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Amin
Topic Author
Posts: 1614
Joined: July 14th, 2002, 3:00 am

### Re: Breakthrough in the theory of stochastic differential equations and their simulation

Amin wrote:
Instead of Black Scholes, I was a little more bold and tried solution of CEV noise fokker planck pde. I have given initial steps and I will complete full numerical analysis tomorrow.

We try to find the solution of following fokker planck equation
$\frac{\partial p}{\partial t}=.5 \sigma ^2\frac{\partial ^2\left[x^{2\gamma }p\right]}{\partial x^2}=.5 \sigma ^2x^{2\gamma }\frac{\partial ^2p}{\partial x^2}+2\gamma \sigma ^2x^{2\gamma -1}\frac{\partial p}{\partial x}+\gamma \sigma ^2(2\gamma -1)x^{2\gamma -2}p$ Equation(1)

We change coordinates from p(x,t) to $\text{w(\zeta ,t)}$ such that   $\text{p(x,t)=w(\zeta ,t)}$  We have after the change of coordinates
$\frac{\partial p}{\partial x}=\frac{\partial w}{\partial \zeta }\frac{\partial \zeta }{\partial x}$
$\frac{\partial p}{\partial t}=\frac{\partial w}{\partial \zeta }\frac{\partial \zeta }{\partial t}+\frac{\partial w}{\partial t}$
$\frac{\partial ^2p}{\partial x^2}=\frac{\partial ^2w}{\partial \zeta ^2}\left(\frac{\partial \zeta }{\partial x}\right)^2+\frac{\partial w}{\partial \zeta }\frac{\partial ^2\zeta }{\partial x^2}$  Equations(2)

Substituting Equations(2) in Equation(1), we get
$\frac{\partial w}{\partial t}=.5 \sigma ^2x^{2\gamma }(\frac{\partial \zeta }{\partial x})^2\frac{\partial ^2w}{\partial \zeta ^2}+\left.(.5 \sigma ^2x^{2\gamma }\frac{\partial ^2\zeta }{\partial x^2}-\frac{\partial \zeta }{\partial t}+2\gamma \sigma ^2x^{2\gamma -1}\frac{\partial \zeta }{\partial x}\right)\frac{\partial w}{\partial \zeta }$
$+\gamma \sigma ^2(2\gamma -1)x^{2\gamma -2}w$ Equation(3)

We want to convert the transformed equation(3) into standard heat equation in new coordinates. Assuming we can successfully do that we will get a standard brownian motion as the solution to heat equation in new coordinates.
For that to happen, we will have to impose the conditions

$.5 \sigma ^2x^{2\gamma }(\frac{\partial \zeta }{\partial x})^2=1$  Equation(4)
$\left.(.5 \sigma ^2x^{2\gamma }\frac{\partial ^2\zeta }{\partial x^2}-\frac{\partial \zeta }{\partial t}+2\gamma \sigma ^2x^{2\gamma -1}\frac{\partial \zeta }{\partial x}\right)\frac{\partial w}{\partial \zeta }+\gamma \sigma ^2(2\gamma -1)x^{2\gamma -2}w=0$ Equation(5)

But we also know once we have satisfied the above conditions, as we stated earlier we will get a standard brownian motion as the solution to heat equation in new coordinates.
$w=\frac{1}{\sqrt{4\text{pi} t}}\text{Exp}\left[-\frac{\zeta ^2}{4t}\right]$  Equation(6)
$\frac{\partial w}{\partial \zeta }=\frac{1}{\sqrt{4\text{pi} t}}\text{Exp}\left[-\frac{\zeta ^2}{4t}\right]\left(-\frac{\zeta }{2t}\right)$  Equation(7)

From Equation (4) , we get
$\frac{\partial \zeta }{\partial x}=\frac{1}{\sqrt{.5 }\text{\sigma x}^{\gamma }}$ Equation(8)
From above equation
$\zeta (x)=\frac{x^{1-\gamma }}{\sqrt{.5 }\sigma (1-\gamma )}-\frac{x_0{}^{1-\gamma }}{\sqrt{.5 }\sigma (1-\gamma )}$  Equation(9)
$\frac{\partial ^2\zeta }{\partial x^2}=\frac{(-\gamma )}{\sqrt{.5 }\text{\sigma x}^{\gamma +1}}$ Equation(10)
Putting Equations (6) and (7) in Equaiton(5) and simplifying to cancel the exponential, we get
$\left.(.5 \sigma ^2x^{2\gamma }\frac{\partial ^2\zeta }{\partial x^2}-\frac{\partial \zeta }{\partial t}+2\gamma \sigma ^2x^{2\gamma -1}\frac{\partial \zeta }{\partial x}\right)\left(-\frac{\zeta }{2t}\right)+\gamma \sigma ^2(2\gamma -1)x^{2\gamma -2}=0$ Equation(11)
We try to solve the following ODE  which can be solved for both zeta or x.(I will attempt to give the complete solution tomorrow but we will have to convert the eqution into zeta or x because a map between both variables exists as given by equation(9) for example.
$\frac{\partial \zeta }{\partial t}=\left.(.5 \sigma ^2x^{2\gamma }\frac{\partial ^2\zeta }{\partial x^2}+2\gamma \sigma ^2x^{2\gamma -1}\frac{\partial \zeta }{\partial x}\right)-\left(\frac{2t}{\zeta }\right)\gamma \sigma ^2(2\gamma -1)x^{2\gamma -2}$

I will try to give complete numerically worked out solution tomorrow. The PDE can then easily be solved in the light of example I cited in the previous post. Here is the reference again. Please look at equation 4.24 in example 4.7 on page 48 of the book, "Essential partial differential equations: Analytical and computational aspects" by David F. Griffiths, John W. Dold, and David J. Silvester. I followed the reference in the above derivation.

Sorry for late update. I will do numerics tomorrow but quickly, equation(11) is not an ODE, it is again a PDE. zeta is a function of both x and t and the map in equation(9) should be valid for t=0 though equation(8) for d/dx_zeta(x,t) should be valid everywhere in t. We can expand zeta(x,t) by method of iterated integrals as

zeta(x,t)=zeta(x,0)+ integral(0 to t) d/dt_(zeta(x,0)) dt + double-integral(0,t) d2/dt2_(zeta(x,0)) dt dt + triple-integral(0,t) d3/dt3_(zeta(x,0)) dt dt dt +....higher order terms
all of the  d/dt_(zeta(x,0), d2/dt2_(zeta(x,0)) and d3/dt3_(zeta(x,0)) can be found from the equations 8,9,10 for the initial data combined with the equation 11. Sorry for the quick, non-standard rough notation but will put everything in latex tomorrow and try to add numerics.

