Esercizi sui sistemi differenziali

Esercizi sui sistemi differenziali

Risolvere i seguenti problemi di Cauchy con l’uso delle trasformate di Laplace:

1.

(

y₁⁰ + y₁+ y₂ = e^2x

y₂⁰ − y1− 2y2 = (3− x)e^−x , y1(0) = y2(0) = 0.

2.

(y₁⁰ = y₁− 2y2+ e^x

y₂⁰ = y₁− y2− x , y₁(0) = 1, y₂(0) = 0.

3.







x⁰ = x− y y⁰ = x− z z⁰ = x

, x(0) = y(0) = 1, z(0) = 0.

4.







x⁰ = x + y y⁰ =−x + 3y z⁰ =−z

, x(0) = 1, y(0) = 2, z(0) = 1.

5.

(

x⁰⁰− y⁰ = 0

y⁰⁰+ y⁰ + x⁰ = 2e^−tsin t , x(0) =−1, x⁰(0) = y(0) = 2, y⁰(0) =−2.

6.

(

x⁰⁰= y + t

y⁰⁰ = x + 1 , x(0) = x⁰(0) = 0, y(0) = 1, y⁰(0) =−1.

7.

(

x⁰⁰− x⁰− 2y⁰ = e^−t

y⁰⁰− x⁰ = 0 , x(0) = 0, x⁰(0) = y(0) = y⁰(0) = 1.

8.

(

x⁰⁰− 2y⁰ = sin t

y⁰⁰+ x⁰ = 6e^−2t− 2 , x(0) = y(0) = 0, x⁰(0) = 1, y⁰(0) =−2

Svolgimento:

( 1)

sF₁(s) + F₁(s) + F₂(s) = _s−2¹

sF₂(s)− 2F2(s)− F1(s) = _s+1³ − _(s+1)¹ 2

=⇒

((s + 1)F₁(s) + F₂(s) = _s−2¹ (s− 2)F2(s)− F1(s) = _(s+1)^3s+22

=⇒ (

F2(s) = _s−2¹ − (s + 1)F1(s)

1− (s + 1)(s − 2)F1(s)− F1(s) = _(s+1)^3s+22

=⇒ (

. . .

1−_(s+1)^3s+22 = (s²− s − 1)F1(s) =⇒ (. . .

F₁(s) = _(s+1)¹ 2

=⇒

(F₂(s) = _s−2¹ −_s+1¹ F₁(s) = _(s+1)¹ 2

=⇒ (

y₂(x) = e^2x − e^−x y₁(x) = xe^−x.

1

( 2)

sF₁(s)− 1 = F1(s)− 2F2(s) + _s−1¹

sF₂(s) = F₁(s)− F2(s)−_s¹² =⇒ (

(s− 1)F1(s) + 2F₂(s) = 1 + _s−1¹ F₁(s) = (s + 1)F₂(s) + _s¹2

=⇒ (

(s²− 1)F2(s) +^s_s⁻¹2 + 2F₂(s) = _s−1^s F₁(s) = (s + 1)F₂(s) +_s¹2

=⇒ (

F₂(s)(s²+ 1) = _s−1^s −^s_s⁻¹2

F₁(s) = (s + 1)F₂(s) + _s¹2

=⇒ (F₂(s) = _s2¹+1

s³−s²+2s−1 s²(s−1)

. . .

Ponendo _s2¹+1 · ^s3−s_s2(s^2+2s−1)⁻¹ = ^a_s + _s^b2 + _s−1^c +^ds+e_s2+1, si trova b = s³− s²+ 2s− 1

(s²+ 1)(s− 1) |s=0 = 1, c = s³− s²+ 2s− 1

s²(s² + 1) |s=1 = 1 2 quindi

a

s+ds + e

s²+ 1 = 1 s²+ 1

s³− s²+ 2s− 1 s²(s− 1) − 1

s²− 1

2(s− 1) = 2s− s⁴− s²

2s²(s− 1)(s²+ 1) =−s²+ s + 2 2s(s²+ 1). Pertanto,

a =−s²+ s + 2

2(s²+ 1) |s=0=−1, e ds + e

s²+ 1 =−s²+ s + 2 2s(s²+ 1) +1

s = s²− s

2s(s²+ 1) = s− 1 2(s²+ 1). In definitiva,

F₂(s) =−1 s + 1

s² + 1

2(s− 1)+ s− 1

2(s²+ 1) =⇒ y2(x) =−1 + x + 1 2e^x+ 1

2(cos x− sin x).

