- Cuchulainn
**Posts:**51410**Joined:****Location:**Amsterdam-
**Contact:**

.

Last edited by Cuchulainn on May 7th, 2017, 2:39 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.

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

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:**51410**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

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

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

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:**51410**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

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

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

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.

- Cuchulainn
**Posts:**51410**Joined:****Location:**Amsterdam-
**Contact:**

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

Amin wrote: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.

Sorry friends, I got carried away by a lot of other important things. Probably tomorrow or Friday, I will present a detailed derivation of series solution of PDEs given a certain initial data distribution. I will repeatedly use equation (2) in the quoted post above or its simpler equivalents until all the integrands terms are based only on initial data at time t0 to get a two-dimensional Taylor like expansion. I will give a very detailed derivation. We will use the PDE to convert time derivatives into space derivatives of initial data calculated at t0. The whole derivation is very simple and straightforward. I will put all notes on next post here tomorrow or by Friday.

Given [$]\zeta(x(t_0),t_0)[$] with a certain initial distribution of [$]x(t_0)[$], we want to find out the final solution which can be formally written using Green's function inside an integral over the initial data.

[$]\zeta(x(t),t)=\int_{-\infty}^{\infty} \Phi(x(t),t,x(t_0),t_0) \zeta(x(t_0),t_0) dx(t_0)[$] Equation(1)

We would like to recall that when initial data is a Dirac delta distribution, the green's function inside the integral in Equation(1)and the LHS of Equation(1) will coincide. We try to find out the LHS of Equation(1) with Dirac delta initial arguments and try to use it as a solution to Green's function for the particular PDE.

We write Equation(2) from the previous post as

[$]\zeta(x(t),t)=\zeta(x(t_0),t_0)+\int_{t_0}^{t} \frac{\partial \zeta(x(t_0),s)}{\partial s} ds + \int_{x(t_0)}^{x(t)} \frac{\partial \zeta(x(s),t)}{\partial x} dx(s)[$]

Equation(2)

First trying to work with the 2nd term(or first integral) in the RHS of Equation(2), we get

[$] \int_{t_0}^{t} \frac{\partial \zeta(x(t_0),s)}{\partial s} ds \\

=\int_{t_0}^{t} \frac{\partial}{\partial s} \big[ \zeta(x(t_0),t_0) + \int_{t_0}^{s} \frac {\partial \zeta(x(t_0),u)} {\partial u} du \big] ds [$]

Equation(3)

We notice that the integral inside large brackets in RHS of Equation (3) can similarly be further expanded many times to give other versions of Equation(3) as

[$] \int_{t_0}^{t} \frac{\partial \zeta(x(t_0),s)}{\partial s} ds \\

=\int_{t_0}^{t} \frac{\partial}{\partial s} \big[ \zeta(x(t_0),t_0) + \int_{t_0}^{s} \frac {\partial \zeta(x(t_0),u)} {\partial u} du \big] ds \\

=\int_{t_0}^{t} \frac{\partial}{\partial s} \big[ \zeta(x(t_0),t_0) + \int_{t_0}^{s} \frac {\partial } {\partial u} \big[ \zeta(x(t_0),t_0) + \int_{t_0}^{u} \frac {\partial \zeta(x(t_0),v)} {\partial v} dv \big] du \big] ds \\

=\int_{t_0}^{t} \frac{\partial}{\partial s} \big[ \zeta(x(t_0),t_0) + \int_{t_0}^{s} \frac {\partial } {\partial u} \big[ \zeta(x(t_0),t_0) + \int_{t_0}^{u} \frac {\partial} {\partial v} \big[ \zeta(x(t_0),t_0) \\

+ \int_{t_0}^{v} \frac {\partial \zeta(x(t_0),w)} {\partial w} dw \big] dv \big] du \big] ds[$]

Equation(4)

We can continue the expansions till higher orders and any symbolic computing program might actually want to expand several tens of terms but here we stop and substitute [$]\int_{t_0}^{v} \frac {\partial \zeta(x(t_0),t_0)} {\partial w} dw[$] for [$]\int_{t_0}^{v} \frac {\partial \zeta(x(t_0),w)} {\partial w} dw[$] which is obviously an approximation and we get the resulting approximate equation (We have chosen very few terms only so that we can calculate the integrals very easily to show the simple ideas). We copy the first and last equation parts in Equation(4) in Equation(5) and then try to solve the integrals in RHS of resulting equation (5) later.

