Solucoes edps

Problem 1
Solve the heat equation ut = 4uxx for a rod of length π with both ends insulated (zero Neumann boundary conditions) if the initial temperature is given by φ(x) = 2x − 1. First, formulate the mathematical problem and complete the three steps as described.
Mathematical Formulation (5pts)
When both ends of the rod are insulated, this indicates that the heat flux is fixed (at zero), and the heat flux is related to ux . So we can write the boundary conditions at each side, along with the initial temperature and equation, as ut = 4uxx ,

0 < x < π,

(2)

0≤x≤π

u(x, 0) = φ(x) = 2x − 1

(1)

t≥0

ux (0, t) = ux (4, t) = 0,

t>0

(3)

Step 1: Derive an expression for all nontrivial product (separated) solutions including an eigenvalue problem satisfying the boundary conditions (10pts)
The first thing we do is assume that we can separate the variables, i.e. that u(x, t) = X (x)T (t).
Then ut = XT and uxx = X T , so we plug in to (1) to get
XT = 4X T which we can rearrange to get
T
X
=
4T
X
Since both sides depend on different variables, but are equal for all x and t, then must both be constant. We’ll choose −λ as this constant:
T
X
=
= −λ
4T
X
This means that the PDE (1) separates into the two separate ODEs
T + 4T = 0

(4)

X + λX = 0
Now what is left to find our eigenvalue problem is to separate the boundary conditions (2). The first condition gives ux (0, t) = X (0)T (t) = 0
Since T (t) ≡ 0 means that u(x, t) ≡ 0, which is the trivial case, we must have X (0) = 0. Similarly, the second condition gives ux (π, t) = X (π )T (t) = 0
So we must have X (π ) = 0 for the same reason as above. So now we have the eigenvalue problem
X + λX = 0,
X (0) = 0 = X (π )

00

(9)

0 0 is required. Thus πnt πnt
Tn (t) = an cos √ + bn sin √
2
2 and our product solutions are given by un (x, t) = Xn (x)Tn (t) = sin

πnx πnt πnt an cos
+ bn sin
2
2
2

Step 3: Use the superposition principle and Fourier