just consider the FPE of SDEs in Bessel coordinates with constant volatility. I am sure once this works for SDEs in Bessel coordinates, we would be able to extrapolate it to original coordinates. So we write the Above equation(13) of copied post for SDEs in Bessel coordinates as

[$]dX(Z{(\frac{\partial Z}{\partial X})}^2 + \frac{\partial^2 Z}{\partial X^2} )[$]

[$]= -\mu(X) (Z {(\frac{\partial Z}{\partial X})}^2+ \frac{\partial^2 Z}{\partial X^2}) dt[$]

[$]+ .5 {\sigma(X)}^2 ((Z^2-1) {(\frac{\partial Z}{\partial X})}^3 -3 Z \frac{\partial Z}{\partial X} \frac{\partial^2 Z}{\partial X^2} + \frac{\partial^3 Z}{\partial X^3}) dt[$]

[$] - \frac{\partial \mu(X)}{\partial X} |\frac{\partial Z}{\partial X}| dt[$]

Equation(13 A)
I re-write the equation as (I have changed the sign on second line below. It was probably wrong but I will fix this later in previous equations)
[$]dX(Z{(\frac{\partial Z}{\partial X})}^2 + \frac{\partial^2 Z}{\partial X^2} )[$]
[$] -\mu(X) (Z {(\frac{\partial Z}{\partial X})}^2+ \frac{\partial^2 Z}{\partial X^2}) dt[$]
[$]=+ .5 {\sigma(X)}^2 ((Z^2-1) {(\frac{\partial Z}{\partial X})}^3 -3 Z \frac{\partial Z}{\partial X} \frac{\partial^2 Z}{\partial X^2} + \frac{\partial^3 Z}{\partial X^3}) dt[$]
[$] - \frac{\partial \mu(X)}{\partial X} |\frac{\partial Z}{\partial X}| dt[$] Equation(13 B)
I multiply the whole equation with [$]{(\frac{\partial X}{\partial Z})}^3[$] and also re-arrange to get
[$]dX(Z (\frac{\partial X}{\partial Z}) + {(\frac{\partial X}{\partial Z})}^3 \frac{\partial^2 Z}{\partial X^2} )[$]
[$]-\mu(X) (Z (\frac{\partial X}{\partial Z})+ {(\frac{\partial X}{\partial Z})}^3 \frac{\partial^2 Z}{\partial X^2}) dt[$]
[$]- .5 {\sigma(X)}^2 dt[$]
[$]=+ .5 {\sigma(X)}^2 (Z^2 -3 Z {(\frac{\partial X}{\partial Z})}^2 \frac{\partial^2 Z}{\partial X^2} +{(\frac{\partial X}{\partial Z})}^3 \frac{\partial^3 Z}{\partial X^3}) dt[$]
[$] - \frac{\partial \mu(X)}{\partial X} {(\frac{\partial X}{\partial Z})}^2 dt[$] Equation(14A)
We want to convert the first three lines of the above Equation(14 A) into a complete differential of square with following terms and we want to calculate the squared differential below so that we would know what terms have to be added/subtracted to first three lines of Equation(14 A) in order to make it a complete squared differential
[$].5 d\Big[\int_{t_0}^{t} dX(s) - \int_{t_0}^{t} \mu(X) ds + ( Z (\frac{\partial X}{\partial Z}) - \frac{\partial^2 X}{\partial Z^2})\Big]^2 [$] Equation(16A)
We apply the square inside the differential and get the following
[$].5 d\Big[ {[\int_{t_0}^{t} dX]}^2 -2 [\int_{t_0}^{t} dX] [ \int_{t_0}^{t} \mu(X) ds] +2 [\int_{t_0}^{t} dX] ( Z (\frac{\partial X}{\partial Z}) - \frac{\partial^2 X}{\partial Z^2})[$]
[$]+ {[\int_{t_0}^{t} \mu(X) ds]}^2 -2 [\int_{t_0}^{t} \mu(X) ds] ( Z (\frac{\partial X}{\partial Z})- \frac{\partial^2 X}{\partial Z^2})[$]
[$]+ {( Z (\frac{\partial X}{\partial Z}) - \frac{\partial^2 X}{\partial Z^2}^2)}^2 \Big] [$] Equation(16A)
We apply the differential on the above equation to get
[$][\int_{t_0}^{t} dX(s)] dX(s)+ .5 {\sigma(x)}^2 ds- dX(s) [ \int_{t_0}^{t} \mu(X) ds]- [\int_{t_0}^{t} dX(s)] \mu(X) ds +dX(s) ( Z (\frac{\partial X}{\partial Z}) - \frac{\partial^2 X}{\partial Z^2})[$]
[$]+ [\int_{t_0}^{t} \mu(X) ds] \mu(X) ds - \mu(X) ds ( Z (\frac{\partial X}{\partial Z})- \frac{\partial^2 X}{\partial Z^2})[$] Equation(17A)
We only have second, fifth and sixth term present in the equation coming from FPE equation. We will have to add rest of the terms on both sides and then complete the squared differential on LHS.
Considering first term, [$]][\int_{t_0}^{t} dX(s)] dX(s)[$] we have
[$][\int_{t_0}^{t} dX(s)] = \mu(X) (t-t_0) + \sigma(X) (Z(t)-Z(t_0))[$]
and
[$]dX(s)] = \mu(X) ds + \sigma(X) dZ(s)[$]
so we have the first terms in Equation(17A) as
[$][\int_{t_0}^{t} dX(s)] dX(s)= {\mu(X)}^2 (t-t_0) dt + {\sigma(X)}^2 (Z(t)-Z(t_0))dZ(t) +\mu(X) \sigma(X) (t-t_0)dZ(t) +\mu(X) \sigma(X) (Z(t)-Z(t_0))d(t) [$]
and the third term in Equation(17A) as
[$]dX(s) [ \int_{t_0}^{t} \mu(X) ds]= {\mu(X)}^2 (t-t_0) dt + \mu(X) \sigma(X) (t-t_0) dZ(t)[$]\
and fourth term in Equation(17A) as
[$][\int_{t_0}^{t} dX(s)] \mu(X) ds={\mu(X)}^2 (t-t_0) dt + +\mu(X) \sigma(X) (Z(t)-Z(t_0))d(t)[$]
and sixth term in Equation(17A) as
[$] [\int_{t_0}^{t} \mu(X) ds] \mu(X) ds={\mu(X)}^2 (t-t_0) dt[$]
So we could have from Equation() and Equation()
[$].5 d\Big[\int_{t_0}^{t} dX(s) - \int_{t_0}^{t} \mu(X) ds + ( Z (\frac{\partial X}{\partial Z}) - \frac{\partial^2 X}{\partial Z^2})\Big]^2 [$]
[$]= +{\Big[ ( Z (\frac{\partial X}{\partial Z}) - \frac{\partial^2 X}{\partial Z^2})\Big]}^2 [$]

