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578 CHAPTER 16 The Wave Equation
2
2
∂ y ∂ y
5. = 8 for 0 < x < 2π,t > 0 Graph the fortieth partial sum for some values of the
∂t 2 ∂x 2 time.
y(0,t)= y(2π,t) = 0for t ≥ 0
12. Transverse vibrations in a homogeneous rod of length
∂y
y(x,0)= f (x), (x,0) = 0for0 ≤ x ≤ 2π π are modeled by the partial differential equation
∂t
∂ u ∂ u
4 2
3x for 0 ≤ x ≤ π a 4 + = 0for0 < x <π,t > 0.
where f (x) = ∂x 4 ∂t 2
6π − 3x for π< x ≤ 2π.
2
2
∂ y ∂ y Here u(x,t) is the displacement at time t of the cross
6. = 4 for 0 < x < 5,t > 0 section through x perpendicular to the x-axis and
∂t 2 ∂x 2
2
a = EI/ρ A,where E is Young’s modulus, I is the
y(0,t)= y(5,t) = 0for t ≥ 0
moment of inertia of this cross section, ρ is the con-
∂y
y(x,0)= 0, (x,0) = g(x) for 0 ≤ x ≤ 5 stant density and A is the constant cross-sectional
∂t area.

0 for 0 ≤ x < 4
where g(x) = (a) Let u(x,t) = X(x)T (t) to separate the variables.
5 − x for 4 ≤ x ≤ 5.
(b) Solve for values of the separation constant and for
2
2
∂ y ∂ y
7. = 9 for 0 < x < 2,t > 0 X and T in the case of free ends, in which
∂t 2 ∂x 2
3
2
3
2
y(0,t)= y(2,t) = 0for t ≥ 0 ∂ u ∂ u ∂ u ∂ y
(0,t)= (π,t)= (0,t)= (π,t)=0
∂y ∂x 2 ∂x 2 ∂x 3 ∂x 3
y(x,0)= x(x − 2), (x,0) = g(x) for 0 ≤ x ≤ 2
∂t for t > 0.

0 for 0 ≤ x < 1/2and 1 < x ≤ 2 (c) Solve for the separation constant and for X and T
where g(x) = in the case of supported ends in which
3 for 1/2 ≤ x ≤ 1.
2
2
2
2
∂ y ∂ y ∂ u ∂ u
8. = 25 for 0 < x <π,t > 0 u(0,t) = u(π,t) = (0,t) = (π,t) = 0.
∂t 2 ∂x 2 ∂x 2 ∂x 2
y(0,t)= y(π,t) = 0for t ≥ 0
13. Solve the telegraph equation
∂y
y(x,0)= sin(2x), (x,0) = π − x for 0 ≤ x ≤ π 2 2
∂t ∂ u + A ∂u + Bu = c 2 ∂ u
9. Solve the initial-boundary value problem ∂t 2 ∂t ∂x 2
2
2
∂ y ∂ y for 0 < x < L,t > 0. A and B are positive constants.
= 3 + 2x for 0 < x < 2,t > 0
∂t 2 ∂x 2 The boundary conditions are
y(0,t) = y(2,t) = 0for t ≥ 0
u(0,t) = u(L,t) = 0for t ≥ 0,
∂y
y(x,0) = 0, (x,0) = 0for0 ≤ x ≤ 2. and the initial conditions are
∂t
Graph the fortieth partial sum of the solution for some ∂u
u(x,0) = f (x), (x,0) = 0for0 ≤ x ≤ L.
values of the time. ∂t
10. Solve
2
2
2
2
2
Assume that A L < 4(BL + c π ).
2
2
∂ y ∂ y 14. (a) Write a series solution for
2
= 9 + x for 0 < x < 4,t > 0
∂t 2 ∂x 2
2
2
∂ y ∂ y
3
y(0,t) = y(4,t) = 0for t ≥ 0 = 9 + 5x for 0 < x < 4,t > 0
∂t 2 ∂x 2
∂y
y(x,0) = 0, (x,0) = 0for0 ≤ x ≤ 4. y(0,t) = y(4,t) = 0for t ≥ 0
∂t
∂y
Graph the fortieth partial sum of the solution for y(x,0) = cos(πx), (x,0) = 0for 0 ≤ x ≤ 4.
selected values of t, ∂t
11. Solve (b) Write a series solution when the forcing term 5x 3
2
2
∂ y ∂ y is removed.
= − cos(x) for 0 < x < 2π,t > 0
∂t 2 ∂x 2 (c) In order to gauge the effect of the forcing term
on the motion, graph the fortieth partial sum of
y(0,t) = y(2π,t) = 0for t ≥ 0
the solutions in parts (a) and (b) on the same set
∂y of axes when t = 0.4 seconds. Repeat this for
y(x,0) = 0, (x,0) = 0for 0 ≤ x ≤ 2π.
∂t t = 0.8,1.4, 2,2.5, 3, and 4.
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October 14, 2010 15:23 THM/NEIL Page-578 27410_16_ch16_p563-610