Template:Numerical calculation of Q

Numerical Calculation of [math]\displaystyle{ \mathbf{Q} }[/math]

We begin by truncating to a finite number ([math]\displaystyle{ N }[/math]) of evanescent modes,

[math]\displaystyle{ \mathbf{Q}\left[ f \right] =-\sum_{m=0}^{N}k_{m}\int_{-h}^{0} f\left( s\right) \phi_{m}\left( s\right) \mathrm{d}s \frac{\phi _{m} \left( z\right)}{A_m} . }[/math]

We calculate the integral with the same panels as we used to approximate the integrals of the Green function and its normal derivative . Similarly, we assume that [math]\displaystyle{ f(s) \, }[/math] is constant over each panel and integrate [math]\displaystyle{ \phi _{m}\left( s\right) }[/math] exactly. This gives us the following matrix factorisation of [math]\displaystyle{ \mathbf{Q} }[/math]

[math]\displaystyle{ \mathbf{Q}[f]=\mathbf{S}\,\mathbf{R}\,[f]. }[/math]

The components of the matrices [math]\displaystyle{ \mathbf{S} }[/math] and [math]\displaystyle{ \mathbf{R} }[/math] are

[math]\displaystyle{ s_{jm} = -k_m\phi _{m}\left( z_{j}\right) , }[/math] [math]\displaystyle{ r_{mj} = \frac{1}{A_m} \int_{z_{j}-\Delta x/2}^{z_{j}+\Delta x/2}\phi _{m}\left( s\right) \mathrm{d}s }[/math]

Reflection and Transmission Coefficients

Recall that our Sommerfeld radiation condition can be expressed in the form

[math]\displaystyle{ \lim\limits_{x\rightarrow -\infty }\phi \left( x,z\right) = \phi_0 e^{-k_{0}x} +R\phi_0 e^{k_{0}x}, }[/math]

and that the potential in the region [math]\displaystyle{ \Omega }[/math] is of the form

[math]\displaystyle{ \phi \left( x,z\right) = \phi_0(z)e^{-k_0 x} +\sum_{m=0}^{\infty } a_m \phi _{m}\left( z\right) e^{k_{m}\left( x+l\right) }. }[/math]

Note that for this series, if [math]\displaystyle{ m=0 }[/math], then [math]\displaystyle{ k_m }[/math] is imaginary, giving rise to a propagating wave. For [math]\displaystyle{ m \geq 1 }[/math], there is only a local contribution to this propagating wave -- in the extremes, there is no contribution (evanescent modes).

So when looking at the Reflected and Transmitted waves, we only consider [math]\displaystyle{ m=0 }[/math], that is,

[math]\displaystyle{ \begin{align} \lim\limits_{x\rightarrow -\infty }\phi &= \phi_0 e^{-k_{0}x} + a_0 \phi_0 e^{k_0 (x+l)} ,\\ &= \phi_0 e^{-k_{0}x} +R\phi_0 e^{k_{0}x}. \\ \end{align} }[/math]

Consequently, it is straightforward to see that [math]\displaystyle{ R = a_0 e^{k_0 l} }[/math]. Recall from earlier that

[math]\displaystyle{ a_m +\delta_{m0}e^{k_0 l} = \frac{1}{A_m} \int_{-h}^{0} f(z) \phi_m(z) \mathrm{d}z, }[/math]

therefore,

[math]\displaystyle{ a_0= \left[ \frac{1}{A_0} \int_{-h}^{0} f(z) \phi_0(z) \mathrm{d}z - e^{k_0 l} \right]. }[/math]

However, we make use of the fact that [math]\displaystyle{ \mathbf{Q}[f]=\mathbf{S}\,\mathbf{R}\,[f] }[/math], where the components of the matrix [math]\displaystyle{ \mathbf{R} }[/math] is

[math]\displaystyle{ r_{mj} = \frac{1}{A_m} \int_{z_{j}-\Delta x/2}^{z_{j}+\Delta x/2}\phi _{m}\left( s\right) \mathrm{d}s, }[/math]

which admits the representation

[math]\displaystyle{ a_0= \left[ \sum_{j} r_{0j} f(z_j) - e^{k_0 l} \right]. }[/math]

Consequently,

[math]\displaystyle{ R= \left[ \sum_{j} r_{0j} f(z_j) - e^{k_0 l} \right]e^{k_0 l}. }[/math]

So in summary, if we multiply the potential at the left (after subtracting the incident wave) and the right by [math]\displaystyle{ \mathbf{R} }[/math] we can calculate the coefficients in the eigenfunction expansion, and hence determine the reflection and transmission coefficient. where [math]\displaystyle{ z_{j} }[/math] is the value of the [math]\displaystyle{ z }[/math] coordinate in the centre of the panel and [math]\displaystyle{ \Delta x }[/math] is the panel length.