Solution to AMS 502 Homework­5 (Total: 60 points) 1. (Pg. 94 #4) The partial differential equation Δ
arises in the study of wave propagation in a nonhomogeneous elastic medium: is nonnegative and proportional to the coefficient of elasticity at . a) Define an appropriate notion of energy for solutions; b) Verify the corresponding energy inequality; c) Use the energy method to prove that solutions are uniquely determined by their Cauchy data. | |
Solution: (a) The energy integral is |
|
. We differentiate this function of , we obtain Integration by parts then yields Δ
In other words, must be a constant. (b) For any time 0, , let function: 1
| |
,
2
0 :|
|
. Consider the energy |
0
.
1 We claim that (1) is a nonincreasing function of , i.e. the following energy inequality holds: 0
0
.
2 ,
,
To prove (2), we introduce the following notations: |
Ω
, :|
|
, :|
Notice that Ω
unit normal on Ω is given on On , the normal ,…, ,
unit length condition ,0
,0
. where the unions are disjoint. Moreover, the exterior by 0, … ,0,1 , and on by 0, … ,0, 1 . satisfies ; together with the 1, this implies 1
1
Given a solution , we define the vector field 2
, ,2
If we calculate the divergence, we find ·
2
,
|
. |
2
2
0 2
The divergence theorem therefore implies Page 1 of 5 ·
0 . Ω
Now, on , the following inequality holds: 2
|
√1
We may compute on ·
2
So in particular, 1
|
√1
|
·
|
0 , 0 . We have 0
·
·
|
|
|
|
|
|
, which proves (2). (c) Let both and be the solutions to ,0
for initial conditions ,0
, ,0
,0
0
Thus, we know that Δ
0, ∞ with , we have on . Let ,
0,
,
0
0 and it follows that 1
| |
2
is nonnegative, the third term implies that 0
Since ,
0 and thus ,
, . 2. (Pg. 99 #1) Find dispersive wave solutions of the n‐dimensional linear Klein‐Gordon equation Δ
0. Solution: Consider the solution written as ,
where ,
·
is an ‐vector and ·
· is the inner product. The dispersion relation is | |
. Which defines the frequency ω as a function of the wave number . It represents all waves with the wave number propagate with constant phase velocity . In fact, we obtain ,
·
| |
·
| |
. 3. (Pg. 99 #6) Find the solution of the telegrapher’s equation 0 ,0
0 where is an arbitrary satisfying the initial condition ,0
and function. Solution: By a change of dependent variable, we can reduce this problem to 0 on ,
0. In fact, first introduce the characteristic coordinates and . Page 2 of 5 Thus the equation becomes: 1
4
1
4
0 4
Now let ,
To find that ,
exp
1
4
1
4
satisfies 1
0
4
4
4
Converting back to , coordinates, we have ,
Where ,0
,
,
0 with initial conditions satisfies the equation ,0
,0
and ,
,0
. Thus, we obtain ·
·
And the dispersion relation is 1
4
Actually, we obtain ·
,
·
With the initial conditions ·
,0
1
4
,0
·
·
·
1
2
1
2
·
·
Thus, we obtain 2
1
4
2
and
1
4
1
4
,
Therefore, the solution is ,
2
,
1
4
1
2
1
4. (Pg. 110 #1) Let Ω
,
separation of the variable ·
,
2
·
4
·
:
1
, :0
to solve the Dirichlet problem 1
1,0
2
, and use Page 3 of 5 Δ
0
Ω
for 0
1,
Solution: It is natural to use polar coordinates 1
1
0
1,0
,
0
2
,
and in which the problem becomes 0,
1,
If we write ,
2 .
2 then 0 . Separating the variables, we obtain 0 has solutions . The equation . This means that 1,2, … and ,
ln
superposition we may write But . and cos
0 has solutions ,
cos
sin
. But , must be finite at ,
cos
sin
sin
0, so ; notice ; notice for 0. By , But then 1,
Which shows that the coefficients for . Thus we may write 1
cos
and sin
1) are determined from the Fourier series (
1
cos
, sin
1,2, … and therefore may take an arbitrary value. Notice, however, that is not determined by Moreover, the constant term in the Fourier series for must be zero and then we write 1
2
. 5. (Pg. 110 #3) Prove that the solution of the Robin or third boundary value problem 5 for the Laplace equation is unique when 0 is a constant. Proof: Let and be the solutions of the Robin for the Laplace equation and thus they satisfy ∆
0,
,
Let Ω
Ω
and thus it satisfies ∆
0,
0,
Ω
Ω
Page 4 of 5 Using the Green’s identities, we may write Δ
Ω
·
Δ
Ω
Ω
·
Ω
Then, we have that |
Ω
|
0 Ω
Also, we know that | |
Ω
0
0 Ω
Therefore, we can conclude that 0
Ω and 0
Ω, and thus in Ω. In addition to 0
Ω, we know that 0 in Ω which implies in Ω and thus the uniqueness of the solution is proven. 6. (Pg. 110 #5) Suppose 0 for Ω and consider solutions Ω
Ω of Δ
0 in Ω. Establish uniqueness theorems for (a) the Dirichlet problem, and (b) the Neumann problem. Proof: We show the uniqueness theorems for the Dirichlet problem, Δ
0,
,
in Ω
for
R ,
Ω
0
0,
in Ω
R ,
0
D , And the Neumann problem, Δ
,
Similarly in Pg. 110 #3, let and which satisfies Δ
,
0,
in Ω
on Ω
N , Ω
be the solutions of the problem (D)/(N) and thus let Δ
,
D ;
0,
in Ω
Ω
N , Recalling the Green’s identities, we may write |
Ω
Also, we know that |
0
0 in Ω Ω
0, on Ω in (D) and /
0, on Ω in (N) which show that 0 Ω
This holds in both (D) and (N). Hence for (D) and (N), we have that |
|
0 Ω
Thus, we know that 0 in Ω
in Ω for both (D) and (N). In (D), since 0 on ∂Ω, 0 in Ω. In (N), if 0 in Ω, any two solutions must differ by a constant; if there exists a subset Ω
Ω such that 0 in Ω , solutions are unique in Ω and solutions must differ by a constant everywhere. Page 5 of 5