[$] \int_{t_0}^{t} \frac{\partial \zeta(x(t_0),s)}{\partial s} ds \\

=\int_{t_0}^{t} \frac{\partial}{\partial s} \big[ \zeta(x(t_0),t_0) + \int_{t_0}^{s} \frac {\partial } {\partial u} \big[ \zeta(x(t_0),t_0) + \int_{t_0}^{u} \frac {\partial} {\partial v} \big[ \zeta(x(t_0),t_0) \\

+ \int_{t_0}^{v} \frac {\partial \zeta(x(t_0),t_0)} {\partial w} dw \big] dv \big] du \big] ds[$]

Equation(5)

Solving the integrals in above Equation (5) iteratively one by one we get the equations as

[$] \int_{t_0}^{t} \frac{\partial \zeta(x(t_0),s)}{\partial s} ds \\

=\int_{t_0}^{t} \frac{\partial}{\partial s} \big[ \zeta(x(t_0),t_0) + \int_{t_0}^{s} \frac {\partial } {\partial u} \big[ \zeta(x(t_0),t_0) + \int_{t_0}^{u} \frac {\partial} {\partial v} \big[ \zeta(x(t_0),t_0) \\

+ \int_{t_0}^{v} \frac {\partial \zeta(x(t_0),t_0)} {\partial w} dw \big] dv \big] du \big] ds \\

=\int_{t_0}^{t} \frac{\partial}{\partial s} \big[ \zeta(x(t_0),t_0) + \int_{t_0}^{s} \frac {\partial } {\partial u} \big[ \zeta(x(t_0),t_0) + \int_{t_0}^{u} \frac {\partial} {\partial v} \big[ \zeta(x(t_0),t_0) \\

+ \frac {\partial \zeta(x(t_0),t_0)} {\partial v} |^{v=t_0} (v-t_0) \big] dv \big] du \big] ds \\

=\int_{t_0}^{t} \frac{\partial}{\partial s} \big[ \zeta(x(t_0),t_0) + \int_{t_0}^{s} \frac {\partial } {\partial u} \big[ \zeta(x(t_0),t_0) + \frac {\partial \zeta(x(t_0),t_0)} {\partial u} |^{u=t_0} (u-t_0) \\

+ \frac {\partial^2 \zeta(x(t_0),t_0)} {\partial u^2} |^{u=t_0} \frac{(u-t_0)^2}{2} \big] du \big] ds \\

=\int_{t_0}^{t} \frac{\partial}{\partial s} \big[ \zeta(x(t_0),t_0) + \frac {\partial \zeta(x(t_0),t_0)} {\partial s} |^{s=t_0} (s-t_0) + \frac {\partial^2 \zeta(x(t_0),t_0)} {\partial s^2} |^{s=t_0} \frac{(s-t_0)^2}{2} \\

+\frac {\partial^3 \zeta(x(t_0),t_0)} {\partial s^3} |^{s=t_0} \frac{(s-t_0)^3}{6} \big] ds \\

= \frac{\partial \zeta(x(t_0),t_0)}{\partial t} |^{t=t_0} (t-t_0) + \frac {\partial^2 \zeta(x(t_0),t_0)} {\partial t^2} |^{t=t_0} \frac{(t-t_0)^2}{2} \\

+ \frac {\partial^3 \zeta(x(t_0),t_0)} {\partial t^3} |^{t=t_0} \frac{(t-t_0)^3}{6} + \frac {\partial^4 \zeta(x(t_0),t_0)} {\partial t^4} |^{t=t_0} \frac{(t-t_0)^4}{24}[$]

Equation(6)

In the above I used product rule for derivatives with the fact that [$](s-t_0)=0[$] when [$]s=t_0[$]

Suppose we are given a PDE as

[$]\frac {\partial \zeta} {\partial t} = a \frac{ \partial^2 \zeta}{\partial x^2} + b \frac{ \partial \zeta}{\partial x} + c \zeta[$]

We can use the following formulas in the expansion of Equation(6) and other expansions.

