# Thread: Electric impedance tomography, conductivities, change of variables

1. ## Electric impedance tomography, conductivities, change of variables

Hello.

I'm trying to proof a statement made in Anisotropic conductivities that cannot be detected by EIT:

Let $\displaystyle D\subset \mathbb{R}^n$, n=2 or 3. The equation we are looking at is $\displaystyle \nabla\cdot(\sigma\nabla u) = 0, u|_{\partial D}=f$, where $\displaystyle \sigma=(\sigma_{ij})$ 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 $\displaystyle F\rightarrow D$ be a diffeomorphism, $\displaystyle DF$ it's jacobian.

Under the change of variables $\displaystyle y=F(x)$, we call $\displaystyle F_{*}\sigma = \frac{1}{\det DF}DF\cdot\sigma\cdot DF^T$ the push-forward of $\displaystyle \sigma$, that is

$\displaystyle (F_{*}\sigma)_{ij}(y)=\frac{1}{\det DF}\sum_{k,l=1}^n\frac{\partial F_i}{\partial x_k}(x)\frac{\partial F_j}{\partial x_l}(x)\sigma_{kl}(x)$, where $\displaystyle x=F^{-1}(y)$.

The main idea is, that with $\displaystyle v = u \circ F$, $\displaystyle v$ satisfies $\displaystyle \nabla\cdot(F_{*}\sigma\nabla v) = 0$ - so the same equation as above, just another conductivity - and if $\displaystyle F=id$ on the boundary, the two conductivities give the same boundary measurements (Dirichlet-to-Neumann map).

Now, what I'm trying to proof is:

If $\displaystyle \sigma$ is isotropic, i.e. $\displaystyle \sigma_{ij}(x)=\gamma(x)\delta_{ij}$ with $\displaystyle \gamma(x)>0$, and the push-forward $\displaystyle F_{*}\sigma$ is also isotropic, i.e. $\displaystyle (F_{*}\sigma)_{ij}(y)=\eta(y)\delta_{ij}$, and $\displaystyle F=id$ on the boundary, then $\displaystyle F$ is the identity on $\displaystyle D$.

Using that $\displaystyle \sigma$ is isotropic, I get

$\displaystyle (F_{*}\sigma)_{ij} = c \sum_{k}\frac{\partial F_i}{\partial x_k}\frac{\partial F_j}{\partial x_k}\gamma, c = \det(DF)^{-1}$

and the isotropy of $\displaystyle F_{*}\sigma$ gives

$\displaystyle c \sum_{k}\frac{\partial F_i}{\partial x_k}\frac{\partial F_j}{\partial x_k}\gamma = \eta\delta_{ij}$

Now, I guess what I got to do, is showing that $\displaystyle DF$ is the identity matrix. Anybody got an idea, how to do that?

Thanks.

2. ## n=2

Ok, lets take $\displaystyle n=2$. Then $\displaystyle F = (F_1(x_1, x_2), F_2(x_1, x_2))$ and by the last equation of the previous post for $\displaystyle i\neq j$ we get:

$\displaystyle \frac{\partial F_1}{\partial x_1}\frac{\partial F_2}{\partial x_1} + \frac{\partial F_1}{\partial x_2}\frac{\partial F_2}{\partial x_2} = 0$

So, $\displaystyle F=id$ is definitly a solution, just like any $\displaystyle F = (F_1(x_1), F_2(x_2))$, right?

Now, I haven't used, that $\displaystyle F$ has to be the identity on the boundary?

Still need help!