Paul wrote:
There's a pattern here: list and Amin both ignore the advice of the experts and go merrily their own way, learning nothing in the process.

Do the Black-Scholes PDE with call payoff.

Yes, I will give numerics for Black Scholes lognormal model first and will try to do that tomorrow. Sorry for going the wrong way first.

Amin
Topic Author
Posts: 1614
Joined: July 14th, 2002, 3:00 am

### Re: Breakthrough in the theory of stochastic differential equations and their simulation

In p(x,t) when we start from p(x_0,t_0)  the first argument x is totally general and can change as a function of time and can also change by arbitrary value at the final time cross section. We can write
$x=x_0+ \int _{t_0}^tdx(s)+\int _{x(t)}^xdx$
or
$\text{d\zeta }(x,t)=\frac{\partial }{\partial t}\zeta (x(t),t)+\frac{\partial }{\partial x}\zeta (x(s),s)\frac{\partial x}{\partial s}+\frac{\partial }{\partial x}\zeta (x,t)$
We can also write
$\zeta (x,t)=\zeta \left(x_0,t_0\right)+\int_{t_0}^t \frac{\partial }{\partial s}\zeta (x(s),s) \, ds+\int _{t_0}^t\frac{\partial }{\partial x}\zeta (x(s),s)\frac{\partial x}{\partial s}ds$
$+\int_{x(t)}^x \frac{\partial }{\partial x}\zeta (x,t) \, dx$
From integration by parts
$\int _{t_0}^t\frac{\partial }{\partial x}\zeta (x(s),s)\frac{\partial x}{\partial s}ds=x(t)\frac{\partial }{\partial x}\zeta (x(t),t)-x\left(t_0\right)\frac{\partial }{\partial x}\zeta \left(x\left(t_0\right),t_0\right)$
$-\int _{t_0}^tx(s)\frac{\partial }{\partial x}\left[\frac{\partial }{\partial s}\zeta (x(s),s)\right]ds$
where $\frac{\partial \zeta (x(s),s)}{\partial s}$ is given by the last equation in post (478) and $\frac{\partial \zeta }{\partial x}$ is given by equation (8) in post (478).
Sorry, I was unable to get back to work for past few days but will try to complete the numerical work for Black Scholes quickly in a few days.

Amin
Topic Author
Posts: 1614
Joined: July 14th, 2002, 3:00 am

### Re: Breakthrough in the theory of stochastic differential equations and their simulation

Amin wrote:
Instead of Black Scholes, I was a little more bold and tried solution of CEV noise fokker planck pde. I have given initial steps and I will complete full numerical analysis tomorrow.

We try to find the solution of following fokker planck equation
$\frac{\partial p}{\partial t}=.5 \sigma ^2\frac{\partial ^2\left[x^{2\gamma }p\right]}{\partial x^2}=.5 \sigma ^2x^{2\gamma }\frac{\partial ^2p}{\partial x^2}+2\gamma \sigma ^2x^{2\gamma -1}\frac{\partial p}{\partial x}+\gamma \sigma ^2(2\gamma -1)x^{2\gamma -2}p$ Equation(1)

We change coordinates from p(x,t) to $\text{w(\zeta ,t)}$ such that   $\text{p(x,t)=w(\zeta ,t)}$  We have after the change of coordinates
$\frac{\partial p}{\partial x}=\frac{\partial w}{\partial \zeta }\frac{\partial \zeta }{\partial x}$
$\frac{\partial p}{\partial t}=\frac{\partial w}{\partial \zeta }\frac{\partial \zeta }{\partial t}+\frac{\partial w}{\partial t}$
$\frac{\partial ^2p}{\partial x^2}=\frac{\partial ^2w}{\partial \zeta ^2}\left(\frac{\partial \zeta }{\partial x}\right)^2+\frac{\partial w}{\partial \zeta }\frac{\partial ^2\zeta }{\partial x^2}$  Equations(2)

Substituting Equations(2) in Equation(1), we get
$\frac{\partial w}{\partial t}=.5 \sigma ^2x^{2\gamma }(\frac{\partial \zeta }{\partial x})^2\frac{\partial ^2w}{\partial \zeta ^2}+\left.(.5 \sigma ^2x^{2\gamma }\frac{\partial ^2\zeta }{\partial x^2}-\frac{\partial \zeta }{\partial t}+2\gamma \sigma ^2x^{2\gamma -1}\frac{\partial \zeta }{\partial x}\right)\frac{\partial w}{\partial \zeta }$
$+\gamma \sigma ^2(2\gamma -1)x^{2\gamma -2}w$ Equation(3)

We want to convert the transformed equation(3) into standard heat equation in new coordinates. Assuming we can successfully do that we will get a standard brownian motion as the solution to heat equation in new coordinates.
For that to happen, we will have to impose the conditions

$.5 \sigma ^2x^{2\gamma }(\frac{\partial \zeta }{\partial x})^2=1$  Equation(4)
$\left.(.5 \sigma ^2x^{2\gamma }\frac{\partial ^2\zeta }{\partial x^2}-\frac{\partial \zeta }{\partial t}+2\gamma \sigma ^2x^{2\gamma -1}\frac{\partial \zeta }{\partial x}\right)\frac{\partial w}{\partial \zeta }+\gamma \sigma ^2(2\gamma -1)x^{2\gamma -2}w=0$ Equation(5)

But we also know once we have satisfied the above conditions, as we stated earlier we will get a standard brownian motion as the solution to heat equation in new coordinates.
$w=\frac{1}{\sqrt{4\text{pi} t}}\text{Exp}\left[-\frac{\zeta ^2}{4t}\right]$  Equation(6)
$\frac{\partial w}{\partial \zeta }=\frac{1}{\sqrt{4\text{pi} t}}\text{Exp}\left[-\frac{\zeta ^2}{4t}\right]\left(-\frac{\zeta }{2t}\right)$  Equation(7)

