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I have two partial differential equations that I want to solve (for $\ \sigma $) by finite differences: $\ -\frac{\partial \sigma}{\partial x}(x,y,t) -p(x,y,t)\frac{\partial \sigma}{\partial t}(x,y,t) = p(x,y,t)$

$\ -\frac{\partial \sigma}{\partial y}(x,y,t) -q(x,y,t)\frac{\partial \sigma}{\partial t}(x,y,t) = q(x,y,t)$

where p and q are known.

This is an over determined system (one unknown, two equations) so I am seeking an Least Squares Error (LSE) solution.

I know that for the inversion problem:

$\ d = \mathbf{F}m $

, where F is a forward operator, the LSE is:

$\ \mathbf{F}^T\mathbf{F}m = \mathbf{F}^Td$

In my case $\ \mathbf{d}=[p(x,y,t),q(x,y,t)]^T$, $\ \mathbf{m} = [\sigma] $ and $\ \mathbf{F} =[-\frac{\partial}{\partial x}-p\frac{\partial}{\partial t}, -\frac{\partial}{\partial y}-q\frac{\partial}{\partial t}]^T$ ?

How do I solve this by finite differences?

Thanks in advance for any answers!


Thanks to Andrew we can instead solve the easier equation:

$\ qs_x=ps_y $

for each discrete t value.

I have m discrete x values: $\ x_i, 1<=i<=m $ and n discrete y values: $\ y_j, 1<=j<=n $.

My boundary condition is:

$\ s(m/2,n/2)=t $

I could of course just integrate outwards from the center point, but this would yield a different result depending on the path taken.

Therefore I am looking for a LSE solution to this equation. How can I do this?

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up vote 3 down vote accepted

The problem can be simplified a bit. Putting $s(x,y,t)=\sigma(x,y,t)+t$ turns the system into a homogeneous one: $$ -s_x-ps_t=0, $$ $$ -s_y-qs_t=0. $$ Excluding $s_t$ we have $qs_x=ps_y\;$.

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Thanks! That made the equations much simpler. – Andy Jun 26 '12 at 11:53
But how do I find the LSE for this? – Andy Jun 26 '12 at 12:01
There is at least one exact solution $s\equiv0$ or $\sigma\equiv-t$. Besides for concrete $p$, $q$ it is possible to try to find exact solutions (if any) by the characteristics method and then substitute the general solution into one of the equations. – Andrew Jun 26 '12 at 17:34
For $qs_x=ps_y\;$, $s(x,x)=t\;$ the solution is $s(x,y)=t$. – Andrew Jun 27 '12 at 5:26

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