a - Nous commencerons par calculer la dérivée première :
Nous voyons que f(t) peut s'écrire :
f
(
t
)
=
H
(
t
)
.
1
ω
e
−
λ
t
sin
ω
t
{\displaystyle f(t)=H(t).{\frac {1}{\omega }}e^{-\lambda t}\sin {\omega t}}
On a alors :
f
′
(
t
)
=
δ
(
t
)
.
f
(
0
)
+
H
(
t
)
.
1
ω
(
−
λ
e
−
λ
t
sin
ω
t
+
ω
e
−
λ
t
cos
ω
t
)
=
H
(
t
)
.
e
−
λ
t
ω
(
−
λ
sin
ω
t
+
ω
cos
ω
t
)
{\displaystyle f'(t)=\delta (t).f(0)+H(t).{\frac {1}{\omega }}\left(-\lambda e^{-\lambda t}\sin {\omega t}+\omega e^{-\lambda t}\cos {\omega t}\right)=H(t).{\frac {e^{-\lambda t}}{\omega }}\left(-\lambda \sin {\omega t}+\omega \cos {\omega t}\right)}
f
″
(
t
)
=
δ
(
t
)
.
e
−
λ
t
ω
(
−
λ
sin
ω
t
+
ω
cos
ω
t
)
+
H
(
x
)
1
ω
[
−
λ
e
−
λ
t
(
−
λ
sin
ω
t
+
ω
cos
ω
t
)
+
e
−
λ
t
(
−
λ
ω
cos
ω
t
−
ω
2
sin
ω
t
)
]
=
δ
(
t
)
+
H
(
x
)
e
−
λ
t
ω
[
λ
2
sin
ω
t
−
2
λ
ω
cos
ω
t
−
ω
2
sin
ω
t
]
{\displaystyle {\begin{aligned}f''(t)&=\delta (t).{\frac {e^{-\lambda t}}{\omega }}\left(-\lambda \sin {\omega t}+\omega \cos {\omega t}\right)+H(x){\frac {1}{\omega }}\left[-\lambda e^{-\lambda t}\left(-\lambda \sin {\omega t}+\omega \cos {\omega t}\right)+e^{-\lambda t}\left(-\lambda \omega \cos {\omega t}-\omega ^{2}\sin {\omega t}\right)\right]\\&=\delta (t)+H(x){\frac {e^{-\lambda t}}{\omega }}\left[\lambda ^{2}\sin {\omega t}-2\lambda \omega \cos {\omega t}-\omega ^{2}\sin {\omega t}\right]\end{aligned}}}
b - On a :
L
.
f
(
p
)
=
∫
0
∞
e
−
p
.
t
f
(
t
)
d
t
=
∫
0
∞
e
−
p
.
t
1
ω
e
−
λ
t
sin
ω
t
d
t
=
1
ω
∫
0
∞
e
−
(
p
+
λ
)
.
t
sin
ω
t
d
t
{\displaystyle {\mathcal {L}}.f(p)=\int _{0}^{\infty }e^{-p.t}f(t)\,\mathrm {d} t=\int _{0}^{\infty }e^{-p.t}{\frac {1}{\omega }}e^{-\lambda t}\sin {\omega t}\,\mathrm {d} t={\frac {1}{\omega }}\int _{0}^{\infty }e^{-(p+\lambda ).t}\sin {\omega t}\,\mathrm {d} t}
Faisons successivement deux intégrations par partie :
L
.
f
(
p
)
=
1
ω
∫
0
∞
e
−
(
p
+
λ
)
.
t
sin
ω
t
d
t
=
−
1
ω
(
p
+
λ
)
[
e
−
(
p
+
λ
)
.
t
sin
ω
t
]
0
∞
+
1
p
+
λ
∫
0
∞
e
−
(
p
+
λ
)
.
t
cos
ω
t
d
t
=
1
p
+
λ
∫
0
∞
e
−
(
p
+
λ
)
.
t
cos
ω
t
d
t
=
−
1
(
p
+
λ
)
2
[
e
−
(
p
+
λ
)
.
t
cos
ω
t
]
0
∞
−
ω
(
p
+
λ
)
2
∫
0
∞
e
−
(
p
+
λ
)
.