From Equation (4) , we get
$\frac{\partial \zeta }{\partial x}=\frac{1}{\sqrt{.5 }\text{\sigma x}^{\gamma }}$ Equation(8)
From above equation
$\zeta (x)=\frac{x^{1-\gamma }}{\sqrt{.5 }\sigma (1-\gamma )}-\frac{x_0{}^{1-\gamma }}{\sqrt{.5 }\sigma (1-\gamma )}$  Equation(9)
$\frac{\partial ^2\zeta }{\partial x^2}=\frac{(-\gamma )}{\sqrt{.5 }\text{\sigma x}^{\gamma +1}}$ Equation(10)
Putting Equations (6) and (7) in Equaiton(5) and simplifying to cancel the exponential, we get
$\left.(.5 \sigma ^2x^{2\gamma }\frac{\partial ^2\zeta }{\partial x^2}-\frac{\partial \zeta }{\partial t}+2\gamma \sigma ^2x^{2\gamma -1}\frac{\partial \zeta }{\partial x}\right)\left(-\frac{\zeta }{2t}\right)+\gamma \sigma ^2(2\gamma -1)x^{2\gamma -2}=0$ Equation(11)
We try to solve the following ODE  which can be solved for both zeta or x.(I will attempt to give the complete solution tomorrow but we will have to convert the eqution into zeta or x because a map between both variables exists as given by equation(9) for example.
$\frac{\partial \zeta }{\partial t}=\left.(.5 \sigma ^2x^{2\gamma }\frac{\partial ^2\zeta }{\partial x^2}+2\gamma \sigma ^2x^{2\gamma -1}\frac{\partial \zeta }{\partial x}\right)-\left(\frac{2t}{\zeta }\right)\gamma \sigma ^2(2\gamma -1)x^{2\gamma -2}$

I will try to give complete numerically worked out solution tomorrow. The PDE can then easily be solved in the light of example I cited in the previous post. Here is the reference again. Please look at equation 4.24 in example 4.7 on page 48 of the book, "Essential partial differential equations: Analytical and computational aspects" by David F. Griffiths, John W. Dold, and David J. Silvester. I followed the reference in the above derivation.

OK, I tried to do the above work for driftless lognormal diffusion. We try to find the solution of following fokker planck equation
$\frac{\partial p}{\partial t}= \sigma ^2\frac{\partial ^2\left[x^2p\right]}{\partial x^2}= \sigma ^2x^2\frac{\partial ^2p}{\partial x^2}+4\sigma ^2x\frac{\partial p}{\partial x}+2\sigma ^2p$ Equation (1)

I changed coordinates from p(x,t) to w(zeta,t) such that   p(x,t)=w(zeta,t)  We have after the change of coordinates (as I did in the quoted post)
$\frac{\partial w}{\partial t}=\sigma ^2x^2(\frac{\partial \zeta }{\partial x})^2\frac{\partial ^2w}{\partial \zeta ^2}+\left.(\sigma ^2x^2\frac{\partial ^2\zeta }{\partial x^2}-\frac{\partial \zeta }{\partial t}+4\sigma ^2x\frac{\partial \zeta }{\partial x}\right)\frac{\partial w}{\partial \zeta }+2\sigma ^2w$  Equation(3)

We want to convert the transformed equation into standard heat equation in new coordinates. Assuming we can successfully do that we will get a standard brownian motion as the solution to heat equation in new coordinates.
For that to happen, we will have to impose the conditions

$\sigma ^2x^2\left(\frac{\partial \zeta }{\partial x}\right)^2=1$  Equation(4)
$\left.( \sigma ^2x^2\frac{\partial ^2\zeta }{\partial x^2}-\frac{\partial \zeta }{\partial t}+4\sigma ^2x\frac{\partial \zeta }{\partial x}\right)\frac{\partial w}{\partial \zeta }+2\sigma ^2w=0$  Equation(5)
But we also know once we have satisfied the above conditions,
$w=\frac{1}{\sqrt{4\text{pi} t}}\text{Exp}\left[-\frac{\zeta ^2}{4t}\right]$ Equation(6)
$\frac{\partial w}{\partial \zeta }=\frac{1}{\sqrt{4\text{pi} t}}\text{Exp}\left[-\frac{\zeta ^2}{4t}\right]\left(-\frac{\zeta }{2t}\right)$ Equation(7)

From Equation (3) , we get
$\frac{\partial \zeta }{\partial x}=\frac{1}{\text{\sigma x}}$ Equation(8)
From above equation
$\zeta (x,0)=\frac{\ln (x)}{\sigma }-\frac{\ln \left(x_0\right)}{\sigma }$ Equation(9)
$\frac{\partial ^2\zeta }{\partial x^2}=\frac{-1}{\text{\sigma x}^2}$ Equation(10)
After substituting Equations(8) and (10) in equation (5), we get
$\left.(\sigma ^2x^2\frac{-1}{\text{\sigma x}^2}-\frac{\partial \zeta }{\partial t}+4\sigma ^2x\frac{1}{\text{\sigma x}}\right)\left(-\frac{\zeta }{2t}\right)+2\sigma ^2=0$ Equation(11)
$\left.(- \sigma -\frac{\partial \zeta }{\partial t}+4\sigma \right)-\left(\frac{2t}{\zeta }\right)2\sigma ^2=0$ Equation(12)
$\frac{\partial \zeta }{\partial t}=3\sigma -\frac{4t}{\zeta }\sigma ^2$ Equation(13)

The solution to the above ode is $\zeta (t)=\zeta (0)+3\text{\sigma t}$, however, we do realize that if we are tracking the mean of the distribution, the first derivative has to be zero in equation (5)  while the second derivative in equation (5) will not be zero because change of mean should be equal to drift in the diffusion which is zero in the case of driftless lognormal SDE we used to get the fokker planck. I would be able to write a new version of equation (5) for the mean of the lognormal driftless diffusion when mean does not change as in our case
$\left.( \sigma ^2x^2\frac{\partial ^2\zeta }{\partial x^2}-\frac{\partial \zeta }{\partial t}\right)\frac{\partial w}{\partial \zeta }+2\sigma ^2w=0$
which has the solution $\zeta (t)=\zeta (0)-\text{\sigma t}$  and using equation(9), we get the new equation as
$\zeta (x,t)=\frac{\ln (x)}{\sigma }-\frac{\ln \left(x_0\right)}{\sigma }-\text{\sigma t}$ which gives us the solution to the lognormal SDE model.
Last edited by Amin on February 15th, 2017, 1:34 pm

Amin
Topic Author
Posts: 1614
Joined: July 14th, 2002, 3:00 am

### Re: Breakthrough in the theory of stochastic differential equations and their simulation

Amin wrote:
Amin wrote:
Instead of Black Scholes, I was a little more bold and tried solution of CEV noise fokker planck pde. I have given initial steps and I will complete full numerical analysis tomorrow.

