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Tutorato di Analisi 3 - Dipartimento di Matematica

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( √1<br />

3. F (x 1 , x 2 , y 1 , y 2 ) = + x1 − 1 − e −y1 + cos y 2 , log(cosh x 1 ) − sin(x 1x 2 )<br />

1 + x 2 1 + y1<br />

2 + arctan y 2<br />

).<br />

(a) F è <strong>di</strong> classe C 1 in<br />

(<br />

un intorno dell’origine, inoltre<br />

)∣ F (0, 0, 0, 0) = (0, 0)<br />

e ∂F<br />

∂y (0, 0, 0, 0) = e −y1 − sin y 2<br />

2y 1 sin(x 1x 2) 1<br />

=<br />

(1+y1) 2 2 1+y2<br />

∣∣∣∣(x1,x 2<br />

( )<br />

( ) 2,y 1,y 2)=(0,0,0,0)<br />

−1 ( )<br />

1 0<br />

∂F<br />

1 0<br />

= è invertibile (con T =<br />

0 1<br />

∂y (0, 0, 0, 0) = ),<br />

0 1<br />

dunque per il teorema della funzione implicita ∃ r, ρ > 0 e g ∈<br />

∈ C 1 (B r ((0, 0)), B ρ ((0, 0))) tale che F (x 1 , x 2 , g 1 (x), g 2 (x)) ≡ 0 ∀x ∈ B r ((0, 0)).<br />

(b) Supponendo r ≤ 1 2 e ρ ≤ 1 si ha che ‖F (x 1, x 2 , 0, 0)‖ =<br />

√ (√1<br />

= + x1 − 1 ) 2<br />

+ (log(cosh x 1 ) − sin(x 1 x 2 ))<br />

1 + x 2 =<br />

√ 2<br />

(√1<br />

= + x1 − 1 + 1 − 1 ) 2<br />

+ (log(1 + (cosh(x 1 ) − 1)) − sin(x 1 x 2 ))<br />

1 + x 2 ≤<br />

√ 2<br />

(∣∣<br />

√<br />

≤ 1 + x1 − 1 ∣ +<br />

∣ 1 − 1 ∣) ∣∣∣ 2<br />

+ (|log(1 + (cosh(x 1 ) − 1))| + |sin(x 1 x 2 )|)<br />

1 + x 2<br />

2 (∣ ∣∣∣∣ (√<br />

≤<br />

√ 1 + x1 − 1 ) (√ 1 + x 1 + 1 )<br />

∣ ∣) 2 ∣∣∣<br />

√ 1 + x1 + 1 ∣ + x 2 ∣∣∣<br />

+ (2| cosh(x 1 ) − 1| + |x 1 x 2 |) 2 ≤<br />

1 + x 2<br />

√ (<br />

|x 1 |<br />

≤ √ 1 + x1 + 1 + |x ) 2 (<br />

2|<br />

+ e<br />

1 − 1 + 2 |ex1 −x1 − 2|<br />

+ x2 1 + ) 2 x2 2<br />

≤<br />

2<br />

2<br />

√<br />

2<br />

(<br />

) 2<br />

√<br />

(<br />

≤ (|x 1 | + 2|x 2 |) 2 + |e x1 − 1| + |e −x1 − 1| + r2<br />

≤ (3r) 2 + 6|x 1 | + r 2<br />

≤<br />

2<br />

2)<br />

√<br />

(<br />

≤ 9r 2 + 6r + r )<br />

√<br />

2 36<br />

=<br />

2 4 r2 + 169<br />

√<br />

205<br />

4 r2 = r ≤ 15 r, dunque per<br />

2 2<br />

avere sup ‖F (x 1 , x 2 , 0, 0)‖ ≤ 15<br />

x∈B r(0,0)<br />

2 r ≤ ρ 4 = ρ<br />

≤<br />

ρ<br />

4‖T ‖ ∞ 2‖T ‖ è sufficiente<br />

prendere ρ = r<br />

30 ; inoltre, I 2 − T ∂F<br />

∂y (x 1, x 2 , y 1 , y 2 ) =<br />

( 1 − e<br />

−y 1<br />

)<br />

sin y 2<br />

= 2y1 sin(x1x2)<br />

− 1 − 1 , dunque essendo ∣ ∣<br />

∣1 − e −y1 ≤ 3|y 1 | ≤<br />

(1+y1) 2 2 1+y2<br />

2 ∣ ≤ 3ρ, | sin(y 2 )| ≤ |y 2 | ≤ ρ,<br />

∣ −2y 1 sin(x 1 x 2 ) ∣∣∣∣<br />

≤ 2|y 1 || sin(x 1 x 2 )| ≤<br />

(1 + y1 2)2 ≤ 2ρ|x 1 x 2 | ≤ 2ρ x2 1 + x 2 2<br />

≤ ρr 2 ≤ ρr ≤ ρ2<br />

2<br />

30 ≤ ρ ∣ ∣∣∣<br />

30 e 1 − 1<br />

1 + y2<br />

2 ∣ =<br />

= y2 2<br />

1 + y 2 2<br />

≤ y2 2 ≤ ρ 2 ≤ ρ, allora per avere<br />

sup<br />

∥ I 2 − T ∂F<br />

∂y (x 1, x 2 , y 1 , y 2 )<br />

∥ ≤<br />

sup<br />

∥ I 2 − T ∂F<br />

∂y (x 1, x 2 , y 1 , y 2 )<br />

∥ ≤ 6ρ ≤<br />

∞<br />

(x 1,x 2,y 1,y 2)∈B r((0,0))×B ρ((0,0))<br />

≤ 2<br />

(x 1,x 2,y 1,y 2)∈B r((0,0))×B ρ((0,0))

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