Engineering Mathematics 2

Engineering Mathematics 2

Lecture 19 Yong Sung Park

On exams

• Grade for Exam 2 is available on eTL

• You can get your answer book from TA (35-202)

• Exam 3 is on 9^th December and covers Fourier Series & PDE

• Exam 3 is online (same format as in Exam 1)

Previously, we discussed

• Wave equation

𝜕²𝑢

𝜕𝑡² = 𝑐² 𝜕²𝑢

𝜕𝑥²

with BC: 𝑢 0, 𝑡 = 𝑢 𝐿, 𝑡 = 0 and IC: 𝑢 𝑥, 0 = 𝑓 𝑥 , 𝑢_𝑡 𝑥, 0 = 𝑔 𝑥

• We used the method of separation of variables followed by Fourier series,

• or D’Alembert’s solution, that is, 𝑢 𝑥, 𝑡 = 𝜙 𝑥 + 𝑐𝑡 + 𝜓 𝑥 − 𝑐𝑡 , which results in

𝑢 𝑥, 𝑡 = 1

2 𝑓 𝑥 + 𝑐𝑡 + 𝑓 𝑥 − 𝑐𝑡 + 1 2𝑐 න

𝑥−𝑐𝑡 𝑥+𝑐𝑡

𝑔 𝑠 𝑑𝑠

12.5 Heat equation

• (Time rate of change of total heat in a body) = (net outward flux of heat)

• ^𝜕𝑢

𝜕𝑡 = 𝑐²𝛻²𝑢,

• 𝑐² = 𝐾/𝜌𝜎

is the thermal diffusivity, in which 𝐾 is the thermal conductivity, 𝜌 the density, and 𝜎 the specific heat of the material of the body.

12.6 1D solution and 2D steady solution of the heat equation

• The 1-D heat equation

𝜕𝑢

𝜕𝑡 = 𝑐² 𝜕²𝑢

𝜕𝑥²

with the boundary conditions: 𝑢 0, 𝑡 = 𝑢 𝐿, 𝑡 = 0 for all 𝑡 ≥ 0, and the initial condition: 𝑢 𝑥, 0 = 𝑓 𝑥 for 0 ≤ 𝑥 ≤ 𝐿

• Solution of 1D heat equation, which is a parabolic type, using (1) method of separation of variables, followed by

(2) Fourier series (to satisfy the initial condition).

• Example 4:

What are the boundary conditions for a 1-D bar with insulated ends as well as insulated body.

• The 2-D heat equation:

𝜕𝑢

𝜕𝑡 = 𝑐² 𝜕²𝑢

𝜕𝑥² + 𝜕²𝑢

𝜕𝑦²

• At steady state, ^𝜕𝑢

𝜕𝑡 = 0, that is

𝜕²𝑢

𝜕𝑥² + 𝜕²𝑢

𝜕𝑦² = 0

• Boundary value problem

First BVP or Dirichlet problem 𝑢 is prescribed on the boundary 𝐶

Second BVP or Neumann Problem 𝜕𝑢/𝜕𝑛 is prescribed on 𝐶

Third BVP or Robin mixed Problem 𝑢 is prescribed on some part of 𝐶 and

𝜕𝑢/𝜕𝑛 for the rest

• Solve the 2-D steady heat equation with the following BCs:

• Solution of steady 2D heat equation, which is an elliptic type, using (1) method of separation of variables, followed by

(2) Fourier series (to satisfy the other boundary condition).

0 a

b

𝑢 = 0

𝑢 = 0 𝑢 = 0

𝑢 = 𝑓 𝑥

Before we begin,

• Recall

(1) ׬₀^∞ 𝑒^−𝑥2𝑑𝑥 = ^𝜋

2

(2) ׬₀^∞ 𝑒^−𝑥2 cos 2𝑏𝑥 𝑑𝑥 = ^𝜋

2 𝑒^−𝑏2

(3) Finally, show that ℱ 𝑒^−𝑎𝑥2 = ¹

2𝑎 𝑒^{−𝑤2/4𝑎}

12.7 Heat equation: modeling very long bars

• Want to solve the heat equation

𝜕𝑢

𝜕𝑡 = 𝑐² 𝜕²𝑢

𝜕𝑥² satisfying the initial condition

𝑢 𝑥, 0 = 𝑓 𝑥 over the entire 𝑥 axis −∞ < 𝑥 < ∞ .

