Fourier Series and Partial Differential Equations Lecture Notes

Chapter 4. The wave equation 43
Example 4.5 A string occupies −∞ < x ≤ 0 and is fixed at x = 0. A wave y(x,t) =
F(x−ct) is incident from x < 0. Find the reflected wave.
Figure 15
The solution of the wave equation is
y = F(x−ct) +G(x+ct) ,
(4.95)
incident reflected
where G is to be found. The condition y(0,t) = 0 is to be satisfied for all t. Hence
F(−ct)+G(ct) = 0, for all t, and so G(θ) = −F(−θ) for all θ. Thus
y(x,t) = F(x−ct) −F(−x−ct) .
(4.96)
incident reflected
4.7 Uniqueness of an IBVP for a finite string
We consider the wave equation
∂2y ∂t2 = c2∂2 y
∂x2, for 0 < x < L and t > 0, (4.97)
and prove a uniqueness theorem based on energy considerations. The kinetic energy of
the string is
L 2 1 ∂y
ρ dx.
2 ∂t
(4.98)
0
The stress energy is the product of the tension and the extension, where the extension is
L 2
L ∂y
1+ dx−L = 1+
0 ∂x
0
1
2
∂y
+... dx−L, (4.99)
2 ∂x
L
2 ∂y
≈ dx. (4.100)
∂x
Thus
0
0
E(t) := 1
L 2
2
∂y ∂y
ρ +T dx, (4.101)
2 ∂t ∂x
is the energy of the string. The energy appears to depend upon the time but in important
cases it is actually constant.
Lemma 4.1 If y(x,t) is a solution of the wave equation (4.97) and satisfies the boundary
conditions
y(0,t) = 0 and y(L,t) = 0 for t ≥ 0, (4.102)
then E(t) is constant for t ≥ 0.