We try to find the solution of following fokker planck equation
$\frac{\partial p}{\partial t}=.5 \sigma ^2\frac{\partial ^2\left[x^{2\gamma }p\right]}{\partial x^2}=.5 \sigma ^2x^{2\gamma }\frac{\partial ^2p}{\partial x^2}+2\gamma \sigma ^2x^{2\gamma -1}\frac{\partial p}{\partial x}+\gamma \sigma ^2(2\gamma -1)x^{2\gamma -2}p$ Equation(1)

We change coordinates from p(x,t) to $\text{w(\zeta ,t)}$ such that   $\text{p(x,t)=w(\zeta ,t)}$  We have after the change of coordinates
$\frac{\partial p}{\partial x}=\frac{\partial w}{\partial \zeta }\frac{\partial \zeta }{\partial x}$
$\frac{\partial p}{\partial t}=\frac{\partial w}{\partial \zeta }\frac{\partial \zeta }{\partial t}+\frac{\partial w}{\partial t}$
$\frac{\partial ^2p}{\partial x^2}=\frac{\partial ^2w}{\partial \zeta ^2}\left(\frac{\partial \zeta }{\partial x}\right)^2+\frac{\partial w}{\partial \zeta }\frac{\partial ^2\zeta }{\partial x^2}$  Equations(2)

Substituting Equations(2) in Equation(1), we get
$\frac{\partial w}{\partial t}=.5 \sigma ^2x^{2\gamma }(\frac{\partial \zeta }{\partial x})^2\frac{\partial ^2w}{\partial \zeta ^2}+\left.(.5 \sigma ^2x^{2\gamma }\frac{\partial ^2\zeta }{\partial x^2}-\frac{\partial \zeta }{\partial t}+2\gamma \sigma ^2x^{2\gamma -1}\frac{\partial \zeta }{\partial x}\right)\frac{\partial w}{\partial \zeta }$
$+\gamma \sigma ^2(2\gamma -1)x^{2\gamma -2}w$ Equation(3)

We want to convert the transformed equation(3) into standard heat equation in new coordinates. Assuming we can successfully do that we will get a standard brownian motion as the solution to heat equation in new coordinates.
For that to happen, we will have to impose the conditions

$.5 \sigma ^2x^{2\gamma }(\frac{\partial \zeta }{\partial x})^2=1$  Equation(4)
$\left.(.5 \sigma ^2x^{2\gamma }\frac{\partial ^2\zeta }{\partial x^2}-\frac{\partial \zeta }{\partial t}+2\gamma \sigma ^2x^{2\gamma -1}\frac{\partial \zeta }{\partial x}\right)\frac{\partial w}{\partial \zeta }+\gamma \sigma ^2(2\gamma -1)x^{2\gamma -2}w=0$ Equation(5)

But we also know once we have satisfied the above conditions, as we stated earlier we will get a standard brownian motion as the solution to heat equation in new coordinates.
$w=\frac{1}{\sqrt{4\text{pi} t}}\text{Exp}\left[-\frac{\zeta ^2}{4t}\right]$  Equation(6)
$\frac{\partial w}{\partial \zeta }=\frac{1}{\sqrt{4\text{pi} t}}\text{Exp}\left[-\frac{\zeta ^2}{4t}\right]\left(-\frac{\zeta }{2t}\right)$  Equation(7)

From Equation (4) , we get
$\frac{\partial \zeta }{\partial x}=\frac{1}{\sqrt{.5 }\text{\sigma x}^{\gamma }}$ Equation(8)
From above equation
$\zeta (x)=\frac{x^{1-\gamma }}{\sqrt{.5 }\sigma (1-\gamma )}-\frac{x_0{}^{1-\gamma }}{\sqrt{.5 }\sigma (1-\gamma )}$  Equation(9)
$\frac{\partial ^2\zeta }{\partial x^2}=\frac{(-\gamma )}{\sqrt{.5 }\text{\sigma x}^{\gamma +1}}$ Equation(10)
Putting Equations (6) and (7) in Equaiton(5) and simplifying to cancel the exponential, we get
$\left.(.5 \sigma ^2x^{2\gamma }\frac{\partial ^2\zeta }{\partial x^2}-\frac{\partial \zeta }{\partial t}+2\gamma \sigma ^2x^{2\gamma -1}\frac{\partial \zeta }{\partial x}\right)\left(-\frac{\zeta }{2t}\right)+\gamma \sigma ^2(2\gamma -1)x^{2\gamma -2}=0$ Equation(11)
We try to solve the following ODE  which can be solved for both zeta or x.(I will attempt to give the complete solution tomorrow but we will have to convert the eqution into zeta or x because a map between both variables exists as given by equation(9) for example.
$\frac{\partial \zeta }{\partial t}=\left.(.5 \sigma ^2x^{2\gamma }\frac{\partial ^2\zeta }{\partial x^2}+2\gamma \sigma ^2x^{2\gamma -1}\frac{\partial \zeta }{\partial x}\right)-\left(\frac{2t}{\zeta }\right)\gamma \sigma ^2(2\gamma -1)x^{2\gamma -2}$

I will try to give complete numerically worked out solution tomorrow. The PDE can then easily be solved in the light of example I cited in the previous post. Here is the reference again. Please look at equation 4.24 in example 4.7 on page 48 of the book, "Essential partial differential equations: Analytical and computational aspects" by David F. Griffiths, John W. Dold, and David J. Silvester. I followed the reference in the above derivation.

