Creed Reilly, Sophomore, Engineering
Advisor: Professor Anna Mazzucato
Graduate Student: Yajie Zhang
Diffusion coefficient c jumps at x=1/2 (the interface). Impose transmission conditions at
interface. Solve equation in [0,1]. Impose Dirichlet boundary conditions at x=0,1. Initial
condition is sin(πx).
General Heat Equation in 1 Dimension with Transmission Condition
Ω = [0,1]
1
Ω1 = [0, )
2
1
Ω2 = ( , 1]
2
𝑇 = [0,1]
1
𝑃 = {𝑥 = }
2
𝑈𝑡 = 𝐶12 𝑈𝑥𝑥
𝐹𝑜𝑟 𝑥 𝑖𝑛 Ω1 , 𝑡 > 0
𝑈𝑡 = 𝐶22 𝑈𝑥𝑥
𝐹𝑜𝑟 𝑥 𝑖𝑛 Ω2 , 𝑡 > 0
𝑈 𝑡, 𝑥 = 0
𝐹𝑜𝑟 𝑥 = 0,1 , 𝑡 > 0
𝑈 1, 𝑥 = 𝑔 = sin 𝜋𝑥
𝐹𝑜𝑟 0 ≤ 𝑥 ≤ 1
𝑈+ = 𝑈−
𝑎𝑡 𝑥 = 𝑃
𝐶1 𝑈𝑥+ = 𝐶2 𝑈𝑥−
𝑎𝑡 𝑥 = 𝑃
Model Composite Materials:
 This is the simplest (explicit) first-order finite difference method to solve the heat
equation.
 First order Taylor expansion was used for the time derivative (Ut)
 The center-difference method was used for the second space derivative (Uxx)
 Because this is an explicit method, a convergence condition had to be observed:
𝐶 2 ∆𝑡
1 − 2𝜎 ≥ 0, 𝑤ℎ𝑒𝑟𝑒 𝜎 =
∆𝑥 2
1 𝐶 2 ∆𝑡
≥
2
∆𝑥 2
∆𝑥 2
∴ Δ𝑡 ≤
2𝐶 2
Discretization of the Exact Solution for 𝑥 = 0,
1
2
∪
1
2
1
,1 :
Discretization of the Exact Solution for 𝑥 = :
𝑈𝑖𝑛+1 − 𝑈𝑖𝑛
𝑈𝑡 ≈
∆𝑡
𝑈𝑥𝑥
𝑛
𝑛
𝐶 2 (𝑈𝑖+1
− 2𝑈𝑖𝑛 + 𝑈𝑖−1
)
≈
∆𝑥 2
𝑛
𝑛
𝑈𝑖𝑛+1 − 𝑈𝑖𝑛 𝑈𝑖+1
− 2𝑈𝑖𝑛 + 𝑈𝑖−1
≈
∆𝑡
∆𝑥 2
𝑈𝑖𝑛+1
𝐶 2 ∆𝑡 𝑛
2𝐶 2 ∆𝑡 𝑛 𝐶 2 ∆𝑡 𝑛
≈
𝑈 + 1−
𝑈𝑖 +
𝑈
∆𝑥 2 𝑖+1
∆𝑥 2
∆𝑥 2 𝑖−1
𝐶 2 ∆𝑡
𝑛
𝑛
𝜎=
; 𝑈𝑖𝑛+1 ≈ 𝜎𝑈𝑖+1
+ 1 − 2𝜎 𝑈𝑖𝑛 + 𝜎𝑈𝑖−1
2
∆𝑥
2
𝐶1 𝑈𝑥+ = 𝐶2 𝑈𝑥−
𝑛
𝑛
𝑈𝑖+1
− 𝑈𝑖𝑛
𝑈𝑖𝑛 − 𝑈𝑖−1
𝐶1 (
) ≈ 𝐶2 (
)
Δ𝑥
Δ𝑥
𝑛
𝑛
𝐶1 𝑈𝑖+1
+ 𝐶2 𝑈𝑖−1
≈ 𝐶1 + 𝐶2 𝑈𝑖𝑛
𝑈𝑖𝑛
𝑛
𝑛
𝐶1 𝑈𝑖+1
+ 𝐶2 𝑈𝑖−1
≈
𝐶1 + 𝐶2
•
The L2 Error Calculation is shown below:
𝑥1 − 𝑥2
•
•
2
=
𝑖
𝑥1𝑖 − 𝑥2𝑖
2
The program used is seen to converge as long as the L2 error decreases as the
displacement step decreases.
Since the error change is on a logarithmic scale, the equation Log𝐸 =
𝐿2(1)
log 2 𝐿2(2) should approach the value α of approximately 1.
CL=1
CR=2
Δx=0.1
CL=1
CR=2
Δx=0.025
Δx
0.1
0.05
0.025
0.0125
0.00625
0.003125
L2(1)
3.03E-09
4.05E-09
3.29E-09
2.09E-09
1.17E-09
6.23E-10
Linf(1)
4.51E-09
6.02E-09
4.89E-09
3.10E-09
1.74E-09
9.24E-10
L2(2)
4.05E-09
3.29E-09
2.09E-09
1.17E-09
6.23E-10
No Mem
Linf(2)
6.02E-09
4.89E-09
3.10E-09
1.74E-09
9.24E-10
No Mem
LogE
-0.42206
0.302807
0.65612
0.829307
0.915078
N/A
Table 1: L2 and L∞ error for various displacement steps
Graph 1: Diffusion of energy when the left
half has a C=1 and the right has a C=2.
Graph 2: Diffusion of energy when the left
half has a C=1 and the right has a C=100.