Ok, lets take . Then and by the last equation of the previous post for we get:
So, is definitly a solution, just like any , right?
Now, I haven't used, that has to be the identity on the boundary?
Still need help!
Hello.
I'm trying to proof a statement made in Anisotropic conductivities that cannot be detected by EIT:
Let , n=2 or 3. The equation we are looking at is , where is the conductivity (symmetric, positive definite). In the paper, anisotropic conductivities are constructed, which can't be detected by measurements on the boundary.
They do this by using a change of variables and say, that the anisotropy of the conductivity is crucial, because otherwise, the construction imposes, that the only valid change of variables is the identity. That's what I'm trying to understand.
So, let \rightarrow D" alt="F\rightarrow D" /> be a diffeomorphism, it's jacobian.
Under the change of variables , we call the push-forward of , that is
, where .
The main idea is, that with , satisfies - so the same equation as above, just another conductivity - and if on the boundary, the two conductivities give the same boundary measurements (Dirichlet-to-Neumann map).
Now, what I'm trying to proof is:
If is isotropic, i.e. with , and the push-forward is also isotropic, i.e. , and on the boundary, then is the identity on .
Using that is isotropic, I get
and the isotropy of gives
Now, I guess what I got to do, is showing that is the identity matrix. Anybody got an idea, how to do that?
Thanks.
Ok, lets take . Then and by the last equation of the previous post for we get:
So, is definitly a solution, just like any , right?
Now, I haven't used, that has to be the identity on the boundary?
Still need help!