find the deflection of a vibrating string of unit length having fixed ends with initial velocity zero and intial deflection f(x)=2(sin(3x)-sin(2x))
Answers
Step-by-step explanation:
Let w(x,t)w(x,t) be the function describing deflection of the string. The function satisfies wave equation
w_{tt}=a^2w_{xx}w
tt
=a
2
w
xx
and from the condition of the question we have
w(0,t)=w(1,t)=0w(0,t)=w(1,t)=0
as the ends are fixed and initial conditions are
w(x,0)=0w(x,0)=0
w_t(x,0)=u(x)w
t
(x,0)=u(x)
In order to solve the PDE we’ll use separation of variables, so w(x,t)=X(x)T(t)w(x,t)=X(x)T(t). If we put this in the equation, we’ll obtain \frac{T''}{a^2T}=\frac{X''}{X}=-\lambda
a
2
T
T
′′
=
X
X
′′
=−λ. So we have Sturm-Liouville problem \begin{cases} X''+\lambda X=0\\ X(0)=X(1)=0 \end{cases}{
X
′′
+λX=0
X(0)=X(1)=0
. The problem has infinitely many non-trivial solutions X_n=\sin(\pi n x)X
n
=sin(πnx) and eigenvalues \lambda_n=\pi^2n^2λ
n
=π
2
n
2
. Then T_n''+ \lambda_n a^2 T_n=0T
n
′′
+λ
n
a
2
T
n
=0 hence T_n=C_{1n}\cos(\pi n at)+C_{2n}\sin(\pi n a t)T
n
=C
1n
cos(πnat)+C
2n
sin(πnat) where C_{1n}, C_{2n}C
1n
,C
2n
are arbitrary constants. So
w(x,t)=\sum_{n=1}^{\infty}T_n(t)X_n(x)=\sum_{n=1}^{\infty}(C_{1n}\cos(\pi n at)+C_{2n}\sin(\pi n a t))\sin(\pi n x)w(x,t)=∑
n=1
∞
T
n
(t)X
n
(x)=∑
n=1
∞
(C
1n
cos(πnat)+C
2n
sin(πnat))sin(πnx)
In order to find C_{1n}, C_{2n}C
1n
,C
2n
we’ll use initial conditions. w(x,0)=\sum_{n=1}^{\infty}C_{1n}\sin(\pi n x)=0w(x,0)=∑
n=1
∞
C
1n
sin(πnx)=0, so all C_{1n}=0C
1n
=0
w_t(x,0)=\sum_{n=1}^{\infty}C_{2n}\pi n a\sin(\pi n x)=u(x)w
t
(x,0)=∑
n=1
∞
C
2n
πnasin(πnx)=u(x)
let’s expand u(x)u(x) into Fourier sine series on interval 0<x<10<x<1.
u(x)=\sum_{n=1}^{\infty}b_n \sin(\pi n x)u(x)=∑
n=1
∞
b
n
sin(πnx)
b_n=2 \int_0^{1} u(x) \sin(\pi n x)\, dx=\frac{4\sin(\frac{\pi n}{2})}{\pi^2 n^2}b
n
=2∫
0
1
u(x)sin(πnx)dx=
π
2
n
2
4sin(
2
πn
)
So
u(x)=\sum_{n=1}^{\infty}\frac{4\sin(\frac{\pi n}{2})}{\pi^2 n^2} \sin(\pi n x)u(x)=∑
n=1
∞
π
2
n
2
4sin(
2
πn
)
sin(πnx)
and equating coefficients in the sums we obtain C_{2n}=\frac{4\sin(\frac{\pi n}{2})}{\pi^3n^3a}C
2n
=
π
3
n
3
a
4sin(
2
πn
)
. So
w(x,t)=\sum_{n=1}^{\infty}\frac{4\sin(\frac{\pi n}{2})}{\pi^3n^3a}\sin(\pi n a t)\sin(\pi n x)w(x,t)=∑
n=1
∞
π
3
n
3
a
4sin(
2
πn
)
sin(πnat)sin(πnx)
or it can be rewritten as
w(x,t)=\sum_{n=1}^{\infty}(-1)^{n+1}\frac{4}{\pi^3(2n-1)^3a}\sin(\pi (2n-1) a t)\sin(\pi (2n-1) x)w(x,t)=∑
n=1
∞
(−1)
n+1
π
3
(2n−1)
3
a
4
sin(π(2n−1)at)sin(π(2n−1)x)