AMATH 353
Partial Differential Equations and Waves
Weston Barger
Summer 2016
Homework 7
Due: Wednesday, August 16, 2016
1. Consider the initial value problem
(
ut + tuux = 0,
–∞ < x < ∞, 0 < t
u(x , 0) = sin(x )
(0.1)
Solve this problem for t ≥ 0. Plot the characteristic lines. Using your plot, determine (approximately)
for how long the solutions are valid i.e. use your plot to graphically determine tb . Upload the code
used for generating your plots.
Solution. Using the characteristic methods, we get the ODE
dx (t )
= tu,
dt
x (0) = x0 .
Now, since the PDE in (0.3) is homogeneous, u along our characteristic curve (x (t ), t ) is constant.
Therefore,
u(x (t ), t ) = u(x (0), 0) = u0 (x0 ) = sin(x0 )
We are left with the ODE
dx (t )
= t sin(x0 )
dt
x (0) = x0 ,
which has the solution
x (t ) =
1
sin(x0 )t 2 + x0 .
2
(0.2)
The solution to (0.1) is u(x , t ) = sin(x0 (x , t )), where x0 is the solution to (0.2). These characteristics
are plotted in Figure 1. It looks as though the breaking time is tb ≈ 1.7.
1
4
3
2
1
-10
-5
5
10
Figure 1: The characteristics (0.2).
2. Consider the initial value problem
(
ut + uux = 0,
–∞ < x < ∞, 0 < t
u(x , 0) = –x
(0.3)
Solve this problem for t ≥ 0. Plot the characteristic lines. Using your plot, determine (approximately)
for how long the solutions are valid i.e. use your plot to graphically determine tb . Upload the code
used for generating your plots.
Solution. Using the characteristic methods, we get the ODE
dx (t )
= u,
dt
x (0) = x0 .
Now, since the PDE in (0.3) is homogeneous, u along our characteristic curve (x (t ), t ) is constant.
Therefore,
u(x (t ), t ) = u(x (0), 0) = u0 (x0 ) = –x0 .
We are left with the ODE
dx (t )
= –x0 ,
dt
x (0) = x0 ,
which has the solution
x (t ) = –x0 t + x0 = x0 (1 – t ) .
2
(0.4)
Here, we can explicitly solve for x0 giving
x0 =
x
.
1–t
Therefore, the solution to (0.3) is
u(x , t ) = u0 (x0 (x , t )) = –x0 (x , t ) =
x
.
t–1
The characteristics (0.4) are plotted in Figure 2. It looks as though the breaking time is tb = 1.
2.0
1.5
1.0
0.5
-2
-1
1
2
Figure 2: The characteristics (0.4).
Bonus. Exercise 17.8
Solution. As the equation x = e –1/x0 t + x0 cannot be solved for x0 explicitly, we must resort to
numerical methods. For the point (x , t ) = (1, 2), get 1 = 2e –1/x0 + x0 . We can rearrange this equation
to get f (x0 ) := e –1/x0 (1 – x0 ) – 2. Finding the roots for x0 > 0 yields the single root of x̂0 ≈ 0.610874.
Plugging this into our initial data yields
u(1, 2) ≈ u(x̂0 , 0) = e –1/x̂0 = 0.194563.
3