OK, I tried to do the above work for driftless lognormal diffusion. We try to find the solution of following fokker planck equation
$\frac{\partial p}{\partial t}= \sigma ^2\frac{\partial ^2\left[x^2p\right]}{\partial x^2}= \sigma ^2x^2\frac{\partial ^2p}{\partial x^2}+4\sigma ^2x\frac{\partial p}{\partial x}+2\sigma ^2p$ Equation (1)

I changed coordinates from p(x,t) to w(zeta,t) such that   p(x,t)=w(zeta,t)  We have after the change of coordinates (as I did in the quoted post)
$\frac{\partial w}{\partial t}=\sigma ^2x^2(\frac{\partial \zeta }{\partial x})^2\frac{\partial ^2w}{\partial \zeta ^2}+\left.(\sigma ^2x^2\frac{\partial ^2\zeta }{\partial x^2}-\frac{\partial \zeta }{\partial t}+4\sigma ^2x\frac{\partial \zeta }{\partial x}\right)\frac{\partial w}{\partial \zeta }+2\sigma ^2w$  Equation(3)

We want to convert the transformed equation into standard heat equation in new coordinates. Assuming we can successfully do that we will get a standard brownian motion as the solution to heat equation in new coordinates.
For that to happen, we will have to impose the conditions

$\sigma ^2x^2\left(\frac{\partial \zeta }{\partial x}\right)^2=1$  Equation(4)
$\left.( \sigma ^2x^2\frac{\partial ^2\zeta }{\partial x^2}-\frac{\partial \zeta }{\partial t}+4\sigma ^2x\frac{\partial \zeta }{\partial x}\right)\frac{\partial w}{\partial \zeta }+2\sigma ^2w=0$  Equation(5)
But we also know once we have satisfied the above conditions,
$w=\frac{1}{\sqrt{4\text{pi} t}}\text{Exp}\left[-\frac{\zeta ^2}{4t}\right]$ Equation(6)
$\frac{\partial w}{\partial \zeta }=\frac{1}{\sqrt{4\text{pi} t}}\text{Exp}\left[-\frac{\zeta ^2}{4t}\right]\left(-\frac{\zeta }{2t}\right)$ Equation(7)

From Equation (3) , we get
$\frac{\partial \zeta }{\partial x}=\frac{1}{\text{\sigma x}}$ Equation(8)
From above equation
$\zeta (x,0)=\frac{\ln (x)}{\sigma }-\frac{\ln \left(x_0\right)}{\sigma }$ Equation(9)
$\frac{\partial ^2\zeta }{\partial x^2}=\frac{-1}{\text{\sigma x}^2}$ Equation(10)
After substituting Equations(8) and (10) in equation (3), we get
$\left.(\sigma ^2x^2\frac{-1}{\text{\sigma x}^2}-\frac{\partial \zeta }{\partial t}+4\sigma ^2x\frac{1}{\text{\sigma x}}\right)\left(-\frac{\zeta }{2t}\right)+2\sigma ^2=0$ Equation(11)
$\left.(- \sigma -\frac{\partial \zeta }{\partial t}+4\sigma \right)-\left(\frac{2t}{\zeta }\right)2\sigma ^2=0$ Equation(12)
$\frac{\partial \zeta }{\partial t}=3\sigma -\frac{4t}{\zeta }\sigma ^2$ Equation(13)

The solution to the above ode is $\zeta (t)=\zeta (0)+3\text{\sigma t}$, however, we do realize that if we are tracking the median of the diffusion, the first derivative has to be zero while the second derivative will not be zero because change of median should be equal to drift in the diffusion which is zero in the case of driftless lognormal SDE we used to get the fokker planck. We would be able to write a new version of equation (5) for the median of the lognormal driftless diffusion when median does not change as in our case
$\left.( \sigma ^2x^2\frac{\partial ^2\zeta }{\partial x^2}-\frac{\partial \zeta }{\partial t}\right)\frac{\partial w}{\partial \zeta }+2\sigma ^2w=0$
which has the solution $\zeta (t)=\zeta (0)-\text{\sigma t}$  and using equation(9), we get the new equation as
$\zeta (x,t)=\frac{\ln (x)}{\sigma }-\frac{\ln \left(x_0\right)}{\sigma }-\text{\sigma t}$ which gives us the solution to the lognormal SDE model.

Sorry the equation (3) in the above post is

$\frac{\partial w}{\partial t}=\sigma ^2x^2(\frac{\partial \zeta }{\partial x})^2\frac{\partial ^2w}{\partial \zeta ^2}+\left.(\sigma ^2x^2\frac{\partial ^2\zeta }{\partial x^2}-\frac{\partial \zeta }{\partial t}+4\sigma ^2x\frac{\partial \zeta }{\partial x}\right)\frac{\partial w}{\partial \zeta }+2\sigma ^2w$

For the solution to ODE in Equation(13) in previous post and for the later ODE when we took change in mean to be zero, I used the mathematica program given in example 4, page 11-12 of my paper. I used no initial condition and omitted the line 6 in the code there where initial condition is numerically specified. As you can see the solution to ODE is independent of the initial conditions. You can download the paper from https://papers.ssrn.com/sol3/papers2.cfm?abstract_id=2872598

Amin
Topic Author
Posts: 1614
Joined: July 14th, 2002, 3:00 am

### Re: Breakthrough in the theory of stochastic differential equations and their simulation

Amin wrote:
Instead of Black Scholes, I was a little more bold and tried solution of CEV noise fokker planck pde. I have given initial steps and I will complete full numerical analysis tomorrow.

We try to find the solution of following fokker planck equation
$\frac{\partial p}{\partial t}=.5 \sigma ^2\frac{\partial ^2\left[x^{2\gamma }p\right]}{\partial x^2}=.5 \sigma ^2x^{2\gamma }\frac{\partial ^2p}{\partial x^2}+2\gamma \sigma ^2x^{2\gamma -1}\frac{\partial p}{\partial x}+\gamma \sigma ^2(2\gamma -1)x^{2\gamma -2}p$ Equation(1)

We change coordinates from p(x,t) to $\text{w(\zeta ,t)}$ such that   $\text{p(x,t)=w(\zeta ,t)}$  We have after the change of coordinates
$\frac{\partial p}{\partial x}=\frac{\partial w}{\partial \zeta }\frac{\partial \zeta }{\partial x}$
$\frac{\partial p}{\partial t}=\frac{\partial w}{\partial \zeta }\frac{\partial \zeta }{\partial t}+\frac{\partial w}{\partial t}$
$\frac{\partial ^2p}{\partial x^2}=\frac{\partial ^2w}{\partial \zeta ^2}\left(\frac{\partial \zeta }{\partial x}\right)^2+\frac{\partial w}{\partial \zeta }\frac{\partial ^2\zeta }{\partial x^2}$  Equations(2)

Substituting Equations(2) in Equation(1), we get
$\frac{\partial w}{\partial t}=.5 \sigma ^2x^{2\gamma }(\frac{\partial \zeta }{\partial x})^2\frac{\partial ^2w}{\partial \zeta ^2}+\left.(.5 \sigma ^2x^{2\gamma }\frac{\partial ^2\zeta }{\partial x^2}-\frac{\partial \zeta }{\partial t}+2\gamma \sigma ^2x^{2\gamma -1}\frac{\partial \zeta }{\partial x}\right)\frac{\partial w}{\partial \zeta }$
$+\gamma \sigma ^2(2\gamma -1)x^{2\gamma -2}w$ Equation(3)