t
sin
ω
t
d
t
=
1
(
p
+
λ
)
2
−
ω
2
(
p
+
λ
)
2
L
.
f
(
p
)
{\displaystyle {\begin{aligned}{\mathcal {L}}.f(p)&={\frac {1}{\omega }}\int _{0}^{\infty }e^{-(p+\lambda ).t}\sin {\omega t}\,\mathrm {d} t\\&=-{\frac {1}{\omega (p+\lambda )}}\left[e^{-(p+\lambda ).t}\sin {\omega t}\right]_{0}^{\infty }+{\frac {1}{p+\lambda }}\int _{0}^{\infty }e^{-(p+\lambda ).t}\cos {\omega t}\,\mathrm {d} t\\&={\frac {1}{p+\lambda }}\int _{0}^{\infty }e^{-(p+\lambda ).t}\cos {\omega t}\,\mathrm {d} t\\&=-{\frac {1}{(p+\lambda )^{2}}}\left[e^{-(p+\lambda ).t}\cos {\omega t}\right]_{0}^{\infty }-{\frac {\omega }{(p+\lambda )^{2}}}\int _{0}^{\infty }e^{-(p+\lambda ).t}\sin {\omega t}\,\mathrm {d} t\\&={\frac {1}{(p+\lambda )^{2}}}-{\frac {\omega ^{2}}{(p+\lambda )^{2}}}{\mathcal {L}}.f(p)\\\end{aligned}}}
Nous avons obtenu :
L
.
f
(
p
)
+
ω
2
(
p
+
λ
)
2
L
.
f
(
p
)
=
1
(
p
+
λ
)
2
{\displaystyle {\mathcal {L}}.f(p)+{\frac {\omega ^{2}}{(p+\lambda )^{2}}}{\mathcal {L}}.f(p)={\frac {1}{(p+\lambda )^{2}}}}
En multipliant les deux membres par (p + λ)2 , nous avons :
L
.
f
(
p
)
(
p
+
λ
)
2
+
ω
2
L
.
f
(
p
)
=
1
{\displaystyle {\mathcal {L}}.f(p)(p+\lambda )^{2}+\omega ^{2}{\mathcal {L}}.f(p)=1}
Soit finalement :
L
.
f
(
p
)
=
1
(
p
+
λ
)
2
+
ω
2
{\displaystyle {\mathcal {L}}.f(p)={\frac {1}{(p+\lambda )^{2}+\omega ^{2}}}}
c - Dériver une distribution revient à multiplier sa transformée de Laplace par p. On remarque que :
G
(
p
)
=
p
2
(
p
+
λ
)
2
+
ω
2
{\displaystyle G(p)={\frac {p^{2}}{(p+\lambda )^{2}+\omega ^{2}}}}
est la multiplication par p2 de la transformée de Laplace de f, donc correspond à la transformée de Laplace de la dérivée seconde de f qui est :
f
″
(
t
)
=
δ
(
t
)
+
H
(
x
)
e
−
λ
t
ω
[
λ
2
sin
ω
t
−
2
λ
ω
cos
ω
t
−
ω
2
sin
ω
t
]
{\displaystyle f''(t)=\delta (t)+H(x){\frac {e^{-\lambda t}}{\omega }}\left[\lambda ^{2}\sin {\omega t}-2\lambda \omega \cos {\omega t}-\omega ^{2}\sin {\omega t}\right]}
d - En passant au transformée de Laplace, on obtient :
p
2
(
p
+
λ
)
2
+
ω
2
+
2
p
(
p
+
λ
)
2
+
ω
2
+
4
1
(
p
+
λ
)
2
+
ω
2
=
1
{\displaystyle {\frac {p^{2}}{(p+\lambda )^{2}+\omega ^{2}}}+2{\frac {p}{(p+\lambda )^{2}+\omega ^{2}}}+4{\frac {1}{(p+\lambda )^{2}+\omega ^{2}}}=1}
qui s'écrit :
p
2
+
2
p
+
4
=
(
p
+
λ
)
2
+
ω
2
{\displaystyle p^{2}+2p+4=(p+\lambda )^{2}+\omega ^{2}}
Développons le second membre :
p
2
+
2
p
+
4
=
p
2
+
2
λ
p
+
λ
2
+
ω
2
{\displaystyle p^{2}+2p+4=p^{2}+2\lambda p+\lambda ^{2}+\omega ^{2}}
Par identification, on obtient :
{
λ
=
1
ω
=
3
{\displaystyle {\begin{cases}\lambda =1\\\omega ={\sqrt {3}}\end{cases}}}
d - D'après la question précédente
f
(
t
)
=
H
(
t
)
1
3
e
−
t
sin
(
t
3
)
{\displaystyle f(t)=H(t){\frac {1}{\sqrt {3}}}e^{-t}\sin {\left(t{\sqrt {3}}\right)}}
est la solution de :
f
″
+
2
f
′
+
4
f
=
δ
{\displaystyle f''+2f'+4f=\delta }
qui peut aussi s'écrire :
(
δ
″
+
2
δ
′
+
4
δ
)
⋆
f
=
δ
{\displaystyle (\delta ''+2\delta '+4\delta )\star f=\delta }
u
″
+
2
u
′
+
4
u
=
H
(
t
)
.