• Using the method of separation of variables, we can find 𝑢 𝑥, 𝑡; 𝑝 = 𝐴 cos 𝑝𝑥 + 𝐵 sin 𝑝𝑥 𝑒^{−𝑐2𝑝2𝑡} in which 𝑝 is the parameter.

• In order to satisfy the initial condition, we use Fourier integral:

𝑢 𝑥, 𝑡 = න

0

∞

𝑢 𝑥, 𝑡; 𝑝 𝑑𝑝 = න

0

∞

𝐴 cos 𝑝𝑥 + 𝐵 sin 𝑝𝑥 𝑒^{−𝑐2𝑝2𝑡} 𝑑𝑝

• The initial condition

𝑢 𝑥, 0 = ׬₀^∞ 𝐴 𝑝 cos 𝑝𝑥 + 𝐵 𝑝 sin 𝑝𝑥 𝑑𝑝 = 𝑓 𝑥

where

𝐴 𝑝 = ¹

𝜋 ׬_−∞^∞ 𝑓 𝑣 cos 𝑝𝑣 𝑑𝑣, 𝐵 𝑝 = ¹

𝜋 ׬_−∞^∞ 𝑓 𝑣 sin 𝑝𝑣 𝑑𝑣

Substituting 𝐴 𝑝 and 𝐵 𝑝 into the solution for the heat equation and using

න

0

∞

𝑒^−𝑥2 cos 2𝑏𝑥 𝑑𝑥 = 𝜋

2 𝑒^−𝑏2

the solution reduces to 𝑢 𝑥, 𝑡 = 1

2𝑐 𝜋𝑡 න

−∞

∞

𝑓 𝑣 exp − 𝑥 − 𝑣 ²

4𝑐²𝑡 𝑑𝑣

• Using Fourier transform

ℱ 𝑢_𝑡 = 𝑐²ℱ 𝑢_𝑥𝑥 ,

𝜕 ො𝑢

𝜕𝑡 = −𝑐²𝑤²𝑢ො We can get 𝑢 𝑥, 𝑡 = ¹

2𝜋 ׬_−∞^∞ 𝑓 𝑤 𝑒መ ^{−𝑐2𝑤2𝑡}𝑒^𝑖𝑤𝑥𝑑𝑤, in which 𝑓 𝑤 =መ ¹

2𝜋 ׬_−∞^∞ 𝑓 𝑣 𝑒^{−𝑖𝑤𝑣}𝑑𝑣

• Using convolution First, recall

𝑓 ∗ 𝑔 𝑥 = න

−∞

∞

𝑓 𝑝 𝑔 𝑥 − 𝑝 𝑑𝑝 = න

−∞

∞

𝑓 𝑥 − 𝑝 𝑔 𝑥 𝑑𝑝 𝑓 ∗ 𝑔 𝑥 = ׬_−∞^∞ 𝑓 𝑤 ොመ 𝑔 𝑤 𝑒^𝑖𝑤𝑥𝑑𝑤

• Exercise: find 𝑔 𝑤ො that makes 𝑢 𝑥, 𝑡 = ¹

2𝜋 ׬_−∞^∞ 𝑓 𝑤 𝑒መ ^{−𝑐2𝑤2𝑡}𝑒^𝑖𝑤𝑥𝑑𝑤 into the form

𝑢 𝑥, 𝑡 = 𝑓 ∗ 𝑔 𝑥 = න

−∞

∞ 𝑓 𝑤 ොመ 𝑔 𝑤 𝑒^𝑖𝑤𝑥𝑑𝑤

• Exercise: starting from

ℱ 𝑒^−𝑎𝑥2 = 1

2𝑎 𝑒^{−𝑤2/4𝑎}

find 𝑔 𝑥 = ℱ⁻¹ 𝑔 𝑤ො ,

thus arrive at 𝑢 𝑥, 𝑡 = ¹

2𝑐 𝜋𝑡 ׬_−∞^∞ 𝑓 𝑣 exp − ^{𝑥−𝑣 2}

4𝑐²𝑡 𝑑𝑣.