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THESE de DOCTORAT - cerfacs

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3.3 Solving the system Ax = b 47<br />

[<br />

Ai,j + B i,j<br />

] ˆpo = C j ⇒ A i,j ˆp o = C j (3.15)<br />

in which B i,j = 0. Finally, there is another boundary condition that imposes a relation between<br />

the acoustic velocity normal to the surface û · n and the acoustic pressure ˆp. Injecting the<br />

<strong>de</strong>finition of the acoustic impedance Ẑ = ˆp/iωû · n into Eq. (3.14) leads to:<br />

Ẑ∇ ˆp ·⃗n − iω ˆp = 0 (3.16)<br />

This type of boundary condition is known as a ‘Robin’ boundary condition. Note that B i,j ̸= 0<br />

as soon as Ẑ ̸= 0 and Ẑ ̸= ∞ over some piece of the boundaries of the flow domain.<br />

3.3 Solving the system Ax = b<br />

From Eq. (3.9) it can be seen that the algebraic system to solve is of the form Ax = b. Several<br />

approaches can be used to solve linear systems; they are divi<strong>de</strong>d mainly in two groups: noniterative<br />

or ‘direct’ methods and iterative methods. Within ‘direct’ methods, two approaches are<br />

most often used: the LU (or Gauss Elimination) and the QR <strong>de</strong>composition. Both of them have<br />

the same philosophy: “divi<strong>de</strong> and conquer". In mathematical terms it means<br />

A = LU or A = QR (3.17)<br />

where A is a n × n matrix, L is a lower triangular n × n matrix, U is an upper triangular n × n<br />

matrix, Q is an orthonormal 3 n × n matrix and R is an upper triangular n × n matrix. When<br />

solving the linear system by the LU <strong>de</strong>composition, three steps are followed<br />

1. Ax = b → LUx = b<br />

2. solve Lw = b for the unknown w<br />

3. solve Ux = w for the unknown x.<br />

If a QR <strong>de</strong>composition is preferred, three steps are also consi<strong>de</strong>red<br />

1. Ax = b → QRx = b<br />

2. solve Qw = b for the unknown w<br />

3 An orthornormal matrix is a matrix with n orthogonal columns in which the norm of each column is equal to<br />

one.

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