[$]\frac {\partial \zeta(x(t_0),t_0)} {\partial t} = a \frac{ \partial^2 \zeta(x(t_0),t_0)}{\partial x^2} + b \frac{ \partial \zeta(x(t_0),t_0)}{\partial x} + c \zeta(x(t_0),t_0)[$]

and

[$]\frac {\partial^2 \zeta(x(t_0),t_0)} {\partial t^2} = (a \frac{ \partial^2}{\partial x^2} + b \frac{ \partial}{\partial x} + c)^2 [\zeta(x(t_0),t_0)][$]

and

[$]\frac {\partial^n \zeta(x(t_0),t_0)} {\partial t^n} = (a \frac{ \partial^2}{\partial x^2} + b \frac{ \partial}{\partial x} + c)^n [\zeta(x(t_0),t_0)][$]

In the above attempt at explanation I have only expanded the first integral in Equation(2), the expansion for second integral(or third term) in Equation(2) though very similar but a little bit more tedious. I will write that in detail tomorrow and complete the solution expansion and then try to follow with numerical examples over the weekend.

[$]\zeta(x(t),t)=\int_{-\infty}^{\infty} \Phi(x(t),t,x(t_0),t_0) \zeta(x(t_0),t_0) dx(t_0)[$] Equation(1)

We would like to recall that when initial data is a Dirac delta distribution, the green's function inside the integral in Equation(1)and the LHS of Equation(1) will coincide. We try to find out the LHS of Equation(1) with Dirac delta initial arguments and try to use it as a solution to Green's function for the particular PDE.

We write Equation(2) from the previous post as

[$]\zeta(x(t),t)=\zeta(x(t_0),t_0)+\int_{t_0}^{t} \frac{\partial \zeta(x(t_0),s)}{\partial s} ds + \int_{x(t_0)}^{x(t)} \frac{\partial \zeta(x(s),t)}{\partial x} dx(s)[$]

Equation(2)

First trying to work with the 2nd term(or first integral) in the RHS of Equation(2), we get

[$] \int_{t_0}^{t} \frac{\partial \zeta(x(t_0),s)}{\partial s} ds \\

=\int_{t_0}^{t} \frac{\partial}{\partial s} \big[ \zeta(x(t_0),t_0) + \int_{t_0}^{s} \frac {\partial \zeta(x(t_0),u)} {\partial u} du \big] ds [$]

Equation(3)

We notice that the integral inside large brackets in RHS of Equation (3) can similarly be further expanded many times to give other versions of Equation(3) as

[$] \int_{t_0}^{t} \frac{\partial \zeta(x(t_0),s)}{\partial s} ds \\

=\int_{t_0}^{t} \frac{\partial}{\partial s} \big[ \zeta(x(t_0),t_0) + \int_{t_0}^{s} \frac {\partial \zeta(x(t_0),u)} {\partial u} du \big] ds \\

=\int_{t_0}^{t} \frac{\partial}{\partial s} \big[ \zeta(x(t_0),t_0) + \int_{t_0}^{s} \frac {\partial } {\partial u} \big[ \zeta(x(t_0),t_0) + \int_{t_0}^{u} \frac {\partial \zeta(x(t_0),v)} {\partial v} dv \big] du \big] ds \\

=\int_{t_0}^{t} \frac{\partial}{\partial s} \big[ \zeta(x(t_0),t_0) + \int_{t_0}^{s} \frac {\partial } {\partial u} \big[ \zeta(x(t_0),t_0) + \int_{t_0}^{u} \frac {\partial} {\partial v} \big[ \zeta(x(t_0),t_0) \\

+ \int_{t_0}^{v} \frac {\partial \zeta(x(t_0),w)} {\partial w} dw \big] dv \big] du \big] ds[$]

Equation(4)

We can continue the expansions till higher orders and any symbolic computing program might actually want to expand several tens of terms but here we stop and substitute [$]\int_{t_0}^{v} \frac {\partial \zeta(x(t_0),t_0)} {\partial w} dw[$] for [$]\int_{t_0}^{v} \frac {\partial \zeta(x(t_0),w)} {\partial w} dw[$] which is obviously an approximation and we get the resulting approximate equation (We have chosen very few terms only so that we can calculate the integrals very easily to show the simple ideas). We copy the first and last equation parts in Equation(4) in Equation(5) and then try to solve the integrals in RHS of resulting equation (5) later.

