HOMEWORK 2 SOLUTIONS
Section 1.3
1.
We assume the vertical position of the string as a function u(x, t), and consider
a segment of the string from x0 to x1 . We make use of Newton’s second law and
we consider only the motion in the vertical direction. The combined force in the
vertical direction through the end points is:
−T ux (x0 , t) + T ux (x1 , t),
the force that comes in due to resistance is
Z x1
ux (x, t)dx,
−k
x0
where k is the resistance constant. Hence the total force is
Z x1
ux (x, t)dx.
−T ux (x0 , t) + T ux (x1 , t) − k
x0
This should be equal to the vertical acceleration
Z x1
−
utt (x, t)ρ(x, t)dx,
x0
giving us
Z
x1
Z
x1
ux (x, t)dx = −
−T ux (x0 , t) + T ux (x1 , t) − k
utt (x, t)ρ(x, t)dx.
x0
x0
Now we use rectangle rule to approximate the integrals, we get
−T ux (x0 , t) + T ux (x1 , t) − kux (x, t)(x1 − x0 ) = −utt (x, t)ρ(x, t)(x1 − x0 ).
Divide both sides by (x1 − x0 ), we get
T
ux (x1 , t) − ux (x0 , t)
− kux (x, t) = −utt (x, t)ρ(x, t).
(x1 − x0 )
Notice that
ux (x1 , t) − ux (x0 , t)
= utt (x0 , t),
x1 →x0
(x1 − x0 )
we arrive at the partial differential equation
lim
T uxx − kux = ρutt .
1
3.
For this problem, we assume the temperature of the rod at position x time t
to be u(x, t), and we consider a segment of the rod from position x0 to x1 . We
establish the PDE based on conservation of heat, that is, the total change of the
amount of heat within the segment
Z x1
dM
Acρ(x, t)ut (x, t)dx
=
dt
x0
is equal to the net flux of heat into the segment, where A is the cross-section area
of the rod, c is the heat capacity. The flux through the two end points of the rod
segment satisfies Fick’s law, so this part of the flux can be written as
Akux (x1 , t) − Akux (x0 , t),
where k is some constant. The flux through the sides of the rod satisfies Newton’s
law of cooling, and can be written as
Z x1
Lµ(T0 − u(x, t))dx,
x0
where L is the perimeter of the rod and µ is some other constant. Thus we arrive
at
Z x1
Z x1
Lµ(T0 − u(x, t))dx.
Acρ(x, t)ut (x, t)dx = Akux (x1 , t) − Akux (x0 , t) +
x0
x0
Now use rectangle rule to approximate the integral, and divide both sides by (x1 −
x0 ), we get
Acρ(x, t)ut (x, t) = Ak
ux (x1 , t) − ux (x0 , t)
+ Lµ(T0 − u(x, t)).
x1 − x 0
Taking the limit as x1 → x0 we get
Acρ(x, t)ut (x, t) = Akuxx (x, t) + Lµ(T0 − u(x, t)).
Divide both sides by Acρ we get
ut =
k
Lµ
uxx +
(T − u)
cρ
Acρ 0
5.
We assume the concentration of the diffusion substance to be u(x, t), again we
look at a segment from x0 to x1 , the change of mass in this section is
Z x1
ut (x, t)dx,
x0
2
which should equal to the influx minus the outflux. The combined influx from the
two ends of the section according to Fick’s law is
−kux (x0 , t) + kux (x1 , t),
besides, there is other influx due to the moving of the medium, which is
V u(x0 , t) − V u(x1 , t),
hence the net flux would just be
−kux (x0 , t) + kux (x1 , t) + V u(x0 , t) − V u(x1 , t),
and we get the equation
Z x1
ut (x, t)dx = −kux (x0 , t) + kux (x1 , t) + V u(x0 , t) − V u(x1 , t),
x0
next we use rectangle rule, divide both sides by x1 − x0 , and apply the limit to get
the derivatives, just like what we did from the previous two problems, we arrive at
ut = kuxx − V ux .
6. (extra problem)
This problem is definitely not required. If you can’t do it, don’t be discouraged.
But if you hope to understand it, I can help you with that as well.
Here for the solution, I’m not going to write down all the details, but I will
write down some key steps. First the 3D heat equation takes the form
ut = uxx + uyy + uzz ,
the right hand side can also be written as ∆u. What we want to do here is to
find the expression for ∆u in cylindrical coordinates. You need to write down the
change between the two coordinations x = r cos θ, y = r sin θ, z = z. After that,
you need to write down the Jacobi matrix
∂(r, θ, z)
.
