Partial Differential Equations, Winter 2015
Homework #1
Due: Wednesday, January 28th, 2015 in tutorial.
1. (Chapter 1.1, 3)
For each of the following, state the order of the equation and whether it is nonlinear,
linear homogeneous, or linear inhomogeneous.
(a)
(b)
(c)
(d)
(e)
(f)
(g)
ut − uxx + 1 = 0 order 2, linear inhomogeneous
ut − uxx + xu = 0 order 2, linear homogeneous
ut − uxxt + uux = 0 order 3, nonlinear
utt − uxx + x2 = 0 order 2, linear inhomogeneous
ux (1 + u2x )−1/2 + uy (1 + u2y )−1/2 = 0 order 1, nonlinear
ux + ey uy = 0 order 1, linear homogeneous
√
ut + uxxxx + 1 + u = 0 order 4, nonlinear
2. (Chapter 1.1, 4)
Let u1 and u2 be two solutions of the inhomogeneous linear equation Lu = g. Show
that u1 − u2 is a solution of the homogeneous equation Lu = 0.
Solution:
If Lu1 = g = Lu2 , then by linearity
L(u1 − u2 ) = Lu1 − Lu2 = g − g ≡ 0
3. (Chapter 1.1, 12)
Verify by direct substitution that
un (x, y) = sin(nx) sinh(ny)
is a solution of uxx + uyy = 0 for every n > 0.
Solution:
d2
sinh(ny) = n2 sinh(ny)
dy 2
d2
sin(nx) = −n2 sin(nx)
dx2
1
And so (un )yy (x, y) = n2 u(x, y) and (un )xx (x, y) = −n2 u(x, y).
4. (Chapter 1.2, 2)
Solve the equation 3uy + uxy = 0. (Hint: let v = uy . What is the PDE for v?)
Solution:
The PDE for v = vy is
3v + vx = 0
∂
(exp(3x) · v) = 0
=⇒
∂x
=⇒ exp(3x) · v = f (y)
=⇒ v(x, y) = f (y) exp(−3x),
where we have used the integrating factor exp(3x) and f is a function of one variable.
Since v = uy , we integrate with respect to y and get
u(x, y) = F (y) exp(−3x) + g(x),
where F 0 = f and g is a function of one variable.
5. (Chapter 1.2, 10)
Solve ux + uy + u = ex+2y with u(x, 0) = 0.
Hint: use the change of coordinates: ζ = x + y, η = x − y.
Solution:
Make the change of variable ζ = x + y and η = x − y, then
ux = uζ + uη
uy = uζ − uη
So ux + uy = 2uζ and x + 2y = (3ζ − η)/2. The PDE, ux + uy + u = ex+2y becomes
2uζ + u = exp(3ζ/2 − η/2)
1
exp(−η/2)
uζ + u =
exp(3ζ/2)
2
2
Note that exp(−η/2) is a constant with respect to ζ so this is really an ODE in ζ. We
use the integrating factor exp(ζ/2) and get
exp(−η/2)
d
ueζ/2 =
exp(2ζ)
dζ
2
exp(−η/2)
=⇒ ueζ/2 =
exp(2ζ) + f (η)
4
exp(−η/2)
=⇒ u =
exp(3ζ/2) + f (η) exp(−ζ/2)
4
where f is an arbitrary function of one variable. Changing back to x and y variables
and skipping some algebra,
u(x, y) =
exp(x + 2y)
+ f (x − y) exp(−(x + y)/2).
4
Finally we plug in the initial condition, u(x, 0) = 0 which gives,
0 = exp(x)/4 + f (x) exp(−x/2)
3x
=⇒ f (x) = − exp
/4.
2
Summing up,
3x
exp(x + 2y) 1
− exp
exp(−(x + y)/2)
u(x, y) =
4
4
2
exp(x + 2y) exp ((x − y)/2)
=
−
.