[$] \int_{t_0}^{t} \frac{\partial \zeta(x(t_0),s)}{\partial s} ds \\

=\int_{t_0}^{t} \frac{\partial}{\partial s} \big[ \zeta(x(t_0),t_0) + \int_{t_0}^{s} \frac {\partial } {\partial u} \big[ \zeta(x(t_0),t_0) + \int_{t_0}^{u} \frac {\partial} {\partial v} \big[ \zeta(x(t_0),t_0) \\

+ \int_{t_0}^{v} \frac {\partial \zeta(x(t_0),t_0)} {\partial w} dw \big] dv \big] du \big] ds[$]

Equation(5)

Solving the integrals in above Equation (5) iteratively one by one we get the equations as

[$] \int_{t_0}^{t} \frac{\partial \zeta(x(t_0),s)}{\partial s} ds \\

=\int_{t_0}^{t} \frac{\partial}{\partial s} \big[ \zeta(x(t_0),t_0) + \int_{t_0}^{s} \frac {\partial } {\partial u} \big[ \zeta(x(t_0),t_0) + \int_{t_0}^{u} \frac {\partial} {\partial v} \big[ \zeta(x(t_0),t_0) \\

+ \int_{t_0}^{v} \frac {\partial \zeta(x(t_0),t_0)} {\partial w} dw \big] dv \big] du \big] ds \\

=\int_{t_0}^{t} \frac{\partial}{\partial s} \big[ \zeta(x(t_0),t_0) + \int_{t_0}^{s} \frac {\partial } {\partial u} \big[ \zeta(x(t_0),t_0) + \int_{t_0}^{u} \frac {\partial} {\partial v} \big[ \zeta(x(t_0),t_0) \\

+ \frac {\partial \zeta(x(t_0),t_0)} {\partial v} |^{v=t_0} (v-t_0) \big] dv \big] du \big] ds \\

=\int_{t_0}^{t} \frac{\partial}{\partial s} \big[ \zeta(x(t_0),t_0) + \int_{t_0}^{s} \frac {\partial } {\partial u} \big[ \zeta(x(t_0),t_0) + \frac {\partial \zeta(x(t_0),t_0)} {\partial u} |^{u=t_0} (u-t_0) \\

+ \frac {\partial^2 \zeta(x(t_0),t_0)} {\partial u^2} |^{u=t_0} \frac{(u-t_0)^2}{2} \big] du \big] ds \\

=\int_{t_0}^{t} \frac{\partial}{\partial s} \big[ \zeta(x(t_0),t_0) + \frac {\partial \zeta(x(t_0),t_0)} {\partial s} |^{s=t_0} (s-t_0) + \frac {\partial^2 \zeta(x(t_0),t_0)} {\partial s^2} |^{s=t_0} \frac{(s-t_0)^2}{2} \\

+\frac {\partial^3 \zeta(x(t_0),t_0)} {\partial s^3} |^{s=t_0} \frac{(s-t_0)^3}{6} \big] ds \\

= \frac{\partial \zeta(x(t_0),t_0)}{\partial t} |^{t=t_0} (t-t_0) + \frac {\partial^2 \zeta(x(t_0),t_0)} {\partial t^2} |^{t=t_0} \frac{(t-t_0)^2}{2} \\

+ \frac {\partial^3 \zeta(x(t_0),t_0)} {\partial t^3} |^{t=t_0} \frac{(t-t_0)^3}{6} + \frac {\partial^4 \zeta(x(t_0),t_0)} {\partial t^4} |^{t=t_0} \frac{(t-t_0)^4}{24}[$]

Equation(6)

In the above I used product rule for derivatives with the fact that [$](s-t_0)=0[$] when [$]s=t_0[$]

Suppose we are given a PDE as

[$]\frac {\partial \zeta} {\partial t} = a \frac{ \partial^2 \zeta}{\partial x^2} + b \frac{ \partial \zeta}{\partial x} + c \zeta[$]

We can use the following formulas in the expansion of Equation(6) and other expansions.