We want to convert the transformed equation(3) into standard heat equation in new coordinates. Assuming we can successfully do that we will get a standard brownian motion as the solution to heat equation in new coordinates.
For that to happen, we will have to impose the conditions

$.5 \sigma ^2x^{2\gamma }(\frac{\partial \zeta }{\partial x})^2=1$  Equation(4)
$\left.(.5 \sigma ^2x^{2\gamma }\frac{\partial ^2\zeta }{\partial x^2}-\frac{\partial \zeta }{\partial t}+2\gamma \sigma ^2x^{2\gamma -1}\frac{\partial \zeta }{\partial x}\right)\frac{\partial w}{\partial \zeta }+\gamma \sigma ^2(2\gamma -1)x^{2\gamma -2}w=0$ Equation(5)

But we also know once we have satisfied the above conditions, as we stated earlier we will get a standard brownian motion as the solution to heat equation in new coordinates.
$w=\frac{1}{\sqrt{4\text{pi} t}}\text{Exp}\left[-\frac{\zeta ^2}{4t}\right]$  Equation(6)
$\frac{\partial w}{\partial \zeta }=\frac{1}{\sqrt{4\text{pi} t}}\text{Exp}\left[-\frac{\zeta ^2}{4t}\right]\left(-\frac{\zeta }{2t}\right)$  Equation(7)

From Equation (4) , we get
$\frac{\partial \zeta }{\partial x}=\frac{1}{\sqrt{.5 }\text{\sigma x}^{\gamma }}$ Equation(8)
From above equation
$\zeta (x)=\frac{x^{1-\gamma }}{\sqrt{.5 }\sigma (1-\gamma )}-\frac{x_0{}^{1-\gamma }}{\sqrt{.5 }\sigma (1-\gamma )}$  Equation(9)
$\frac{\partial ^2\zeta }{\partial x^2}=\frac{(-\gamma )}{\sqrt{.5 }\text{\sigma x}^{\gamma +1}}$ Equation(10)
Putting Equations (6) and (7) in Equaiton(5) and simplifying to cancel the exponential, we get
$\left.(.5 \sigma ^2x^{2\gamma }\frac{\partial ^2\zeta }{\partial x^2}-\frac{\partial \zeta }{\partial t}+2\gamma \sigma ^2x^{2\gamma -1}\frac{\partial \zeta }{\partial x}\right)\left(-\frac{\zeta }{2t}\right)+\gamma \sigma ^2(2\gamma -1)x^{2\gamma -2}=0$ Equation(11)
We try to solve the following ODE  which can be solved for both zeta or x.(I will attempt to give the complete solution tomorrow but we will have to convert the eqution into zeta or x because a map between both variables exists as given by equation(9) for example.
$\frac{\partial \zeta }{\partial t}=\left.(.5 \sigma ^2x^{2\gamma }\frac{\partial ^2\zeta }{\partial x^2}+2\gamma \sigma ^2x^{2\gamma -1}\frac{\partial \zeta }{\partial x}\right)-\left(\frac{2t}{\zeta }\right)\gamma \sigma ^2(2\gamma -1)x^{2\gamma -2}$

I will try to give complete numerically worked out solution tomorrow. The PDE can then easily be solved in the light of example I cited in the previous post. Here is the reference again. Please look at equation 4.24 in example 4.7 on page 48 of the book, "Essential partial differential equations: Analytical and computational aspects" by David F. Griffiths, John W. Dold, and David J. Silvester. I followed the reference in the above derivation.

I have tried to solve the fokker-planck PDE related to driftless CEV diffusion. Here is the steps I have taken for the solution. Unlike the above analysis, I used the standard definition of normal with .5 in the exponential and .5 in the standard heat equation.
The equation (3) in the quoted post above becomes
$\frac{\partial w}{\partial t}=.5 \sigma ^2x^{2\gamma }(\frac{\partial \zeta }{\partial x})^2\frac{\partial ^2w}{\partial \zeta ^2}+\left.(.5 \sigma ^2x^{2\gamma }\frac{\partial ^2\zeta }{\partial x^2}-\frac{\partial \zeta }{\partial t}+2\gamma \sigma ^2x^{2\gamma -1}\frac{\partial \zeta }{\partial x}\right)\frac{\partial w}{\partial \zeta }$
$+\gamma \sigma ^2(2\gamma -1)x^{2\gamma -2}w$   Equation(1)

We want to convert the transformed equation into standard heat equation in new coordinates. Assuming we can successfully do that we will get a standard brownian motion as the solution to heat equation in new coordinates. For that to happen, we will have to impose the conditions
$\sigma ^2x^{2\gamma }\left(\frac{\partial \zeta }{\partial x}\right)^2=1$   Equation(2)
$\left.(.5 \sigma ^2x^{2\gamma }\frac{\partial ^2\zeta }{\partial x^2}-\frac{\partial \zeta }{\partial t}+2\gamma \sigma ^2x^{2\gamma -1}\frac{\partial \zeta }{\partial x}\right)\frac{\partial w}{\partial \zeta }+\gamma \sigma ^2(2\gamma -1)x^{2\gamma -2}w=0$  Equation(3)

But we also know once we have satisfied the above conditions,

$w=\frac{1}{\sqrt{2\text{pi} t}}\text{Exp}\left[-\frac{\zeta ^2}{2t}\right]$ Equation(4)
$\frac{\partial w}{\partial \zeta }=\frac{1}{\sqrt{2\text{pi} t}}\text{Exp}\left[-\frac{\zeta ^2}{2t}\right]\left(-\frac{\zeta }{t}\right)$   Equation(5)

From equation(2) above, we can find the following three equations
$\frac{\partial \zeta }{\partial x}=\frac{1}{\text{\sigma x}^{\gamma }}$  Equation(6)
$\zeta (x,0)=\frac{x^{1-\gamma }}{\sigma (1-\gamma )}-\frac{x_0{}^{1-\gamma }}{\sigma (1-\gamma )}$   Equation(7)
$\frac{\partial ^2\zeta }{\partial x^2}=\frac{(-\gamma )}{\text{\sigma x}^{\gamma +1}}$   Equation(8)

