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Untitled - Cdm.unimo.it

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Ordinary Differential Equations 195<br />

(9.3.18) Bw(ψ,φ) :=<br />

(9.3.19) Fw(φ) :=<br />

<br />

<br />

(ψ −<br />

I<br />

˜ ψ) ′ [(φ − ˜ φ)w] ′ dx +<br />

<br />

+ µ ψ(φ −<br />

I<br />

˜ <br />

φ)w dx + µ<br />

I<br />

f(φ − ˜ φ)w dx +<br />

<br />

I<br />

I<br />

<br />

I<br />

˜ψ ′ ˜ φ ′ dx<br />

ψ ˜ φ dx, ∀ψ,φ ∈ X,<br />

f ˜ φ dx + σ2φ(1) − σ1φ(−1), ∀φ ∈ X.<br />

In the above expressions, for a given function χ ∈ X, we have defined ˜χ ∈ X to be<br />

the function ˜χ(x) := 1<br />

1<br />

2 (1 − x)χ(−1) + 2 (1 + x)χ(1), x ∈ Ī. We note that ˜χ ∈ P1<br />

and χ − ˜χ ∈ H 1 0,w(I). For ν = 0, (9.3.18) and (9.3.19) give respectively (9.3.15) and<br />

(9.3.16). As usual, we are interested in finding the solution U ∈ X of (9.3.4).<br />

W<strong>it</strong>hout entering into details of the proof, we claim that <strong>it</strong> is possible to find a constant<br />

µ ∗ > 0 such that, for any µ ∈]0,µ ∗ ], the bilinear form Bw in (9.3.18) fulfills the<br />

requirements (9.3.5) and (9.3.6). This result is given in funaro(1988) for the Chebyshev<br />

case and can be easily generalized to other Jacobi weights. This implies existence and<br />

uniqueness of a weak solution U. When f ∈ C0 ( Ī), we can relate U to problem (9.1.5)<br />

by arguing as follows. Considering that ˜ φ ′ ∈ P0 is a constant function, we wr<strong>it</strong>e<br />

(9.3.20)<br />

=<br />

<br />

<br />

Ũ<br />

I<br />

′ φ ˜′ dx = (Ũ(1) − Ũ(−1))˜ φ ′ = (U(1) − U(−1)) ˜ φ ′<br />

U<br />

I<br />

′ φ ˜′ ′<br />

dx = U (1) φ(1) ˜ ′<br />

− U (−1) φ(−1) ˜ −<br />

Therefore, since Ũ ′′ ≡ 0 and ˜ φ(±1) = φ(±1), one has<br />

<br />

(9.3.21) Bw(U,φ) = (−U ′′ + µU)(φ − ˜ φ)w dx +<br />

I<br />

<br />

I<br />

<br />

I<br />

U ′′ ˜ φ dx.<br />

(−U ′′ + µU) ˜ φ dx<br />

+ U ′ (1)φ(1) − U ′ (−1)φ(−1) = Fw(φ), ∀φ ∈ X.<br />

One first uses test functions φ belonging to H 1 0,w(I) (therefore ˜ φ ≡ 0). This implies<br />

the differential equation −U ′′ + µU = f in I. Finally, we eliminate the integrals and<br />

obtain the boundary cond<strong>it</strong>ions.

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