[$]\frac {\partial \zeta(x(t_0),t_0)} {\partial t} = a \frac{ \partial^2 \zeta(x(t_0),t_0)}{\partial x^2} + b \frac{ \partial \zeta(x(t_0),t_0)}{\partial x} + c \zeta(x(t_0),t_0)[$]

and

[$]\frac {\partial^2 \zeta(x(t_0),t_0)} {\partial t^2} = (a \frac{ \partial^2}{\partial x^2} + b \frac{ \partial}{\partial x} + c)^2 [\zeta(x(t_0),t_0)][$]

and

[$]\frac {\partial^n \zeta(x(t_0),t_0)} {\partial t^n} = (a \frac{ \partial^2}{\partial x^2} + b \frac{ \partial}{\partial x} + c)^n [\zeta(x(t_0),t_0)][$]

In the above attempt at explanation I have only expanded the first integral in Equation(2), the expansion for second integral(or third term) in Equation(2) though very similar but a little bit more tedious. I will write that in detail tomorrow and complete the solution expansion and then try to follow with numerical examples over the weekend.

Amin wrote:Given [$]\zeta(x(t_0),t_0)[$] with a certain initial distribution of [$]x(t_0)[$], we want to find out the final solution which can be formally written using Green's function inside an integral over the initial data.

[$]\zeta(x(t),t)=\int_{-\infty}^{\infty} \Phi(x(t),t,x(t_0),t_0) \zeta(x(t_0),t_0) dx(t_0)[$] Equation(1)

We would like to recall that when initial data is a Dirac delta distribution, the green's function inside the integral in Equation(1)and the LHS of Equation(1) will coincide. We try to find out the LHS of Equation(1) with Dirac delta initial arguments and try to use it as a solution to Green's function for the particular PDE.

We write Equation(2) from the previous post as

[$]\zeta(x(t),t)=\zeta(x(t_0),t_0)+\int_{t_0}^{t} \frac{\partial \zeta(x(t_0),s)}{\partial s} ds + \int_{x(t_0)}^{x(t)} \frac{\partial \zeta(x(s),t)}{\partial x} dx(s)[$]

Equation(2)

First trying to work with the 2nd term(or first integral) in the RHS of Equation(2), we get

[$] \int_{t_0}^{t} \frac{\partial \zeta(x(t_0),s)}{\partial s} ds \\

=\int_{t_0}^{t} \frac{\partial}{\partial s} \big[ \zeta(x(t_0),t_0) + \int_{t_0}^{s} \frac {\partial \zeta(x(t_0),u)} {\partial u} du \big] ds [$]

Equation(3)

We notice that the integral inside large brackets in RHS of Equation (3) can similarly be further expanded many times to give other versions of Equation(3) as

[$] \int_{t_0}^{t} \frac{\partial \zeta(x(t_0),s)}{\partial s} ds \\

=\int_{t_0}^{t} \frac{\partial}{\partial s} \big[ \zeta(x(t_0),t_0) + \int_{t_0}^{s} \frac {\partial \zeta(x(t_0),u)} {\partial u} du \big] ds \\

=\int_{t_0}^{t} \frac{\partial}{\partial s} \big[ \zeta(x(t_0),t_0) + \int_{t_0}^{s} \frac {\partial } {\partial u} \big[ \zeta(x(t_0),t_0) + \int_{t_0}^{u} \frac {\partial \zeta(x(t_0),v)} {\partial v} dv \big] du \big] ds \\