∂(x, y, z)
However, according to the form of coordinate transfer, you can only get


cos θ −r sin θ 0
∂(x, y, z) 
sin θ r cos θ 0  ,
=
∂(r, θ, z)
0
0
1
however this matrix is very easy to invert, and you invert the matrix to get


cos θ
sin θ 0
∂(r, θ, z)
=  − 1r sin θ 1r cos θ 0  .
∂(x, y, z)
0
0
1
3
Then, notice that the Laplacian operator can be written as
∆ = (∂x , ∂y , ∂z ) · (∂x , ∂y , ∂z ),
we then have
2
∂r
∂(r, θ, z) 
∂θ  u.
·
∆u = (∂x , ∂y , ∂z ) · (∂x , ∂y , ∂z )u = 
∂(x, y, z)
∂z


The rest is just direct calculation. In performing the calculation, there are many
steps you can simplify due to the independence of θ and z.
7. (extra problem)
This problem follows similar path as the previous one. The only thing that is
different is the Jacobi matrix, but the way and the idea are the same. This is not
required of course, so don’t be discouraged if you can’t understand anything of it.
Section 1.5
4. (extra problem)
(a) If u(x, y, z is a solution, then we can add an arbitrary constant c to it to get
another solution.
(b) The PDE is
∆u = f (x, y, z),
integrate on both sides
ZZZ
ZZZ
∆udxdydz =
f (x, y, z)dxdydz.
D
D
The left hand side can be viewed as
ZZZ
div∇udxdydz,
D
and we apply divergence theorem to get
ZZZ
ZZ
div∇udxdydz =
D
n · ∇uds,
∂D
where n is the unit outer normal vector. The Neumann boundary condition says
that
∂u
= 0.
n · ∇u =
∂n
4
Hence
ZZ
n · ∇uds = 0,
∂D
and thus
ZZZ
f (x, y, z)dxdydz = 0.
D
(c) Neumann boundary condition basically means no heat exchange from boundary, or no flux at the boundary. That means the domain is insulated at the boundary. Thus assuming the temperature inside D is u(x, t), then you can add a constant
to it to make it u(x, t) + c, it forms a solution as well, and the domain boundary is
still insulated.
Section 1.6
1.
(a) The PDE can be written as
uxx − 4uxy + uyy + 2uy + 4u = 0.
Here a = 1, b = −4, c = 1, so b2 − 4ac = 16 − 4 = 12 > 0, so it is Elliptic.
(b) Here a = 9, b = 6, and c = 1, hence b2 −4ac = 36−36 = 0, so it is Parabolic.
4.
Here a = 1, b = −4 and c = 4, so b2 − 4ac = 16 − 16 = 0, it is Parabolic.
Now u(x, y) = f (y + 2x) + xg(y + 2x), then
ux = 2f 0 (y + 2x) + g(y + 2x) + 2xg 0 (y + 2x),
and
uy = f 0 (y + 2x) + xg 0 (y + 2x).
Furthermore,
uxx = 4f 00 (y + 2x) + 4g 0 (y + 2x) + 4xg 00 (y + 2x),
uxy = 2f 00 (y + 2x) + g 0 (y + 2x) + 2xg 00 (y + 2x),
uyy = f 00 (y + 2x) + xg 00 (y + 2x).
Substituting this into uxx − 4uxy + 4uyy we get
uxx − 4uxy + 4uyy
=4f 00 + 4g 0 + 4xg 00 − 4(2f 00 + g 0 + 2xg 00 ) + 4(f 00 + xg 00 )
=0.
Hence we have verified that u(x, y) = f (y + 2x) + xg(y + 2x) is a solution.
5
5. (extra problem)
Let u(x, y) = v(x, y)eαx+βy , then we have
ux = vx eαx+βy + αveαx+βy
uy = vy eαx+βy + βveαx+βy
uxx = vxx eαx+βy + 2αvx eαx+βy + α2 veαx+βy
uyy = vyy eαx+βy + 2βvy eαx+βy + β 2 veαx+βy .
Then we have
uxx + 3uyy − 2ux + 24uy + 5u e−(αx+βy)
=vxx + 3vyy + (2α − 2)vx + (6β + 24)vy + (α2 + 3β 2 − 2α + 24β + 5)v.
In order to make the coefficient in front of vx and vy zero, we have to have
β = −4.
