Test 2
18.303 Linear Partial Differential Equations
Matthew J. Hancock
Dec. 1, 2004
1 Rules
You may only use pencils, pens, erasers, and straight edges. No calculators, notes,
books or other aides are permitted. Scrap paper will be provided.
Be sure to show a few key intermediate steps and make statements in words when
deriving results -answers only will not get full marks.
2 Given
The Jacobian determinant of the mapping (r, s) → (x, t)is
3 Questions
Total points: 49
1
Figure 1: Plot of f (x).
3.1
Part A
Consider the following quasi-linear PDE,
where the initial condition is
1. [3 points] Sketch f (x)vs. x. Solution: See Figure 1.
2. [6 points] Find the parametric solution. First write down the relevant ODEs
for dx/dr, dt/dr, du/dr. Please take the initial conditions t = 0 and x = s at r = 0.
What is the initial condition (i.e. at r =0) for u? Solve for t, u and x (in that order)
as functions of r, s.
Solution: We can write the PDE as
Thus the parametric solution is defined by the ODEs
with initial conditions at r =0,
t =0,
x = s,
u = u (x, 0) = u (s, 0) = f (s)
2
Integrating the ODEs and imposing the ICs gives
t = r
u = f (s)
x = (1 + f (s)) r + s =(1 + f (s)) t + s
3. [5 points] At what time ts and position xs does a shock first form? Hint: you
might need to consider negative values of f’ (s).
Solution: The Jacobian is
Shocks occur (the solution breaks down) where J = 0, i.e. where
The first shock occurs at
where min f’ ( s) = −1. Since f’ (1/ 2) = −1, the s =1/ 2 characteristic can be used
to find the shock location at t = ts =1,
4. [6 points] Write down x in terms of t, s and f (s). For each of s = −1, 0, 1,
write down x as a function of t and plotitinthe xt-plane up to the shock time t = ts
you found in 3. Put all three curves in the same plot. Label where the shock occurs.
You have just plotted the three important characteristics.
Solution: Note that the s = −1, 0, 1 characteristics are given by
s = −1: x = (1 + f (−1)) t − 1= 2t − 1
s
=
0 : x =(1 + f (0)) t +0 = 3t
s
=
1 : x =(1 + f (1)) t +1 = 2t +1
These are plotted in Figure 2.
5. [4.5 points] Fill in the tables below:
3
Figure 2: Plot of characteristics for question 4.
Figure 3: Plot of u(x, t0) for part A for t0 =0, 0.5and 1
4
5. [3 points] Plot the three points (x, u) from the table for t =1/2, in the xu-plane
and connect the dots. You have just plotted u (x, t0)at t0 =1/2.
Solution: See the middle plot in Figure 3.
6. [3 points] Plot the three points (x, u) in the table for t = ts , in the xu-plane
and connect the dots. Label the shock. You have just plotted u (x, ts).
Solution: See the bottom plot in Figure 3.
3.2 Part B
We’re going to do questions 2-6 over again, for a VERY similar PDE,
The only difference between this PDE and the one in part A is the (1 −u) instead of
(1 + u). The initial condition is the SAME as part A,
7. [6 points] Find the parametric solution. First write down the relevant ODEs
for dx / dr, dt / dr, du / dr. Please take the initial conditions t =0 and x = s at r =0.
What is the initial condition (i.e. at r =0) for u? Solve for t, u and x (in that order)
as functions of r, s.
Solution: We can write the PDE as
Thus the parametric solution is defined by the ODEs
with initial conditions at r =0,
t = 0,
x = s,
u = u ( x, 0) = u (s, 0) = f (s)
5
Integrating the ODEs and imposing the ICs gives
t = r
u =
f (s)
x =
(1 − f (s)) r + s =(1 − f (s)) t + s
8. [5 points] At what time ts and position xs does a shock first form?
Solution: The Jacobian is
Shocks occur (the solution breaks down) where J = 0, i.e. where
The first shock occurs at
where max f’ (s) = 1. Since f’ (−1 / 2) = 1, the s = −1 / 2 characteristic can be used
to find the shock location at t = ts =1,
9. [6 points] Write down x in terms of t, s and f ( s ).
For each of s = −1, 0, 1,
write down x as a function of t and plot it in the xt-plane up to the shock time t = ts
you found in 3. Put all three curves in the same plot. Label where the shock occurs.
You have just plotted the three important characteristics.
Solution: Note that the s = −1, 0, 1 characteristics are given by
s
s
s
= −1: x = (1 − f (−1)) t − 1= −1
= 0 : x =(1 − f (0)) t + 0 = −t
= 1 : x =(1 − f (1)) t +1 = 1
These are plotted in Figure 4.
10. [4.5 points] Fill in the tables below:
6
Figure 4: Plot of characteristics for question 9.
11. [3 points] Plot the three points ( x, u ) from the table for t = 1/2, in the
xu-plane and connect the dots. You have just plotted u (x, t0) at t0 =1 / 2.
Solution: See the middle plot in Figure 5.
12. [3 points] Plot the three points ( x, u ) from the table for t = ts, in the
xu-plane and connect the dots. You have just plotted u (x, ts). Label the shock.
Solution: See the bottom plot in Figure 5.
7
Figure 5: Plot of u(x, t0) for part B for t0 = 0, 0.5and 1.
8