Solution to AMS502 Homework­4 Total: 50 points 1. (P71 #3) Consider the first‐order equation 0 a) If , show that ,
is a weak solution. b) Can you find any discontinuous weak solutions” c) Is there a transmission condition for a weak solution with jump discontinuity along the characteristic ? Solution: (a) A weak solution must satisfy 0 for all Ω . By changing variables , to , and , we have that ,
and , we have 2 . . Also 2
,
0
for all
R Ω
So ,
(b) Take is a weak solution. with discontinuous as 1,
0,
0
, 0
Therefore, this discontinuous function is a weak solution. (c) No. The transmission condition is given by 0,
0,
0,
0 0. So there is no a weak solution with jump discontinuity along 2. (P82 #2) Solve the initial/boundary value problem 0
,0
0,
,
0,
0
,0
,
0
0
0
0
0.
(a) Find a Fourier series solution, and sum the series in regions bounded by characteristics. Do you think the solution is unique? (b) Use the parallelogram rule to solve this problem; is the resulting solution unique? Continuous? ? Solution: (a) If we look for a Fourier series solution, we can try to find , in the form ,
Formally, ,
we have that sin
cos
satisfies the boundary conditions 0,
0
. Then, , can be written as ,
sin
,
0 for
0 and thus Page 1 of 5 0, we find the 0, whose Moreover, we substitute it in the partial differential equation
functions must satisfy the ordinary differential equations general solution is · sin
· cos
are determined by the initial conditions; namely, we use and The constants ,0
sin
,0
0 · sin
1 Hence, we can integrate to find 2
2
4
0 · sin
0
2
1 · sin
1
1
2 1
,
cos
0,
Therefore, we find the Fourier series solution as 4
,
2
1
2
,
sin 2
1
1
cos
sin 2
sin
1
sin
(b) We use the parallelogram rule to piece together the solution and the domain decomposition is as plotted in the figures. In region C1, the solution is defined by d’Alembert’s formula and the solution is 1
0
2
,
,
In region L1, let ,
0
1·
rule we find that , . ,
In region R1, let ,
0,
. Using the parallelogram rule we ,
find that in L1 and thus ,
and 1
2
and ,
in R1 and thus ,
,
, . Using the parallelogram . Page 2 of 5 ,
In region C2, let ,
in C2 and thus ,
, ,
and . ,
Using the parallelogram rule we find that . ,
Generally, we can find that in region Ci: let ,
,
, ,
and thus and . Using the parallelogram, we find that ,
In region Li, let and thus ,
,
0,
,
and ,
. ,
Using the parallelogram rule, we find that . In region Ri, let , ,
and thus ,
, . Using the parallelogram, we find that ,
and ,
Therefore, we find the general solution in the various regions is written as 1
1
,
,
,
,
1
1
The line separating from 1
is . 1
. In region the 1 and in region the solution is 1
. Thus, we see that solution is 1
the solution is continuous if letting . Similarly, the line separating from is 1
in region the solution is letting . The line separating from 1
1 and . In region the solution is 1
. Thus, we see that the solution is continuous if 1 is 1
1
. In region 1 and in region 1 the solution is 1
. Thus, the solution is 1
we see that the solution is continuous between and 1. Similarly, the line separating from 1 is 2
1 . In region the solution is 1 and in region 1 the solution is 1
the solution is continuous between and 1. In addition, we can see in region , if letting 0, ,
, ,
0. If region C1, since ,
, if letting 1
. Thus, we see that 0 and in region ,
0. 0, , if letting Therefore, the whole solution is 0,
0, ∞ . However, it is easily seen that the . derivatives across region boundaries are not continuous which implies ,
3. (P82 #4) Consider the initial boundary value problem Page 3 of 5 ,0
0
,0
,
0,
,
0
0 0
0,
Where 0
0
0 . If we extend g and h as odd functions on ∞
∞, show that d’Alembert’s formula (6) gives the solution. Solution: by extending g and h as odd functions on ∞
∞, we try to convert the initial boundary value problem to an initial value problem. 0
,
0
,0
,0
Where, G(x) and H(x) are odd functions defined as ,
,
0
0
,
,
0
0
Using d’Alembert’s formula, we obtain the solution of the initial value problem as ,
1
2
1
2
To show this solution also solves the original initial boundary value problem, we note that 1) Since , satisfies 0
,
0 , thus it also satisfies 0
,
0. ,0
0 thus the 2)
,0
0 and initial values are statisfied. 3) Lastly, we test the boundary value: 0,
1
2
1
2
1
2
1
2
1
2
0 Thus, we have shown that the d’Alembert’s formula gives the solution. 4. (P90 #1) (a) If is a ‐function of the one‐variable s, find a condition on the vector so that , ,
, , ,
is a solution of (26). (Such solutions are called plane waves and are constant on the planes ·
.) (b) Find the relationship which must hold between the initial data and for a plane wave solution. (c) Find all plane wave solutions of (26) with the initial condition , , ,0
1. Solution: (a) since , , ,
, we have ·
Due to (26) Δ
0 , we have that 1
1,2,3 0 which implies the Page 4 of 5 1/| |. ,0
condition is (b)
·
,0
and ·
·
Which implies that (c) Since . Also, we have 1,2,3 . 1, we have 1
1
0
1/ and thus 2/
From (a), we see that ,
√2 . Using Kirchhoff’s formula, we have 1
4 |
4
| |
1
4
1/ which implies |
1
1
4
1
√2
4
| |
1
1
1 0
1
| |
√2
| |
√2 5. (P90 #3) Use Duhamel’s principle to find the solution of the nonhomogeneous wave equation for three space dimensions Δ
, with initial conditions
,0
0
, 0 . What regularity in , is required for the solution to be ? Solution: we consider the nonhomogeneous wave equation with homogeneous initial conditions: Δ
,
,0
,0
0
By Duhamel principle, we reduce the problem to the special homogeneous equations with nonhomogeneous initial conditions: Δ
, 0,
, 0,
0
,
,
0
0
0
,
,
0
Then ,
,
,
solves the nonhomogeneous wave equation. In the three space dimensions, using Kirchhoff’s formula, we know that , ,
1
4
0·
4
| |
,
·
4
| |
,
| |
·
Hence, we have that ,
,
,
1
4
So, we will need ,
to be 4
,
,
| |
in x and ·
| |
·
in t for the solution to be . Page 5 of 5