4
4
6. (Chapter 1.4, 2) Heat conduction is also described by the diffusion equation
ut = κuxx
where u(x, t) is the temperature at x at time t and κ is some constant. A homogeneous
thin pipe occupying the region 0 < x < ` is completely insulated. Its initial temperature is f (x). After a sufficiently long time, the temperature of the object eventually
reaches a steady (or equilibrium) state. We call this the steady state temperature.
Find a formula for the steady-state temperature in terms of f .
Hints:
• The steady state temperature does not change in time. What does this mean for
ut ?
• The ammount of heat is constant. None is lost or gained.
Solution:
We are trying to solve for u(x, t) which solves the diffusion equation with the above
conditions but which does not change in time. This means that ut = 0 and so u(x, t) =
g(x) for some arbitrary function of g. Ie., u does not depend on t. From the diffusion
equation 0 = ut = κuxx and so since κ 6= 0,
g 00 (x) = uxx (x, t) = 0
=⇒ g(x) = ax + b
where a and b are constants. Also, since the pipe is insulated at the ends, this means
there is no heat flow in or out so by Fick’s law, ux (0, t) = 0 = ux (`, t) for all t,
(Neumann condition) which gives that a = 0 and u(x, t) = b. Finally, we have that
heat is conserved. The total heat in the bar at time t is
Z `
u(x, t)dx.
H(t) =
0
We differentiate this with respect to t,
Z `
0
H (t) =
ut (x, t)dx
0
Z `
κuxx (x, t)dx
=
0
= κ(ux (`, t) − ux (0, t))
= 0.
So H(t) is constant. H(0) =
R`
0
f (x)dx while H(t) = `b, so
Z
1 `
b=
f (x)dx
` 0
7. (Chapter 1.5, 2) Consider the problem
u00 (x) + u0 (x) = f (x)
1
u0 (0) = u(0) = [u0 (l) + u(l)],
2
with f (x) a given function.
(a) Is the solution unique? Explain.
(b) Does a solution necessarily exist, or is there a condition that f (x) must satisfy
for existence? Explain.
Hint: Consider w = u + u0 .
Solution:
(a) The solution is not unique. Suppose there is a solution u to the above ODE. Let’s
find a solution to the homogeneous ODE
w00 (x) + w0 (x) = 0
1
w0 (0) = w(0) = [w0 (`) + w(`)].
2
We sub in w(x) = A exp(rx) to the ODE to get
r2 + r = 0 =⇒ r = −1, 0
So the general solution is w(x) = a + b exp(−x). Plugging in the boundary
conditions gives that wb (x) = −2b + b exp(−x) where b is an arbitrary constant.
Now notice that u + w solves the inhomogeneous ODE with the given boundary
conditions. So the solution (if it exists) is not unique. In fact there are infinitely
many solutions because there are infinitely many w that satisfy the homogeneous
ODE (just choose a different b).
(b) Let w = u + u0 . Then the ODE for w is
w0 (x) = f (x)
w(0) = w(`)
Rx
Solving the ODE gives that w(x) = 0 f (s) ds+C. From the boundary conditions,
R`
we must have that 0 f (s) ds = 0. In other words we only have a solution for w
and hence a solution for u when this integral is zero.
8. (Chapter 1.5, 5) Consider the equation
ux + yuy = 0
with the initial condition u(x, 0) = φ(x).
(a) For φ(x) = x, show that no solution exists.
(b) For φ(x) ≡ 1, show that there are many solutions.
Solution:
The ODe for the characteristic curves are given by y 0 (x) = y, whose solution
is y(x) = C exp(x) for C a constant. The characteristic curves are therefore
ye−x = C and so the general solution to the PDE is
u(x, y) = f (e−x y).
Notice that φ(x) = u(x, 0) = f (0). If φ(x) = x we cannot find any f that will
satisfy this equation but if φ(x) ≡ 1 any f with the property that f (0) = 1 will
work. There are infinitely many such functions.