Dalla seconda equazione del sistema si ricava

y₁(x) = y₂⁰(x)+y₂(x)+x = 1+1 2e^x−1

2sin x−1

2cos x−1+x+1 2e^x+1

2cos x−1

2sin x+x = e^x−sin x+2x.

3)





sF₁(s)− 1 = F1(s)− F2(s) sF₂(s)− 1 = F1(s)− F3(s) sF₃(s) = F₁

⇒







s²F₃(s)− 1 = sF3(s)− F2(s) sF₂(s)− 1 = sF3(s)− F3(s) . . .

⇒







(s² − s)F3(s) = 1− F2(s) (s− 1)F3(s) = sF₂(s)− 1 . . .







(s² − s)F3(s) = 1− F2(s) (s² − s)F3(s) = s²F₂(s)− s . . .

⇒





 . . . . . .

1− F2(s) = s²F₂(s)− s

⇒





 . . . . . .

F₂(s) = _s^s+12+1







(s² − s)F3(s) = 1− _s^s+12₊₁ F₂(s) = _s^s+12+1

. . .

⇒







F₃(s) = _s2¹+1

F₂(s) = _s^s+12+1

F₁(s) = _s2^s+1

⇒







z(t) = sin t

y(t) = sin t + cos t x(t) = cos t.

2

4)





sF₁(s)− 1 = F1(s) + F₂(s) sF₂(s)− 2 = −F1(s) + 3F₂(s) sF₃(s)− 1 = −F3

⇒







F₂(s) = (s− 1)F1(s)− 1

s(s− 1)F1(s)− s − 2 = −F1(s) + 3(s− 1)F1(s)− 3 F₃(s) = _s+1¹

⇒





 . . .

(s² − 4s + 4)F1(s) = (s− 1) . . .

⇒





 . . .

F₁(s) = _(s^s₋₂₎⁻¹2 = _s−2¹ +_(s−2)¹ 2

. . .

⇒







F₂(s) = ^(s_(s⁻¹⁾₋₂₎²2 − 1 F1(s) = _s−2¹ +_(s−2)¹ 2

F₃(s) = _s+1¹

⇒







F₂(s) = _(s^2s₋₂₎⁻³2 = _s−2² + _(s−2)¹ 2

F₁(s) = _s−2¹ +_(s−2)¹ 2

F₃(s) = _s+1¹

⇒







y(t) = e^2t(2 + t) x(t) = e^2t(1 + t) z(t) = e^−t.

5) Dato cheL[e^−tsin t](s) =L[sin t](s + 1) = _1+(s+1)¹ 2, si ha (s²F₁(s) + s− 2 − sF2(s) + 2 = 0

s²F₂(s)− 2s + 2 + sF2(s)− 2 + sF1(s) + 1 = _1+(s+1)² 2

⇒ (

F₂(s) = sF₁(s) + 1

(s²+ s)(sF₁(s) + 1) + sF₁(s) = 2s− 1 + _s2+2s+2²

⇒ (

. . .

(s³+ s²+ s)F₁(s) =−s²+ s− 1 + _s2+2s+2²

⇒ (

. . .

F1(s) = _s3+s¹²+s· ^−s_s2⁴+2s+2^−s3−s2 =−_s2+2s+2^s

⇒ (. . .