=\int_{t_0}^{t} \frac{\partial}{\partial s} \big[ \zeta(x(t_0),t_0) + \int_{t_0}^{s} \frac {\partial } {\partial u} \big[ \zeta(x(t_0),t_0) + \int_{t_0}^{u} \frac {\partial} {\partial v} \big[ \zeta(x(t_0),t_0) \\

+ \int_{t_0}^{v} \frac {\partial \zeta(x(t_0),w)} {\partial w} dw \big] dv \big] du \big] ds[$]

Equation(4)

We can continue the expansions till higher orders and any symbolic computing program might actually want to expand several tens of terms but here we stop and substitute [$]\int_{t_0}^{v} \frac {\partial \zeta(x(t_0),t_0)} {\partial w} dw[$] for [$]\int_{t_0}^{v} \frac {\partial \zeta(x(t_0),w)} {\partial w} dw[$] which is obviously an approximation and we get the resulting approximate equation (We have chosen very few terms only so that we can calculate the integrals very easily to show the simple ideas). We copy the first and last equation parts in Equation(4) in Equation(5) and then try to solve the integrals in RHS of resulting equation (5) later.

[$] \int_{t_0}^{t} \frac{\partial \zeta(x(t_0),s)}{\partial s} ds \\

=\int_{t_0}^{t} \frac{\partial}{\partial s} \big[ \zeta(x(t_0),t_0) + \int_{t_0}^{s} \frac {\partial } {\partial u} \big[ \zeta(x(t_0),t_0) + \int_{t_0}^{u} \frac {\partial} {\partial v} \big[ \zeta(x(t_0),t_0) \\

+ \int_{t_0}^{v} \frac {\partial \zeta(x(t_0),t_0)} {\partial w} dw \big] dv \big] du \big] ds[$]

Equation(5)

Solving the integrals in above Equation (5) iteratively one by one we get the equations as

[$] \int_{t_0}^{t} \frac{\partial \zeta(x(t_0),s)}{\partial s} ds \\

=\int_{t_0}^{t} \frac{\partial}{\partial s} \big[ \zeta(x(t_0),t_0) + \int_{t_0}^{s} \frac {\partial } {\partial u} \big[ \zeta(x(t_0),t_0) + \int_{t_0}^{u} \frac {\partial} {\partial v} \big[ \zeta(x(t_0),t_0) \\

+ \int_{t_0}^{v} \frac {\partial \zeta(x(t_0),t_0)} {\partial w} dw \big] dv \big] du \big] ds \\

=\int_{t_0}^{t} \frac{\partial}{\partial s} \big[ \zeta(x(t_0),t_0) + \int_{t_0}^{s} \frac {\partial } {\partial u} \big[ \zeta(x(t_0),t_0) + \int_{t_0}^{u} \frac {\partial} {\partial v} \big[ \zeta(x(t_0),t_0) \\

+ \frac {\partial \zeta(x(t_0),t_0)} {\partial v} |^{v=t_0} (v-t_0) \big] dv \big] du \big] ds \\

=\int_{t_0}^{t} \frac{\partial}{\partial s} \big[ \zeta(x(t_0),t_0) + \int_{t_0}^{s} \frac {\partial } {\partial u} \big[ \zeta(x(t_0),t_0) + \frac {\partial \zeta(x(t_0),t_0)} {\partial u} |^{u=t_0} (u-t_0) \\

+ \frac {\partial^2 \zeta(x(t_0),t_0)} {\partial u^2} |^{u=t_0} \frac{(u-t_0)^2}{2} \big] du \big] ds \\

=\int_{t_0}^{t} \frac{\partial}{\partial s} \big[ \zeta(x(t_0),t_0) + \frac {\partial \zeta(x(t_0),t_0)} {\partial s} |^{s=t_0} (s-t_0) + \frac {\partial^2 \zeta(x(t_0),t_0)} {\partial s^2} |^{s=t_0} \frac{(s-t_0)^2}{2} \\

+\frac {\partial^3 \zeta(x(t_0),t_0)} {\partial s^3} |^{s=t_0} \frac{(s-t_0)^3}{6} \big] ds \\

