- Cuchulainn
**Posts:**50458**Joined:****Location:**Amsterdam-
**Contact:**

http://www.datasimfinancial.com

“The two most important days in your life are the day you are born and the day you find out why.”

“The two most important days in your life are the day you are born and the day you find out why.”

Amin, you are doing a "list." Any reputation you had is rapidly disappearing. My advice is to stop now, and take some appropriate math courses.

- Cuchulainn
**Posts:**50458**Joined:****Location:**Amsterdam-
**Contact:**

This MIT offerring looks good

https://ocw.mit.edu/courses/mathematics ... /calendar/

But i would do this first

https://ocw.mit.edu/courses/mathematics ... -lectures/

And do the exercises/exam. The only way to walk the talk!

https://ocw.mit.edu/courses/mathematics ... /calendar/

But i would do this first

https://ocw.mit.edu/courses/mathematics ... -lectures/

And do the exercises/exam. The only way to walk the talk!

http://www.datasimfinancial.com

“The two most important days in your life are the day you are born and the day you find out why.”

“The two most important days in your life are the day you are born and the day you find out why.”

very very sorry. I confused dp with dp/dt. dp=0 at any given time while obviously dp/dt is not. I would request people to know that under brain control, the brain control agencies continue to take neurotransmitters out of the brain that occasionally results in such lapses. These agencies then ask their favorite pawns to start making fun of the target. From my experience any such lapses are not permanent and my neurotransmitters always continued to come back. Yes it was a lapse as Cuch and Paul suggested. I thank you both for pointing out the error.

- Cuchulainn
**Posts:**50458**Joined:****Location:**Amsterdam-
**Contact:**

I don't agree. As we both say, you need to learn some mathematics. And you can't blame the CIA in this case.

*I confused dp with dp/dt. dp=0 at any given time while obviously dp/dt is not.*

This is not maths. This is like what kids learn in junior cert school.

**Serious question: which PDE books have you studied in the past? **Can you tell us that? And for how many years? In judo, you can see how advanced someone is by belt colour. Are you white, yellow, orange, green, blue, brown or black belt in PDE?

This is not maths. This is like what kids learn in junior cert school.

Last edited by Cuchulainn on March 20th, 2017, 8:22 pm

http://www.datasimfinancial.com

“The two most important days in your life are the day you are born and the day you find out why.”

“The two most important days in your life are the day you are born and the day you find out why.”

I agree with Cuch. You have a very uneven understanding of mathematics. You don't even know where your gaps are. Take some courses.

I will try to present a completely worked out solution for simple diffusion equation in a few days. The idea is very simple. Write the dependent function in terms of integrals that depend on its arguments and then solve the integrals by method of iterated integrals after substituting the PDE for time derivatives. Following is a brief sketch that might change a bit in the final version. I copied the equations from a previous post and some notation might be slightly off.

In p(x,t) when we start from p(x_0,t_0)

We can write

[$]\zeta (x,t)=\zeta \left(x_0,t_0\right)+\int_{t_0}^t \frac{\partial }{\partial s}\zeta (x(s),s) \, ds+\int _{t_0}^t\frac{\partial }{\partial x}\zeta (x(s),s)\frac{\partial x}{\partial s}ds[$]

From integration by parts

[$]\int _{t_0}^t\frac{\partial }{\partial x}\zeta (x(s),s)\frac{\partial x}{\partial s}ds=x(t)\frac{\partial }{\partial x}\zeta (x(t),t)-x\left(t_0\right)\frac{\partial }{\partial x}\zeta \left(x\left(t_0\right),t_0\right)[$]

[$]-\int _{t_0}^tx(s)\frac{\partial }{\partial x}\left[\frac{\partial }{\partial s}\zeta (x(s),s)\right]ds[$]

I really believe that all terms and relevant integrals can be solved by the method of iterated integrals to get the final solution as a two dimensional series. In order to be able to convince Cuch and Paul, I will give a completely worked out numerical example so they can verify analytics with real numbers.

In p(x,t) when we start from p(x_0,t_0)

We can write

[$]\zeta (x,t)=\zeta \left(x_0,t_0\right)+\int_{t_0}^t \frac{\partial }{\partial s}\zeta (x(s),s) \, ds+\int _{t_0}^t\frac{\partial }{\partial x}\zeta (x(s),s)\frac{\partial x}{\partial s}ds[$]

From integration by parts

[$]\int _{t_0}^t\frac{\partial }{\partial x}\zeta (x(s),s)\frac{\partial x}{\partial s}ds=x(t)\frac{\partial }{\partial x}\zeta (x(t),t)-x\left(t_0\right)\frac{\partial }{\partial x}\zeta \left(x\left(t_0\right),t_0\right)[$]

[$]-\int _{t_0}^tx(s)\frac{\partial }{\partial x}\left[\frac{\partial }{\partial s}\zeta (x(s),s)\right]ds[$]

I really believe that all terms and relevant integrals can be solved by the method of iterated integrals to get the final solution as a two dimensional series. In order to be able to convince Cuch and Paul, I will give a completely worked out numerical example so they can verify analytics with real numbers.