e
−
t
.
cos
t
{\displaystyle u''+2u'+4u=H(t).e^{-t}.\cos {t}}
s'écrit aussi :
(
δ
″
+
2
δ
′
+
4
δ
)
⋆
u
=
H
(
t
)
.
e
−
t
.
cos
t
{\displaystyle (\delta ''+2\delta '+4\delta )\star u=H(t).e^{-t}.\cos {t}}
En convolant les deux membre de cette dernière relation par f, on obtient :
(
δ
″
+
2
δ
′
+
4
δ
)
⋆
f
⋆
u
=
f
⋆
H
(
t
)
.
e
−
t
.
cos
t
{\displaystyle (\delta ''+2\delta '+4\delta )\star f\star u=f\star H(t).e^{-t}.\cos {t}}
qui donne :
δ
⋆
u
=
f
⋆
H
(
t
)
.
e
−
t
.
cos
t
{\displaystyle \delta \star u=f\star H(t).e^{-t}.\cos {t}}
soit :
u
(
t
)
=
H
(
t
)
1
3
e
−
t
sin
(
t
3
)
⋆
H
(
t
)
.
e
−
t
.
cos
t
{\displaystyle u(t)=H(t){\frac {1}{\sqrt {3}}}e^{-t}\sin {\left(t{\sqrt {3}}\right)}\star H(t).e^{-t}.\cos {t}}
Calculons ce produit de convolution :
u
(
t
)
=
H
(
t
)
∫
−
∞
∞
H
(
t
−
x
)
1
3
e
−
(
t
−
x
)
sin
(
(
t
−
x
)
3
)
.
H
(
x
)
.
e
−
x
.
cos
x
d
x
=
H
(
t
)
e
−
t
3
∫
0
t
sin
(
(
t
−
x
)
3
)
.
cos
x
d
x
{\displaystyle u(t)=H(t)\int _{-\infty }^{\infty }H(t-x){\frac {1}{\sqrt {3}}}e^{-(t-x)}\sin {\left((t-x){\sqrt {3}}\right)}.H(x).e^{-x}.\cos {x}\,\mathrm {d} x=H(t){\frac {e^{-t}}{\sqrt {3}}}\int _{0}^{t}\sin {\left((t-x){\sqrt {3}}\right)}.\cos {x}\,\mathrm {d} x}
En utilisant la formule trigonométrique :
sin
A
cos
B
=
1
2
(
sin
(
A
−
B
)
+
sin
(
A
+
B
)
)
{\displaystyle \sin A\cos B={\frac {1}{2}}(\sin(A-B)+\sin(A+B))}
on obtient :
u
(
t
)
=
H
(
t
)
e
−
t
2
3
∫
0
t
sin
(
(
t
−
x
)
3
−
x
)
+
sin
(
(
t
−
x
)
3
+
x
)
d
x
=
H
(
t
)
e
−
t
2
3
(
∫
0
t
sin
(
(
1
+
3
)
x
−
t
3
)
d
x
−
∫
0
t
sin
(
(
1
−
3
)
x
+
t
3
)
d
x
)
=
H
(
t
)
e
−
t
2
3
(
1
1
+
3
[
cos
(
(
1
+
3
)
x
−
t
3
)
]
0
t
−
1
1
−
3
[
cos
(
(
1
−
3
)
x
+
t
3
)
]
0
t
)
=
H
(
t
)
e
−
t
2
3
(
cos
(
t
)
−
cos
(
−
t
3
)
1
+
3
−
cos
(
t
)
−
cos
(
t
3
)
1
−
3
)
=
H
(
t
)
e
−
t
(
cos
(
t
)
−
cos
(
t
3
)
)
2
3
(
1
1
+
3
−
1
1
−
3
)
=
H
(
t
)
e
−
t
(
cos
(
t
)
−
cos
(
t
3
)
)
2
3
×
−
2
3
1
−
3
=
1
2
H
(
t
)
e
−
t
(
cos
(
t
)
−
cos
(
t
3
)
)