= \frac{\partial \zeta(x(t_0),t_0)}{\partial t} |^{t=t_0} (t-t_0) + \frac {\partial^2 \zeta(x(t_0),t_0)} {\partial t^2} |^{t=t_0} \frac{(t-t_0)^2}{2} \\

+ \frac {\partial^3 \zeta(x(t_0),t_0)} {\partial t^3} |^{t=t_0} \frac{(t-t_0)^3}{6} + \frac {\partial^4 \zeta(x(t_0),t_0)} {\partial t^4} |^{t=t_0} \frac{(t-t_0)^4}{24}[$]

Equation(6)

In the above I used product rule for derivatives with the fact that [$](s-t_0)=0[$] when [$]s=t_0[$]

Suppose we are given a PDE as

[$]\frac {\partial \zeta} {\partial t} = a \frac{ \partial^2 \zeta}{\partial x^2} + b \frac{ \partial \zeta}{\partial x} + c \zeta[$]

We can use the following formulas in the expansion of Equation(6) and other expansions.

[$]\frac {\partial \zeta(x(t_0),t_0)} {\partial t} = a \frac{ \partial^2 \zeta(x(t_0),t_0)}{\partial x^2} + b \frac{ \partial \zeta(x(t_0),t_0)}{\partial x} + c \zeta(x(t_0),t_0)[$]

and

[$]\frac {\partial^2 \zeta(x(t_0),t_0)} {\partial t^2} = (a \frac{ \partial^2}{\partial x^2} + b \frac{ \partial}{\partial x} + c)^2 [\zeta(x(t_0),t_0)][$]

and

[$]\frac {\partial^n \zeta(x(t_0),t_0)} {\partial t^n} = (a \frac{ \partial^2}{\partial x^2} + b \frac{ \partial}{\partial x} + c)^n [\zeta(x(t_0),t_0)][$]

In the above attempt at explanation I have only expanded the first integral in Equation(2), the expansion for second integral(or third term) in Equation(2) though very similar but a little bit more tedious. I will write that in detail tomorrow and complete the solution expansion and then try to follow with numerical examples over the weekend.

We again write Equation(2) from the previous posts as

[$]\zeta(x(t),t)=\zeta(x(t_0),t_0)+\int_{t_0}^{t} \frac{\partial \zeta(x(t_0),s)}{\partial s} ds + \int_{x(t_0)}^{x(t)} \frac{\partial \zeta(x(s),t)}{\partial x} dx(s)[$]

Equation(2)

and try to solve the second integral.

[$]\int_{x(t_0)}^{x(t)} \frac{\partial \zeta(x(s),t)}{\partial x} dx(s)[$]

[$]=\int_{x(t_0)}^{x(t)} \frac{\partial}{\partial x} \big [ \zeta(x(t_0),t_0) + \int_{t_0}^{t} \frac {\partial \zeta(x(t_0),s)}{\partial s} ds \\

+ \int_{x(t_0)}^{x(s)} \frac {\partial \zeta(x(u),t)}{\partial x} dx(u) \big ] dx(s)[$]

[$]=\int_{x(t_0)}^{x(t)} \frac{\partial}{\partial x} \big [ \zeta(x(t_0),t_0) + \int_{t_0}^{t} \frac {\partial \zeta(x(t_0),s)}{\partial s} ds \\

+ \int_{x(t_0)}^{x(s)} \frac{\partial}{\partial x} \big [ \zeta(x(t_0),t_0) + \int_{t_0}^{t} \frac {\partial \zeta(x(t_0),u)}{\partial u} du \\

+ \int_{x(t_0)}^{x(u)} \frac {\partial \zeta(x(v),t)}{\partial x} dx(v) \big ] dx(u) \big ] dx(s)[$]

[$]=\int_{x(t_0)}^{x(t)} \frac{\partial}{\partial x} \big [ \zeta(x(t_0),t_0) + \int_{t_0}^{t} \frac {\partial \zeta(x(t_0),s)}{\partial s} ds \\

+ \int_{x(t_0)}^{x(s)} \frac{\partial}{\partial x} \big [ \zeta(x(t_0),t_0) + \int_{t_0}^{t} \frac {\partial \zeta(x(t_0),u)}{\partial u} du \\

