Additional problems, week 6
Example 15. Solve the wave equation for u(x,0) =ex, ut(x,0) = sin(x).
Example 16. Solve the wave equation for u(x,0) = ln(1 +x2),ut(x,0) = x−4.
Example 17. Solve the wave equation utt −uxx = 0 for 0 < x < 1, t > 0 for the initial conditionsu(x,0) = sin(πx), ut(x,0) = sin(2πx).
Example 18. Solve the wave equation
utt−∆u= 0, t >0, u(x,0) = 0,
ut(x,0) =p(x).
forx∈R3 with p(x1, x2, x3) =x1x2 using Kirchhoff’s formula.
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Solutions for the additional problems Solution 15. We get
u(x, t) = ex+t+ex−t
2 +1
2 Z x+t
x−t
sin(ξ) dξ
= 1
2 ex+t+ex−t+ (−cos(ξ))|x+tx−t
= 1
2 ex+t+ex−t−cos(x+t) + cos(x−t) .
Solution 16. We get
u(x, t) = 1
2 ln(1 + (x+t)2) + ln(1 + (x−t)2) +1
2 Z x+t
x−t
ξ−4 dξ u(x, t) = 1
2ln (1 + (x+t)2)(1 + (x−t)2) +1
2 ξ2
2 −4ξ
x+t
x−t
| {z }
(∗)
.
Computation of the last part
(∗) = 1 2
(x+t)2
2 −4(x+t)− (x−t)2
2 + 4(x−t)
= 1 2
1
24xt−8t
=xt−4t.
The final solution reads u(x, t) = 1
2ln (1 + (x+t)2)(1 + (x−t)2)
+xt−4t.
-4 1 -2
0.5 2 0
0 0 -2
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Solution 17. We get
u(x, t) = sin(π(x+t)) + sin(π(x−t))
2 +1
2 Z x+t
x−t
sin(2πξ) dξ
= sin(π(x+t)) + sin(π(x−t))
2 +1
2
−cos(2πξ) 2π
x+t
x−t
= sin(π(x+t)) + sin(π(x−t))
2 − 1
4π (cos(2π(x+t))−cos(2π(x−t))).
-1 2
1.5 0
1 1
0.5 0
1
0 -1
Solution 18. In order to solve the wave equation in R3 we use Kirchhoff’s formula. For the sperical coordinates
y1 =x1+tcos(ϕ) sin(θ), y2 =x2 +tsin(ϕ) sin(θ), y3 =x3+tcos(θ)
the solution can be computed as
u(x, t) = 1 4πt
Z π 0
Z 2π 0
(p(y1, y2)t2sin(θ) dϕdθ.
The integration t2sin(θ) dϕdθ is caused by the transformation from cartesian to sperical coordinates. In particular the part t2sin(θ) comes by computing the Jacobean-matrix of the coordinate transformation from
x
S(x,t)
p(y) dσy,
where S(x, t) denotes the sphere with radius t and center x in the three-dimensional space, i.e., S(x, t) = {y∈R3 : |y−x|=t}to
Z π
0
Z 2π
0
p(y1, y2t2sin(θ) dϕdθ.
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Let therefore be (x, y, z) =g(t, ϕ, θ). The Jacobean-matrix reads
Jg =
cos(ϕ) sin(θ) tcos(ϕ) cos(θ) −rsin(ϕ) sin(θ) sin(ϕ) sin(θ) tsin(ϕ) cos(θ) rcos(ϕ) sin(θ)
cos(θ) −rsin(θ) 0
with orthogonal columns and its determinant reads det(Jg) =grgϕgθ = 1·t·sin(θ) =t2sin(θ).
For p=x1x2 we get
u(x, t) = 1 4πt
Z π 0
Z 2π 0
(x1+tcos(ϕ) sin(θ))(x2+tsin(ϕ) sin(θ))t2sin(θ) dϕdθ
= t 4π
Z 2π
0
Z π
0
x1x2sin(θ) +x1tsin(ϕ) sin2(θ)
| R{z }
=0
+x2tcos(ϕ) sin2(θ)
| R{z }
=0
+t2cos(ϕ) sin(ϕ) sin2(θ) dθdϕ
= x1x2t 4π
Z 2π 0
Z π 0
sin(θ) dθdϕ+ t3 4π
Z 2π 0
Z π 0
cos(ϕ) sin(ϕ) sin2(θ)
| {z }
1
2sin(2ϕ) sin2(θ)
dθdϕ
=x1x2t+ 0
and the solution simply reads u(x, t) =x1x2t.
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