α = 1,
Furthermore, let y 0 =
√
3y, we arrive at
vxx + vy0 y0 − 44v = 0
Section 2.1
1.
Use formula (8) from the book, the solution is
Z
1 x+ct
1 x+ct
x−ct
sin sds
u(x, t) = (e
+e
)+
2
2c x−ct
1
= ex cosh ct − (cos (x + ct) − cos (x − ct))
2c
1
x
= e cosh ct + sin x sin ct
c
9.
We factorize the left hand side of the PDE to obtain
∂
∂
∂
∂
uxx − 3uxt − 4utt =
−4
+
u.
∂x
∂t
∂x ∂t
6
Now we want to make a change of coordinates from (x, t) to (x0 , y 0 ), so that
∂x
= 1,
∂x0
∂t
= −4,
∂x0
∂x
= 1,
∂y 0
∂t
= 1.
∂y 0
This gives us
x = x0 + y 0 ,
t = −4x0 + y 0 .
From there we solve for x0 and y 0 we get
y0 =
4x + t
,
5
x0 =
x−t
.
5
It is fine if we rescale x0 and y 0 . It is also fine if we just leave x0 and y 0 there without
rescaling them, we will end up with the same answer. But let us rescale them by
rescaling y 0 with a factor 5/4 and rescaling x0 with a factor 5, we get
1
y 0 = x + t,
4
x0 = x − t.
Apply chain rule, the original PDE becomes
ux0 y0 = 0.
The general solution to that is
1
u(x, t) = f (y 0 ) + g(x0 ) = f (x + t) + g(x − t).
4
From the initial condition that u(x, 0) = φ(x), we get
f (x) + g(x) = φ(x).
(1)
From the general solution, we have that
1
1
ut (x, t) = f 0 (x + t) − g 0 (x − t).
4
4
Putting in the initial condition that ut (x, 0) = ψ(x), we obtain that
1 0
f (x) − g 0 (x) = ψ(x).
4
Integrating it we get
1
f (x) − g(x) =
4
x
Z
ψ(s)ds + A,
0
where A is a constant.
(1) × 1/5 − (2) × 4/5 gives us
1
4
g(x) = φ(x) −
5
5
Z
7
0
x
4
ψ(s)ds − A;
5
(2)
(1) × 4/5 + (2) × 4/5 gives us
4
4
f (x) = φ(x) +
5
5
Z
0
x
4
ψ(s)ds + A.
5
Thus we have
1
u(x, t) =f (x + t) + g(x − t)
4
Z 1
4
1
1
4 x+ 4 t
= φ(x + t) + φ(x − t) +
ψ(s)ds
5
4
5
5 x−t
Now plug in that φ(x) = x2 and ψ(x) = ex , we have that
(x − t)2 4(x + 14 t)2 4ex t/4
u(x, t) =
+
+
(e − e−t ).
5
5
5
8. (extra problem)
(a) Make change of variables v = ru, in another word u = v/r, then we have
utt =
vtt
,
r
ur =
vr
v
− 2,
r
r
urr =
vrr
2v
2v
− 2r + 3 .
r
r
r
Then the spherical wave equation which is
2
2
utt = c urr + ur
r
can be changed to the form
vtt
= c2
r
2vr
2v 2 vr
v
vrr
− 2 + 3 +
− 2
.
r
r
r
r r
r
Work out some simple cancellations we get
vtt = c2 vrr .
Thus we have transferred a 3D spherical wave equation into a 1D wave equation.
(b) From there the solution is
v(r, t) = f (r + ct) + g(r − ct),
where f and g are general functions. And the solution for the spherical wave
equation is hence
1
u(r, t) = (f (r + ct) + g(r − ct)) .
r
(c) Given the initial condition that u(r, 0) = φ(r) and ut (r, 0) = ψ(r), we have
v(r, 0) = ru(r, 0) = rφ(r),
vt (r, 0) = rut (r, 0) = rψ(r).
8
Hence we have
1
v(r, t) = ((r + ct)φ(r + ct) + (r − ct)φ(r − ct)) +
2
Z
r+ct
sψ(s)ds.
r−ct
Hence the solution to the spherical wave equation is
v(r, t)
1
1
u(r, t) =
=
((r + ct)φ(r + ct) + (r − ct)φ(r − ct)) +
r
2r
r
Z
r+ct
sψ(s)ds.
r−ct
10. (extra problem)
Problem 10 is very similar to problem 9, only the numbers get changed, we can
basically follow the same method. Here I don’t repeat the routine calculations.
9