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Untitled - Cdm.unimo.it

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Time-Dependent Problems 239<br />

Since (10.5.4) is in a diagonal form, V and W are also called characteristic variables.<br />

From (10.5.2), the two unknowns are coupled via the boundary relations<br />

(10.5.5)<br />

where R = L = −1.<br />

⎧<br />

⎨V<br />

(−1,t) = L W(−1,t)<br />

⎩<br />

W(1,t) = R V (1,t)<br />

∀t ∈]0,T],<br />

The physical interpretation of formula (10.5.5) is that the travelling waves are reflected<br />

at the endpoints of the vibrating string. First-order systems, such as (10.5.4), are often<br />

encountered in fluid dynamics models (see kreiss and lorenz(1989), section 7.6).<br />

For spectral type approximations, results are generally obtained for hyperbolic<br />

systems wr<strong>it</strong>ten in terms of characteristic variables which are combined together at the<br />

boundary points by (10.5.5), w<strong>it</strong>h R and L satisfying |RL| < 1 (note that for the wave<br />

equation one has instead |RL| = 1). In this s<strong>it</strong>uation, the ampl<strong>it</strong>ude of the oscillations<br />

decays for t → +∞ (energy dissipation due to friction effects).<br />

The collocation method is obtained in the following way. We are concerned w<strong>it</strong>h<br />

finding pn(·,t) ∈ Pn, qn(·,t) ∈ Pn, n ≥ 1, t ∈]0,T], such that<br />

(10.5.6)<br />

⎧<br />

⎪⎨<br />

⎪⎩<br />

∂pn<br />

∂t (η(n)<br />

i ,t) = −ζ ∂pn<br />

∂x (η(n)<br />

i ,t) 1 ≤ i ≤ n, t ∈]0,T],<br />

∂qn<br />

∂t (η(n)<br />

i ,t) = ζ ∂qn<br />

∂x (η(n)<br />

i ,t) 0 ≤ i ≤ n − 1, t ∈]0,T].<br />

At the boundaries, we require that<br />

⎧<br />

⎨pn(−1,t)<br />

= L qn(−1,t)<br />

(10.5.7)<br />

⎩<br />

qn(1,t) = R pn(1,t)<br />

∀t ∈]0,T].<br />

When R and L in (10.5.7) satisfy |LR| < 1, the polynomials pn(·,t) and qn(·,t)<br />

decay to zero when t → +∞ (see lustman(1986)). It is an easy exercise to set up<br />

the 2n × 2n system of ordinary differential equations corresponding to (10.5.6). We<br />

note that two unknowns can be eliminated by virtue of (10.5.7). For n = 2, we get for<br />

example

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