Extra Credit Assignment
The official description for the course says that we should do an introduction to PDE, however, I’m sure that we
won’t have time for that. So what I am doing to rectify the situation is offering this extra credit assignment to
both make up for our lack of an itro to PDE and to give people a way to make up a littl bit for the somewhat
dismal scores on the second exam. You can receive a maximum of 10 extra credit points toward Test 2. This
extra credit assignment is due at the end of the Final Exam for the class, and will not be accepted after that
point.
Definition 1 A partial differential equation (PDE) is an equation involving an unknown function of two or more
variables and certain of its partial derivatives.
In this exercise we’ll consider functions of two variables, such as u(x, t) where x is a spatial variable and t is time,
∂u
or u(x, y), where both x and y are spatial variables. We will use the notation ux = ∂u
∂x , ut = ∂t , etc.
One of the simplest PDEs is the transport or one-way wave equation
ut + bux = 0,
where b ∈ R. Just like with ODEs, we have to specify an inital condition in order to get a unique solution, so we
require that
u(x, 0) = g(x).
In other words, we specify we specify what the wave looks like at time zero, and then u(x, t) tells us what the
wave looks like at time t, once we have found a formula for u(x, t), that is.
Problem 1 Show that u(x, t) = g(x − tb) is a solution to the one-way wave equation problem
ut + bux = 0
(x, t) ∈ R × (0, ∞)
.
u(x, 0) = g(x) x ∈ R
First, it is easy to see that u(x, t) = g(x − tb) satisfies u(x, 0) = g(x − 0 · b) = g(x). Also, we have that
ut (x, t) = −bg ′ (x − tb), and ux (x, t) = g ′ (x − tb), so ut + bux = −bg ′ (x − tb) + bg ′ (x − tb) = 0.
Problem 2 Find a formula for the solution of the nonhomogeneous problem
ut + bux = f (x, t) (x, t) ∈ R × (0, ∞)
.
u(x, 0) = g(x)
x∈R
Steps:
1. Let z(s) = u(x + bs, t + s). Write z ′ (s) in terms of f .
2. Evaluate
Z
0
z ′ (s) ds,
−t
then set the result equal to the integral of the expression you found in step 1 for z ′ (s).
3. Sove for u(x, t) in the resulting expression.
4. Make the change of variables τ = t + s in the integral.
Step 1: Since u is a soluton to the PDE, we have
z ′ (s) = ux (x + bs, t + s)b + ut (x + bs, t + s) = f (x + bs, t + s).
Step 2: Since z ′ (s) is a derivative, it is easy to evaluate
Z
0
z ′ (s) ds = z(0) − z(−t) = u(x, t) − u(x − tb, 0) = u(x, t) − g(x − tb).
−t
1
Therefore,
Z
0
f (x + bs, t + s) ds = u(x, t) − g(x − tb).
−t
Step 3: Rearranging gives
u(x, t) =
Z
0
f (x + bs, t + s) ds + g(x − tb).
−t
Step 4: Making the substitution τ = t + s in the integration gives
Z t
f (x + (τ − t)b, τ ) ds + g(x − tb).
u(x, t) =
0
This step isn’t strictly necessary, but it’s just more satisfying to integrate from 0 to t.
One of the most famous and most useful equations of mathematical physics is Laplace’s equation
uxx + uyy = 0.
One nice thing about Laplace’s equation is that in many situations it can be solved by separation of variables.
Problem 3 Use separtaion of variables to find a function u such that uxx + uyy = 0.
Steps:
1. Write u(x,y)=F(x)G(y) and substitute this choice of u into Laplace’s equation.
2. Move everything that depends on x to one side of the equation, and put everything that depends on y on
the other side.
3. Notice that since the two sides of this equation are always equal, but one side is a funciton of x and the
other is a function of y, then both sides of the equation must be constant. Call this constant λ2 . Now write
down two equations: the left hand side equal to λ2 and the right hand side equal to λ2 .
4. When properly rearranged, these two equations are constant coefficient linear homogeneous ODEs. Solve
them to find F (x) and G(y).
Step 1: If u(x, y) = F (x)G(y), and u satisfies Laplace’s equation, then
0 = uxx + uyy = F ′′ (x)G(y) + F (x)G′′ (y).
Step 2: Rearranging as described, we obtain
G′′ (y)
F ′′ (x)
=−
.
F (x)
G(y)
Step 3: Since we have a function of x that is always equal to a function of y, it must be true that both are
constant. Therefore, we can write
G′′ (y)
F ′′ (x)
= λ2 , −
= λ2 .
F (x)
G(y)
Step 4: The previous equation implies that
F ′′ (x) − λ2 F (x) = 0 and G′′ (y) + λ2 G(y) = 0.
The characteristic equations are
r2 − λ2 = 0 and r2 + λ2 = 0.
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The roots of the first characteristic equation are ±λ, and the roots of the second characteristic equation are ±λi.
Therefore,
F (x) = c1 eλx + c2 e−λx and G(y) = c3 cos λy + c4 sin λy,
so
u(x, y) = (c1 eλx + c2 e−λx )(c3 cos λy + c4 sin λy).
The constants would be determined by imposing boundary conditions.
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