+ \int_{x(t_0)}^{x(u)} \frac{\partial}{\partial x} \big [ \zeta(x(t_0),t_0) + \int_{t_0}^{t} \frac {\partial \zeta(x(t_0),v)}{\partial v} dv \\

+ \int_{x(t_0)}^{x(v)} \frac {\partial \zeta(x(w),t)}{\partial x} dx(w) \big ] dx(v) \big ] dx(u) \big ] dx(s)[$]

Equation (8)

We can divide the above series into two series so that integrals in each series are similar as

Series (1) is given as

[$]=\int_{x(t_0)}^{x(t)} \frac{\partial}{\partial x} \big [ \zeta(x(t_0),t_0) + \int_{x(t_0)}^{x(s)} \frac{\partial}{\partial x} \big [ \zeta(x(t_0),t_0) \\

+ \int_{x(t_0)}^{x(u)} \frac{\partial}{\partial x} \big [ \zeta(x(t_0),t_0) + \int_{x(t_0)}^{x(v)} \frac{\partial}{\partial x} \big [ \zeta(x(t_0),t_0) + \dots \big ] dx(w) \big ] dx(v) \big ] dx(u) \big ] dx(s) [$]

Equation (9)

While series (2) is given as

[$]\int_{x(t_0)}^{x(t)} \frac{\partial}{\partial x} \big [ \int_{t_0}^{t} \frac {\partial \zeta(x(t_0),s)}{\partial s} ds + \int_{x(t_0)}^{x(s)} \frac{\partial}{\partial x} \big [ \int_{t_0}^{t} \frac {\partial \zeta(x(t_0),u)}{\partial u} du \\

+ \int_{x(t_0)}^{x(u)} \frac{\partial}{\partial x} \big [ \int_{t_0}^{t} \frac {\partial \zeta(x(t_0),v)}{\partial v} dv + \int_{x(t_0)}^{x(v)} \frac{\partial}{\partial x} \big [ \int_{t_0}^{t} \frac {\partial \zeta(x(t_0),w)}{\partial w} dw + \dots \big ] dx(w) \big ] dx(v) \big ] dx(u) \big ] dx(s) [$]

Equation(10)

Series (1) is solved and results in a simple Taylor series solution in one dimension as we found for time integrals in the previous posts and the solution is given as

[$]= \frac {\partial \zeta(x(t_0),t_0)} {\partial x} (x(t)-x(t_0)) + \frac {\partial^2 \zeta(x(t_0),t_0)} {\partial x^2} \frac {(x(t)-x(t_0))^2}{2} \\

+ \frac {\partial^3 \zeta(x(t_0),t_0)} {\partial x^3} \frac {(x(t)-x(t_0))^3}{6} + \frac {\partial^4 \zeta(x(t_0),t_0)} {\partial x^4} \frac {(x(t)-x(t_0))^4}{24} + \dots [$]

Equation (11)

while the solution to series (2) is given as

[$]= \frac {\partial} {\partial x} \big [ \int_{t_0}^{t} \frac {\partial \zeta(x(t_0),s)}{\partial s} ds \big ] (x(t)-x(t_0)) + \frac {\partial^2} {\partial x^2} \big [ \int_{t_0}^{t} \frac {\partial \zeta(x(t_0),s)}{\partial s} ds \big ] \frac {(x(t)-x(t_0))^2}{2} \\

+ \frac {\partial^3 } {\partial x^3} \big [ \int_{t_0}^{t} \frac {\partial \zeta(x(t_0),s)}{\partial s} ds \big ] \frac {(x(t)-x(t_0))^3}{6} + \frac {\partial^4} {\partial x^4} \big [ \int_{t_0}^{t} \frac {\partial \zeta(x(t_0),s)}{\partial s} ds \big ] \frac {(x(t)-x(t_0))^4}{24} + \dots [$]

Equation (12)

where solution to each of the time integrals inside large brackets in the above equation is given by expansion in Equation (6) of previous post and this series would correspond to cross derivatives in two dimensional Taylor series.