{\displaystyle {\begin{aligned}u(t)&=H(t){\frac {e^{-t}}{2{\sqrt {3}}}}\int _{0}^{t}\sin {\left((t-x){\sqrt {3}}-x\right)}+\sin {\left((t-x){\sqrt {3}}+x\right)}\,\mathrm {d} x\\&=H(t){\frac {e^{-t}}{2{\sqrt {3}}}}\left(\int _{0}^{t}\sin {\left((1+{\sqrt {3}})x-t{\sqrt {3}}\right)}\,\mathrm {d} x-\int _{0}^{t}\sin {\left((1-{\sqrt {3}})x+t{\sqrt {3}}\right)}\,\mathrm {d} x\right)\\&=H(t){\frac {e^{-t}}{2{\sqrt {3}}}}\left({\frac {1}{1+{\sqrt {3}}}}\left[\cos {\left((1+{\sqrt {3}})x-t{\sqrt {3}}\right)}\right]_{0}^{t}-{\frac {1}{1-{\sqrt {3}}}}\left[\cos {\left((1-{\sqrt {3}})x+t{\sqrt {3}}\right)}\right]_{0}^{t}\right)\\&=H(t){\frac {e^{-t}}{2{\sqrt {3}}}}\left({\frac {\cos {\left(t\right)}-\cos {\left(-t{\sqrt {3}}\right)}}{1+{\sqrt {3}}}}-{\frac {\cos {\left(t\right)}-\cos {\left(t{\sqrt {3}}\right)}}{1-{\sqrt {3}}}}\right)\\&=H(t){\frac {e^{-t}\left(\cos {\left(t\right)}-\cos {\left(t{\sqrt {3}}\right)}\right)}{2{\sqrt {3}}}}\left({\frac {1}{1+{\sqrt {3}}}}-{\frac {1}{1-{\sqrt {3}}}}\right)\\&=H(t){\frac {e^{-t}\left(\cos {\left(t\right)}-\cos {\left(t{\sqrt {3}}\right)}\right)}{2{\sqrt {3}}}}\times {\frac {-2{\sqrt {3}}}{1-3}}\\&={\frac {1}{2}}H(t)e^{-t}\left(\cos {\left(t\right)}-\cos {\left(t{\sqrt {3}}\right)}\right)\end{aligned}}}
![{\displaystyle f(t)={\frac {d}{dt}}\left[e^{-t.}H(t)\left(-{\frac {2}{3}}\sin t+{\frac {5}{6}}\sin {2t}\right)\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a5d360a4196361efe20e582e9360fd7b592b2aa4)
![{\displaystyle f(t)=e^{-t}.H(t)\left[{\frac {2}{3}}(\sin t-\cos t)-{\frac {5}{6}}(\sin {2t}-2\cos {2t}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7a9c14c180ee25466cdc31716bd2a8ce2b3731a2)
![{\displaystyle {\begin{aligned}f''(t)&=\delta (t).{\frac {e^{-\lambda t}}{\omega }}\left(-\lambda \sin {\omega t}+\omega \cos {\omega t}\right)+H(x){\frac {1}{\omega }}\left[-\lambda e^{-\lambda t}\left(-\lambda \sin {\omega t}+\omega \cos {\omega t}\right)+e^{-\lambda t}\left(-\lambda \omega \cos {\omega t}-\omega ^{2}\sin {\omega t}\right)\right]\\&=\delta (t)+H(x){\frac {e^{-\lambda t}}{\omega }}\left[\lambda ^{2}\sin {\omega t}-2\lambda \omega \cos {\omega t}-\omega ^{2}\sin {\omega t}\right]\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/dbcc6135d4a06ddc3dae87dbcedc6b4f36aab2fa)
