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Cuchulainn
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### Transforming Two-Factor PDEs containing Mixed Derivatives to Canonical Form

This technique allows us to trasform e.g, Heston, Basket pdes to pdes with no mixed derivatives.

Taking Heston$(x,y)$ the original pde is

$\frac{1}{2} y x^2 \frac{\partial^2 u}{\partial x^2} + \rho\sigma xy\frac{\partial^2 u}{\partial x \partial y} + \frac{1}{2} \sigma^2 y \frac{\partial^2 u}{\partial y^2}$ ... + ... lower-order terms

By a change of coordinates

$\xi = \sigma\rho log(x) - y$ and $\eta = \sigma\sqrt{(1-\rho^2)}log(x)$

we get a Heston$(\xi, \eta)$ with no mixed derivatives for the principal part of the pde (which can't be bad).

$A \frac{\partial^2 u}{\partial \xi^2} + B \frac{\partial^2 u}{\partial \eta^2}$ ... + ... lower-order terms

where $A = B = \frac{1}{2} \sigma^2 y (1 -\rho^2)$ and $y = (\rho/(1-\rho^2)\eta - \xi$

The question now is:
"The original domain is the positive quarter plane; what is the $\xi-\eta$ domain and most importantly how to prescribe numerical boundary conditions?

The answer could be easy or difficult.
Last edited by Cuchulainn on February 12th, 2020, 2:12 pm, edited 5 times in total.
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Jean Piaget

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### Re: Transforming Two-Factor PDEs containing Mixed Derivatives to Canonical Form

Could you also use Gram-Schmidt?

Btw, are you not missing a drift term for the variance?

Cuchulainn
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### Re: Transforming Two-Factor PDEs containing Mixed Derivatives to Canonical Form

Could you also use Gram-Schmidt?

Btw, are you not missing a drift term for the variance?
What is Orthogonalization?
No. GS is linear algebra? Mine is PDE
I don't think so. You mean advection/convection term in pde? It's not important here.

The question is what is the new boundary?
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Every Time We Teach a Child Something, We Keep Him from Inventing It Himself
Jean Piaget

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### Re: Transforming Two-Factor PDEs containing Mixed Derivatives to Canonical Form

I think I posted while you were still editing your question, in particular regarding the boundary condition, so pls ignore my post.

But to clarify: partial derivatives are vectors, the cross term imply the vectors are not orthogonal. You can always do a local coordinate transform such that the cross-terms disappears. This is what I meant with orthogonalisation, and I have actually done this using GS for a two asset basket option.

Cuchulainn
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### Re: Transforming Two-Factor PDEs containing Mixed Derivatives to Canonical Form

I think I posted while you were still editing your question, in particular regarding the boundary condition, so pls ignore my post.

But to clarify: partial derivatives are vectors, the cross term imply the vectors are not orthogonal. You can always do a local coordinate transform such that the cross-terms disappears. This is what I meant with orthogonalisation, and I have actually done this using GS for a two asset basket option.
Fair enough, I don't use vectors with PDEs because the canonical stuff is more suitable and possibly more straightforward, maybe.
The question again is what is the boundary after the transformation.

Here is what I am using. It is a hard requirement. There's are other methods but they are outside the scope.

https://en.wikipedia.org/wiki/Elliptic_ ... l_equation
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Every Time We Teach a Child Something, We Keep Him from Inventing It Himself
Jean Piaget

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### Re: Transforming Two-Factor PDEs containing Mixed Derivatives to Canonical Form

This is likely too simplistic but $\xi$ can be written as $\xi = \log (x^{\rho\sigma} / e^y)$, so $e^y$ will dominate the limit behaviour? I think there are four limits to consider for $\xi$ and also whether the correlation is positive or negative, but in all cases I think the domain will be $( -\infty,\infty)$, also for $\eta$.

On the boundary conditions, doesn't that depend on the original (non-canonical) boundary condition?

Cuchulainn
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### Re: Transforming Two-Factor PDEs containing Mixed Derivatives to Canonical Form

This is likely too simplistic but $\xi$ can be written as $\xi = \log (x^{\rho\sigma} / e^y)$, so $e^y$ will dominate the limit behaviour? I think there are four limits to consider for $\xi$ and also whether the correlation is positive or negative, but in all cases I think the domain will be $( -\infty,\infty)$, also for $\eta$.

On the boundary conditions, doesn't that depend on the original (non-canonical) boundary condition?
1), yes I think that's reasonable.
2) there should be some kind of relationship,  but specifying it is the core question of this thread.
http://www.datasimfinancial.com
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Every Time We Teach a Child Something, We Keep Him from Inventing It Himself
Jean Piaget

frolloos
Posts: 1620
Joined: September 27th, 2007, 5:29 pm
Location: Netherlands

### Re: Transforming Two-Factor PDEs containing Mixed Derivatives to Canonical Form

2) there should be some kind of relationship,  but specifying it is the core question of this thread.

Writing $\{ x^i \} = (x,y)$ and $\{ \tilde{x}^i \} = (\xi,\eta)$.

I am assuming the non-canonical boundary condition can be written as $B_i u(x,y) = 0$ with $B_i = v(x,y) \frac{\partial}{\partial x^i} + w(x,y)$

Maybe I am not fully understanding your question, but since $x^i = x^i (\tilde{x}^j)$ can you not apply the chain rule to write the boundary conditions in the new coordinates?

I read a more elegant formulation / solution (but boils down to the same thing I believe) is given by minimization of the energy functional associated to the elliptic pde. See the answer to the question in this thread.

Cuchulainn
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### Re: Transforming Two-Factor PDEs containing Mixed Derivatives to Canonical Form

This is likely too simplistic but $\xi$ can be written as $\xi = \log (x^{\rho\sigma} / e^y)$, so $e^y$ will dominate the limit behaviour? I think there are four limits to consider for $\xi$ and also whether the correlation is positive or negative, but in all cases I think the domain will be $( -\infty,\infty)$, also for $\eta$.

On the boundary conditions, doesn't that depend on the original (non-canonical) boundary condition?
This sounds reasonable/doable. The new coordinates $(\xi,\eta)$ can take on any positive or negative values, thus boundary conditions are not even defined (BTW I have no problem with that, in fact I prefer it that way). Now I can map them to a pde on  $(0,1)^2$ and then I am on the home straight as far as (numerical) BCs are concerned.

We had a discussion on this a while back (thread got clobbered on the Night of the Big Wind, a shame). With excellent replies by @ppauper.

https://forum.wilmott.com/viewtopic.php?f=34&t=62593&hilit=well+posed&start=15

Just to summarise

1. Mixed derivatives must go (90% of fdm schemes don't really 'work')
2. It must be easy to produce boundary conditions (mathematically) without handwaving, but heuristics support is OK. (in fact, we can show mathematically  that BCs are not allowed by 1) Fichera theory and 2) energy inequalities).
3. Applicable to a range of 2-factor PDEs.
4. Easy to implement, However, 2 successive transformations needed in 2d space $(x,y) \leftrightarrow ( \xi, \eta) \leftrightarrow (z,w)$.

I'll get back.
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Every Time We Teach a Child Something, We Keep Him from Inventing It Himself
Jean Piaget

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