Amin wrote:I will try to present a completely worked out solution for simple diffusion equation in a few days. The idea is very simple. Write the dependent function in terms of integrals that depend on its arguments and then solve the integrals by method of iterated integrals after substituting the PDE for time derivatives. Following is a brief sketch that might change a bit in the final version. I copied the equations from a previous post and some notation might be slightly off.

In p(x,t) when we start from p(x_0,t_0)

We can write

[$]\zeta (x,t)=\zeta \left(x_0,t_0\right)+\int_{t_0}^t \frac{\partial }{\partial s}\zeta (x(s),s) \, ds+\int _{t_0}^t\frac{\partial }{\partial x}\zeta (x(s),s)\frac{\partial x}{\partial s}ds[$]

From integration by parts

[$]\int _{t_0}^t\frac{\partial }{\partial x}\zeta (x(s),s)\frac{\partial x}{\partial s}ds=x(t)\frac{\partial }{\partial x}\zeta (x(t),t)-x\left(t_0\right)\frac{\partial }{\partial x}\zeta \left(x\left(t_0\right),t_0\right)[$]

[$]-\int _{t_0}^tx(s)\frac{\partial }{\partial x}\left[\frac{\partial }{\partial s}\zeta (x(s),s)\right]ds[$]

I really believe that all terms and relevant integrals can be solved by the method of iterated integrals to get the final solution as a two dimensional series. In order to be able to convince Cuch and Paul, I will give a completely worked out numerical example so they can verify analytics with real numbers.

When we find an integral of a function of two variables of the sort we usually find in PDEs, the right expansion for the terms inside the integral should be

[$]\zeta (x(s),s)=\zeta \left(x(t_0),s\right)+\int _{t_0}^s\frac{\partial }{\partial x}\zeta (x(u),s)\frac{\partial x}{\partial u}du[$] Equation(1)

which can be further expanded as

[$]\zeta (x(s),s)=\zeta \left(x(t_0),t_0\right)+\int_{t_0}^s \frac{\partial }{\partial u}\zeta (x(t_0),u) \, du+\int _{t_0}^s\frac{\partial }{\partial x}\zeta (x(u),s)\frac{\partial x}{\partial u}du[$] Equation (2)

We could in fact use this last equation (2) for the solution of first order odes but we preferred to use equation (1) since it was far more convenient there. Obviously equation (2) and equation (1) are equivalent. In case of PDEs, we would have to prefer to use equation (2) in most general cases. We would have to continue to further expand each integrand term in terms of higher derivatives by representing the integrand in terms of time zero value(initial time) and its derivatives just like we did in equation (1) or equation (2). we would also have to substitute actual PDE wherever we encounter time derivatives. Sorry for slightly bad notation but partial derivatives should be replaced by dx/du in equation(2).

Trying to quickly write the right version of last equation in previous post that I have quoted. From integration by parts

[$]\int _{t_0}^s\frac{\partial }{\partial x}\zeta (x(u),s)\frac{\partial x}{\partial u}du=x(s)\frac{\partial }{\partial x}\zeta (x(s),s)-x\left(t_0\right)\frac{\partial }{\partial x}\zeta \left(x\left(t_0\right),s\right)[$]

[$]-\int _{t_0}^s x(u)\frac{\partial }{\partial x}\left[\frac{\partial }{\partial u}\zeta (x(u),s)\right]du[$]

In the final expansion solution all terms will be expanded until the the terms inside the integrals are in the form of [$]\zeta (x(t_0),t_0)[$] or its x derivatives at time t0 if x has dependence on t. If x does not have a dependence on t, the solution could be easier.

Last edited by Amin on March 22nd, 2017, 4:07 pm

Though I will soon follow with application to PDEs, I would recall what many friends were saying that many times it is very difficult to analytically integrate the iterated integrals that arise when solving ODEs using the method I presented in my paper which is based on equation 1 in previous post. In all such cases, we could use equation (2) and calculate initial derivatives with respect to time at starting time t0 and then use only Taylor coefficients that will follow from iterated integrals over constants. But it will increase computational complexity of the problem. I hope this does help many applied mathematicians who want to apply this method towards more difficult but interesting applications.

Many friends here at the forum and applied mathematicians have asked some very simple question, but you are probably being manipulated by brain agents to ignore those questions so that it looks like you are not able to answer them.

Amin, have you looked at polynomial chaos expansion methods?

I do not know very much in detail about those methods. I do recall reading some papers a few years ago. I do think this method is more natural.