[$]+ .5 {\sigma(X)}^2 (Z^2 +3C_2 Z {(\frac{\partial Z}{\partial X})} \frac{\partial^2 X}{\partial Z^2} +C_4 {(\frac{\partial X}{\partial Z})}^3 \frac{\partial^3 Z}{\partial X^3}) dt[$]

[$] - C_3\frac{\partial \mu(X)}{\partial X} {(\frac{\partial X}{\partial Z})}^2 dt[$]

[$]+{\mu(X)}^2 (t-t_0) dt + {\sigma(X)}^2 (Z(t)-Z(t_0))dZ(t) +\mu(X) \sigma(X) (t-t_0)dZ(t) +\mu(X) \sigma(X) (Z(t)-Z(t_0))d(t)[$]

[$]-{\mu(X)}^2 (t-t_0) dt - \mu(X) \sigma(X) (t-t_0) dZ(t)[$]

[$]-{\mu(X)}^2 (t-t_0) dt - \mu(X) \sigma(X) (Z(t)-Z(t_0))d(t)[$]

[$]+{\mu(X)}^2 (t-t_0) dt[$]

After simplification, we have

[$].5 d\Big[\int_{t_0}^{t} dX(s) - \int_{t_0}^{t} \mu(X) ds + ( Z (\frac{\partial X}{\partial Z}) - \frac{\partial^2 X}{\partial Z^2})\Big]^2 [$]

[$]= +{\Big[ ( Z (\frac{\partial X}{\partial Z}) - \frac{\partial^2 X}{\partial Z^2})\Big]}^2 [$]

[$]+ .5 {\sigma(X)}^2 (Z^2 +3C_2 Z {(\frac{\partial Z}{\partial X})} \frac{\partial^2 X}{\partial Z^2} +C_4 {(\frac{\partial X}{\partial Z})}^3 \frac{\partial^3 Z}{\partial X^3}) dt[$]

[$] - C_3\frac{\partial \mu(X)}{\partial X} {(\frac{\partial X}{\partial Z})}^2 dt[$]

[$]+{\sigma(X)}^2 (.5 dH_2(t)-Z(t_0)dZ(t) )[$]

We can apply time integral on both sides and then take a square-root to solve the equation. The above analysis is very preliminary and only suggestive and may have some algebra errors. I will come with a detailed program in a few days.