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Image Reconstruction for 3D Lung Imaging - Department of Systems ...

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Substitution <strong>of</strong> 2.24 into 2.21 yields<br />

�<br />

[∇ · (vσ∇ũ) − σ∇ũ · ∇v]dΩ = 0 (2.25)<br />

Ω<br />

Rearranging gives<br />

�<br />

�<br />

∇ · (vσ∇ũ)dΩ =<br />

Ω<br />

Ω<br />

σ∇ũ · ∇vdΩ (2.26)<br />

Gauss’ Theorem is the older name <strong>for</strong> the Divergence theorem which is a mathematical<br />

statement <strong>of</strong> the physical fact that, in the absence <strong>of</strong> the creation or<br />

destruction <strong>of</strong> matter, the density within a region <strong>of</strong> space can change only by<br />

having it flow into or away from the region through its boundary. Gauss’ Theorem<br />

is valid in any dimension, however in 2D it is �<br />

∇ · FdV = �<br />

FdS where V is<br />

volume and S is surface area.<br />

Invoking Gauss’ Theorem on 2.26 permits the introduction <strong>of</strong> boundary conditions:<br />

�<br />

�<br />

vσ∇ũ · ˆndΓ = σ∇ũ · ∇vdΩ (2.27)<br />

∂Ω<br />

Ω<br />

Note that ∇φ · ˆn = ∂φ<br />

∂ˆn so 2.27 could also be written<br />

�<br />

�<br />

σ∇ũ · ∇vdΩ = vσ ∂ũ<br />

dΓ (2.28)<br />

∂ˆn<br />

Ω<br />

∂Ω<br />

The boundary integral only needs to be carried out <strong>for</strong> elements underneath electrodes. The<br />

left side <strong>of</strong> 2.28 is <strong>for</strong> the entire mesh. When examined <strong>for</strong> a single triangular 2D element,<br />

k, the left side is<br />

�<br />

σk∇ũ · ∇vdΩ (2.29)<br />

Ek<br />

Substituting the definitions <strong>of</strong> the interpolating versions <strong>of</strong> v and u yields<br />

�<br />

Ek<br />

σk∇<br />

3�<br />

uiφi · ∇<br />

i=1<br />

V<br />

3�<br />

wjũj dΩ (2.30)<br />

i=1<br />

If the conductivity, σk, is constant over a single element then the nodal voltages ui, the<br />

coefficients <strong>for</strong> the weighing functions, and the summations can be moved outside <strong>of</strong> the<br />

integral:<br />

3�<br />

3�<br />

�<br />

σk ui wj<br />

i=1 j=1<br />

Ek<br />

∇φi · ∇φj dΩ (2.31)<br />

It is common to break out the integral part <strong>of</strong> the equation as<br />

�<br />

S k ij =<br />

Ek<br />

∇φi · ∇φj dΩ (2.32)<br />

22<br />

∂V

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