From equation(3), we derive the following equation
$\frac{\partial \zeta }{\partial t}=\left.(.5 \sigma ^2x^{2\gamma }\frac{\partial ^2\zeta }{\partial x^2}+2\gamma \sigma ^2x^{2\gamma -1}\frac{\partial \zeta }{\partial x}\right)$
$-\left(\frac{t}{\zeta }\right)\gamma \sigma ^2(2\gamma -1)x^{2\gamma -2}$   Equation(9)
which simplifies in the case of driftless diffusion where we have to account only for second order derivative as
$\frac{\partial \zeta }{\partial t}=.5 \sigma ^2x^{2\gamma }\frac{\partial ^2\zeta }{\partial x^2}-\left(\frac{t}{\zeta }\right)\gamma \sigma ^2(2\gamma -1)x^{2\gamma -2}$ Equation(10)
using equation(6), we get from equation(10)
$\frac{\partial \zeta }{\partial t}=-.5 \sigma \gamma x^{\gamma -1}-\left(\frac{t}{\zeta }\right)\gamma \sigma ^2(2\gamma -1)x^{2\gamma -2}$ Equation(11)
From equation(7), we get the inverse map from zeta to x which is valid at time zero
$x=\left[\sigma (1-\gamma )\left(\zeta (x,0)+\frac{x_0{}^{1-\gamma }}{\sigma (1-\gamma )}\right)\right]{}^{\frac{1}{(1-\gamma )}}$  Equation(12)
using the above inverse map, we get from equation (11), the following ode
$\frac{\partial \zeta }{\partial t}=-.5 \sigma \gamma \left[\sigma (1-\gamma )\left(\zeta (x,0)+\frac{x_0{}^{1-\gamma }}{\sigma (1-\gamma )}\right)\right]{}^{-1}$
$-\left(\frac{t}{\zeta }\right)\gamma \sigma ^2(2\gamma -1)[\sigma (1-\gamma )\left(\zeta (x,0)+\frac{x_0{}^{1-\gamma }}{\sigma (1-\gamma )}\right)]{}^{-2}$
We can very easily solve the above ordinary differential equation with method of iterated integrals in terms of general initial value. I would like the friends to recall that in the method of iterated integrals the series solution only depends on initial values of the dependent variable which are handily available in almost all the cases. The right initial value would usually be $\zeta (x,0)=0$  which will make many terms in the series expansion solution singular but once we change coordinates back to x  from equation (12), after the solution of ODE in zeta without giving numerical value of zeta(0) in the solution from the method of iterated integrals, the right initial value in terms of x would be x(0) which will give us a very valid solution.

Amin
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### Re: Breakthrough in the theory of stochastic differential equations and their simulation

Many friends complained that there are so many ODEs described by more complicated implicit algebraic expressions that cannot directly be written in an integral from and it would be very difficult to find their solution by method of iterated integrals. Here is my contributing attempt at generalizing the method of iterated integrals to the above mentioned case.
Let us consider a first order ODE described by an implicit expression as
$F\left(\frac{\text{dy}}{\text{dt}},y,t\right)=0$  Equation (1)
When we are given initial values t0 and y(t0), it is obvious we can easily solve Equation (1) for dy/dt(y(t0),t0) or possibly for dy/dt(y(t0),t).
Let us take a total derivative of Equation(1), it reads
$\frac{\partial F}{\partial \frac{\text{dy}}{\text{dt}}}+\frac{\partial F}{\partial \frac{\text{dy}}{\text{dt}}}\frac{\partial \frac{\text{dy}}{\text{dt}}}{\partial y}+\frac{\partial F}{\partial \frac{\text{dy}}{\text{dt}}}\frac{\partial \frac{\text{dy}}{\text{dt}}}{\partial t}$
$+\frac{\partial F}{\partial y}+\frac{\partial F}{\partial y}\frac{\text{dy}}{\text{dt}}+\frac{\partial F}{\partial t}=0$   Equation(2)

Please note that I wrote single equation(2) in two separate latex lines to avoid being written out of the right margin on smaller screens.
Equation(2) can be used in many ways after substituting initial values of y(t0) in conjunction with the equations of method of iterated integrals to solve the ODE dy/dt dictated by the implicit equation(1) above. I will write the complete solution tomorrow but I am sure many friends can easily find out the solution.

Amin
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Joined: July 14th, 2002, 3:00 am

### Re: Breakthrough in the theory of stochastic differential equations and their simulation

Here I do the complete analysis as I promised yesterday.

$F\left(\frac{\text{dy}}{\text{dt}},y,t\right)=0$  Equation(1)

When we are given initial values t0 and y(t0), it is obvious we can easily solve for dy/dt(y(t0),t0).
Let us take a total derivative of Equation(1), it reads

$\frac{\partial F}{\partial \frac{\text{dy}}{\text{dt}}}+\frac{\partial F}{\partial \frac{\text{dy}}{\text{dt}}}\frac{\partial \frac{\text{dy}}{\text{dt}}}{\partial y}+\frac{\partial F}{\partial \frac{\text{dy}}{\text{dt}}}\frac{\partial \frac{\text{dy}}{\text{dt}}}{\partial y}\frac{\text{dy}}{\text{dt}}$
$+\frac{\partial F}{\partial \frac{\text{dy}}{\text{dt}}}\frac{\partial \frac{\text{dy}}{\text{dt}}}{\partial t}+\frac{\partial F}{\partial y}+\frac{\partial F}{\partial y}\frac{\text{dy}}{\text{dt}}+\frac{\partial F}{\partial t}=0$     Equation(2)

please notice that I missed a term yesterday when I wrote the equation (2). We make the following observation that time dependent partial derivatives in third and fourth term of equation(2) can be combined to give the expression below.