![{\displaystyle {\begin{aligned}{\mathcal {L}}.f(p)&={\frac {1}{\omega }}\int _{0}^{\infty }e^{-(p+\lambda ).t}\sin {\omega t}\,\mathrm {d} t\\&=-{\frac {1}{\omega (p+\lambda )}}\left[e^{-(p+\lambda ).t}\sin {\omega t}\right]_{0}^{\infty }+{\frac {1}{p+\lambda }}\int _{0}^{\infty }e^{-(p+\lambda ).t}\cos {\omega t}\,\mathrm {d} t\\&={\frac {1}{p+\lambda }}\int _{0}^{\infty }e^{-(p+\lambda ).t}\cos {\omega t}\,\mathrm {d} t\\&=-{\frac {1}{(p+\lambda )^{2}}}\left[e^{-(p+\lambda ).t}\cos {\omega t}\right]_{0}^{\infty }-{\frac {\omega }{(p+\lambda )^{2}}}\int _{0}^{\infty }e^{-(p+\lambda ).t}\sin {\omega t}\,\mathrm {d} t\\&={\frac {1}{(p+\lambda )^{2}}}-{\frac {\omega ^{2}}{(p+\lambda )^{2}}}{\mathcal {L}}.f(p)\\\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/64bf5785617e376a2cf031dc2729ce48024fac0f)
![{\displaystyle f''(t)=\delta (t)+H(x){\frac {e^{-\lambda t}}{\omega }}\left[\lambda ^{2}\sin {\omega t}-2\lambda \omega \cos {\omega t}-\omega ^{2}\sin {\omega t}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/969e4859331ce3a1c8d7373ebade253e1c12bc69)
![{\displaystyle {\begin{aligned}u(t)&=H(t){\frac {e^{-t}}{2{\sqrt {3}}}}\int _{0}^{t}\sin {\left((t-x){\sqrt {3}}-x\right)}+\sin {\left((t-x){\sqrt {3}}+x\right)}\,\mathrm {d} x\\&=H(t){\frac {e^{-t}}{2{\sqrt {3}}}}\left(\int _{0}^{t}\sin {\left((1+{\sqrt {3}})x-t{\sqrt {3}}\right)}\,\mathrm {d} x-\int _{0}^{t}\sin {\left((1-{\sqrt {3}})x+t{\sqrt {3}}\right)}\,\mathrm {d} x\right)\\&=H(t){\frac {e^{-t}}{2{\sqrt {3}}}}\left({\frac {1}{1+{\sqrt {3}}}}\left[\cos {\left((1+{\sqrt {3}})x-t{\sqrt {3}}\right)}\right]_{0}^{t}-{\frac {1}{1-{\sqrt {3}}}}\left[\cos {\left((1-{\sqrt {3}})x+t{\sqrt {3}}\right)}\right]_{0}^{t}\right)\\&=H(t){\frac {e^{-t}}{2{\sqrt {3}}}}\left({\frac {\cos {\left(t\right)}-\cos {\left(-t{\sqrt {3}}\right)}}{1+{\sqrt {3}}}}-{\frac {\cos {\left(t\right)}-\cos {\left(t{\sqrt {3}}\right)}}{1-{\sqrt {3}}}}\right)\\&=H(t){\frac {e^{-t}\left(\cos {\left(t\right)}-\cos {\left(t{\sqrt {3}}\right)}\right)}{2{\sqrt {3}}}}\left({\frac {1}{1+{\sqrt {3}}}}-{\frac {1}{1-{\sqrt {3}}}}\right)\\&=H(t){\frac {e^{-t}\left(\cos {\left(t\right)}-\cos {\left(t{\sqrt {3}}\right)}\right)}{2{\sqrt {3}}}}\times {\frac {-2{\sqrt {3}}}{1-3}}\\&={\frac {1}{2}}H(t)e^{-t}\left(\cos {\left(t\right)}-\cos {\left(t{\sqrt {3}}\right)}\right)\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8820f2186e751ce0f4b9e6de777b279fff598157)