F₁(s) =−_1+(s+1)^s 2 =−_1+(s+1)^s+1 2 +_1+(s+1)¹ 2

⇒ (F₂(s) = 1−_s²_+2s+2^s2 = _s^2(s+1)2+2s+2 = 2_1+(s+1)^s+1 2

F₁(s) =−_1+(s+1)^s+1 2 +_1+(s+1)¹ 2

⇒ (

y(t) = 2e^−tcos t

x(t) = e^−t(sin t− cos t) 6)(

s²F₁(s) = F₂(s) +_s¹2

s²F₂(s)− s + 1 = F1(s) + ¹_s ⇒ (

F₂(s) = s²F₁(s)− _s¹²

s⁴F₁(s)− 1 − s + 1 = F1(s) +¹_s ⇒ (

. . .

(s⁴− 1)F1(s) = s + ¹_s ⇒ (. . .

F₁(s) = _s(s^s24⁺¹−1) = _s(s2¹−1).

Ponendo _s(s2¹−1) = ^a_s +_s−1^b + _s+1^c si ottiene a =−1, b = c = ¹₂, quindi F₁(s) =−1

s +1 2( 1

s− 1+ 1

s + 1 ⇒ x(t) = 1

2(e^t+ e^−t)− 1 = cosh t − 1.

Inoltre, dalla seconda equazione si deduce y(t) = x(t) + 1 = cosh t; pertanto y⁰(t) = sinh t + C e y(t) = cosh t + Ct + K per opportune costanti C, K. Imponendo i dati iniziali di y(t) si deduce C = −1 e K = 0; quindi y(t) = cosh t − t.

Negli svolgimenti dei prossimi due esercizi alcuni passaggi sono stati omessi.

3

7)(

s²F₁(s)− 1 − sF1(s)− 2sF2(s) + 2 = _s+1¹ s²F₂(s)− s − 1 − sF1(s) = 0 ⇒

( . . .

F₁(s) = sF₂(s)− 1 − ¹_s ⇒ (

(s²− s)(sF2(s)− 1 −¹_s)− 2sF2(s) + 1 = _s+1¹

. . . ⇒

(F₂(s) = (s²− 2 + _s+1¹ )_{s(s−2)(s+1)}¹ = _s(s+1)^{s3+s2−2s−1}2(s−2)

. . .

Ponendo _s(s+1)^{s3+s2−2s−1}2(s−2) = ^a_s + _s+1^b +_(s+1)^c 2 + _s−2^d si ottiene prima a = ¹₂, c = ¹₃, d = ₁₈⁷ e poi per differenza b = ¹₉. Quindi

F2(s) = 1 2s +1

9 1 s + 1+ 1

3 1

(s + 1)² − 7 18

1 s− 2, da cui

y(t) = 1 2 +1

9e^−t+ 1

3te^−t+ 7 18e^2t.

Derivando due volte la y(t) e sostituendo nella seconda equazione si trova

x⁰(t) =−5

9e^−t+ 1

3te^−t+14 9 e^2t e quindi essendo x(t) = x(0) +Rt

0 x⁰(τ ) dτ =Rt

0 x⁰(τ ) dτ , si deduce x(t) = 2

9e^−t −1

3te^−t+ 7

9e^2t− 1.

8)(

s²F₁− 1 − 2sF2(s) = _s2¹+1

s²F₂(s) + 2 + sF − 1(s) = _s+2⁶ −²_s ⇒

(F₁(s) = _s¹2 1 + 2sF − 2(s) + _s2¹+1

s²F₂(s) + 2 + ¹_s + 2F₂(s) +_s(s2¹+1) = _s+2⁶ − ²_s (. . .

F₂(s) =−_s(s+2)(s^2s2+s+42+1) =−²_s +_s+1¹ + _s2^s+1

quindi y(t) = e^−2t + cos t − 2. Dalla prima equazione del sistema di ricava x⁰⁰(t) =

− sin t − 4e^−2t, da cui x⁰(t) = x⁰(0) +

Z _t

0

x⁰⁰(τ ) dτ = 1 + [2e^−2τ + cos τ ]^t₀ = 2e^−2t+ cos t− 2 x(t) = x(0) +

Z t 0

x⁰(τ ) dτ = [−e^−2τ + sin τ − 2τ]^t0 =−e^−2t+ sin t− 2t + 1.

4