$\frac{\partial \frac{\text{dy}}{\text{dt}}}{\partial y}\frac{\text{dy}}{\text{dt}}+\frac{\partial \frac{\text{dy}}{\text{dt}}}{\partial t}=\frac{d^2y}{\text{dt}^2}$ Equation(3)
Now we can write the third and fourth term in equation (2) to give us the
$\frac{\partial F}{\partial \frac{\text{dy}}{\text{dt}}}\frac{\partial \frac{\text{dy}}{\text{dt}}}{\partial y}\frac{\text{dy}}{\text{dt}}+\frac{\partial F}{\partial \frac{\text{dy}}{\text{dt}}}\frac{\partial \frac{\text{dy}}{\text{dt}}}{\partial t}=\frac{\partial F}{\partial \frac{\text{dy}}{\text{dt}}}\frac{d^2y}{\text{dt}^2}$ Equation(4)

using the above expression in equation 2, we get
$\frac{\partial F}{\partial \frac{\text{dy}}{\text{dt}}}\frac{d^2y}{\text{dt}^2}+\frac{\partial F}{\partial \frac{\text{dy}}{\text{dt}}}+\frac{\partial F}{\partial \frac{\text{dy}}{\text{dt}}}\frac{\partial \frac{\text{dy}}{\text{dt}}}{\partial y}+\frac{\partial F}{\partial y}+\frac{\partial F}{\partial y}\frac{\text{dy}}{\text{dt}}+\frac{\partial F}{\partial t}=0$ Equation(5)

which can be solved to give us a second order ODE given below.
$\frac{d^2y}{\text{dt}^2}=-\left[\frac{\partial F}{\partial \frac{\text{dy}}{\text{dt}}}\right]^{-1}\left(\frac{\partial F}{\partial \frac{\text{dy}}{\text{dt}}}+\frac{\partial F}{\partial \frac{\text{dy}}{\text{dt}}}\frac{\partial \frac{\text{dy}}{\text{dt}}}{\partial y}+\frac{\partial F}{\partial y}+\frac{\partial F}{\partial y}\frac{\text{dy}}{\text{dt}}+\frac{\partial F}{\partial t}\right)$   Equation(6)

which can easily be solved by the method of iterated integrals if we can integrate the expressions in corresponding integrals.
We notice that at any given time  tn, equation (1) has to be satisfied as
$F\left(\frac{\text{dy}}{\text{dt}}(y(\text{tn}),\text{tn}),y(\text{tn}),\text{tn}\right)=0$
which means that everywhere on the surface F, the following equation would have to be true as well
$\frac{\partial F}{\partial \frac{\text{dy}}{\text{dt}}}+\frac{\partial F}{\partial \frac{\text{dy}}{\text{dt}}}\frac{\partial \frac{\text{dy}}{\text{dt}}}{\partial y}+\frac{\partial F}{\partial y}=0$  Equation(7)
which makes our second order ODE in equation (6) to be solved as
$\frac{d^2y}{\text{dt}^2}=-\left[\frac{\partial F}{\partial \frac{\text{dy}}{\text{dt}}}\right]^{-1}\left(\frac{\partial F}{\partial y}\frac{\text{dy}}{\text{dt}}+\frac{\partial F}{\partial t}\right)$  Equation(8)

Amin
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### Re: Breakthrough in the theory of stochastic differential equations and their simulation

I am sure other mathematicians understand that my mathematical research that I post here will become part of future research papers. When I started this thread I requested people to give me credit for the original work I did but since I naturally believed in academic honesty of other people and wrote the research paper several months later. My belief later created problems so I thought it would be prudent to request people again to give me credit for the original work I have been posting here and follow the established norms of academic honesty when they do further research or present the same research.

Cuchulainn
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### Re: Breakthrough in the theory of stochastic differential equations and their simulation

.
Last edited by Cuchulainn on May 7th, 2017, 2:40 pm
http://www.datasimfinancial.com

I tell the kids, somebody’s gotta win, somebody’s gotta lose. Just don’t fight about it. Just try to get better.

Amin
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### Re: Breakthrough in the theory of stochastic differential equations and their simulation

@Cuch, I think it should be fairly easy. I will try to write a mathematica program that does that in a few days.

@ppauper, I submitted my paper to Numerische Mathematik several weeks ago and they sent me an email that they had forwarded my manuscript for referee comments. Are you aware how long this process takes for Numerische Mathematik? Thank you.

Cuchulainn
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### Re: Breakthrough in the theory of stochastic differential equations and their simulation

I have  fond that programming up the maths help in avoiding typos in the maths.
NM is a heavyweight journal.
http://www.datasimfinancial.com

I tell the kids, somebody’s gotta win, somebody’s gotta lose. Just don’t fight about it. Just try to get better.

Amin
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Joined: July 14th, 2002, 3:00 am

### Re: Breakthrough in the theory of stochastic differential equations and their simulation

I submitted my manuscript to Numerische Methematik on 29th January and I received an email that it has been forwarded for referee review. I sent a few email inquiries about typical duration of referee review period but they went un-responded. I was surprised when on March 3, they sent me an email that the referee process could take a year and then I received another email that editor did not consider the paper worth publication. What would be another good math journal for publication of my paper?

Cuchulainn
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### Re: Breakthrough in the theory of stochastic differential equations and their simulation

.
Last edited by Cuchulainn on May 7th, 2017, 2:43 pm
http://www.datasimfinancial.com

I tell the kids, somebody’s gotta win, somebody’s gotta lose. Just don’t fight about it. Just try to get better.

Amin
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### Re: Breakthrough in the theory of stochastic differential equations and their simulation

Here is cross-reference to another related thread in off-topic for applications of the research in this technical forum thread.  https://forum.wilmott.com/viewtopic.php?f=15&t=100607
I thought some friends would be interested.

amike
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Joined: October 21st, 2005, 12:57 am

### Re: Breakthrough in the theory of stochastic differential equations and their simulation

amike wrote:
Cuchulainn wrote:
...
Likewise, perturbation is based on Taylor expansions which are always scary. Numerically, they are unstable (that's why RK methods were born).
...

Well I for one have never been frightened when doing a perturbation analysis...  :)

This does highlight what the real issue is here (IMHO).  Looking at cases where the ivp breaks down (or the structure of the problem makes the method inapplicable...) is quite interesting, but all Amin needs to do is hire someone to write down for him some conditions under which his method avoids these cases, and he is once again off to the races.

But it should be clear that given the ii method is equivalent to a small time expansion, there are a multitude of (perfectly regular) examples where the method is completely ineffective.  Here is a classic example:

$\frac{dy(t)}{dt}=x(t),\quad \frac{dx(t)}{dt}=-y(t)-\frac{1}{20}x(t),\quad y(0)=0,\quad x(0)=1,\quad t\in[0,100].$

It is of course a damped SHO, probably the most common system in the universe...  (the damping isn't really necessary here).

Runge-Kutta (4/5) happily integrates this without blinking, and gives a solution that agrees very well with the exact solution on the interval.

Amin: How many terms of your expansion are required to give a reasonable approximation to the solution over the whole interval?  Please, please, please  try this.

Here is 20 terms in your expansion (in red) plotted out to t=20, the blue line is the exact solution and rk45 (they overlap):

Amin -- have you tried this?  What did you find?