Engineering Mathematics 2

Engineering Mathematics 2

Lecture 20 Yong Sung Park

On exams

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

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

• Absolutely NO calculators

• You need to show the screen and the desk with your hands all time

• You CANNOT disappear from the zoom at any time during exam

Previously, we discussed

• Heat equation

• 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).

• 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).

• Solution of 1D heat equation for a very long bar (1) method of separation of variables, followed by (2) Fourier integral (to satisfy the initial condition).

𝑢 𝑥, 𝑡 = 1

2𝑐 𝜋𝑡 න

−∞

∞

𝑓 𝑣 exp − 𝑥 − 𝑣 ²

4𝑐²𝑡 𝑑𝑣

12.8 Two-dimensional wave equation

𝜕²𝑢

𝜕𝑡² = 𝑐² 𝜕²𝑢

𝜕𝑥² + 𝜕²𝑢

𝜕𝑦² where 𝑐² = Τ𝑇 𝜌

12.9 Rectangular membrane. Double Fourier Series

• Two-dimensional wave equation

𝜕²𝑢

𝜕𝑡² = 𝑐² 𝜕²𝑢

𝜕𝑥² + 𝜕²𝑢

𝜕𝑦² with 𝑢 = 0 on the boundary and

𝑢 𝑥, 𝑦, 0 = 𝑓 𝑥, 𝑦 , 𝑢_𝑡 𝑥, 𝑦, 0 = 𝑔 𝑥, 𝑦

a b

Using the following separation of variables

𝑢 𝑥, 𝑦, 𝑡 = 𝐹 𝑥, 𝑦 𝐺 𝑡

We can obtain an ODE for time and a PDE (2D Helmholtz equation) for space:

ሷ𝐺 + 𝜆²𝐺 = 0 and 𝐹_𝑥𝑥 + 𝐹_𝑦𝑦 + 𝜐²𝐹 = 0 where 𝜆 = 𝑐𝜐.

Then using the separation of variables

𝐹 𝑥, 𝑦 = 𝐻 𝑥 𝑄 𝑦

Obtain two ODEs from the 2D Helomholtz equation:

𝑑²𝐻

𝑑𝑥² + 𝑘²𝐻 = 0 and ^𝑑2𝑄

𝑑𝑦² + 𝑝²𝑄 = 0 where 𝑝² = 𝜈² − 𝑘²

• Satisfying the boundary conditions, the solutions for Helmholtz eq are

𝐹_𝑚𝑛 𝑥, 𝑦 = sin 𝑚𝜋𝑥

𝑎 sin 𝑛𝜋𝑦 𝑏

• Then, eigenfuctions for 2D wave equation are

𝑢_𝑚𝑛 𝑥, 𝑦, 𝑡 = 𝐵_𝑚𝑛 cos 𝜆_𝑚𝑛𝑡 + 𝐵_𝑚𝑛^∗ sin 𝜆_𝑚𝑛𝑡 sin 𝑚𝜋𝑥

𝑎 sin 𝑛𝜋𝑦 𝑏

• Eigenvalues: 𝜆_𝑚𝑛 = 𝑐𝜋 ^𝑚2

𝑎² + ^𝑛2

𝑏²

• To satisfy the initial conditions

𝑢 𝑥, 𝑦, 𝑡 = ෍

𝑚=1

∞

෍

𝑛=1

∞

𝑢_𝑚𝑛

• 𝑢 𝑥, 𝑦, 0 = σ_𝑚=1^∞ σ_𝑛=1^∞ 𝐵_𝑚𝑛 sin ^𝑚𝜋𝑥

𝑎 sin ^𝑛𝜋𝑦

𝑏 = 𝑓 𝑥, 𝑦

• Double Fourier series:

σ_𝑚=1^∞ 𝐾_𝑚 𝑦 sin ^𝑚𝜋𝑥

𝑎 = 𝑓 𝑥, 𝑦 where,

𝐾_𝑚 𝑦 = σ_𝑛=1^∞ 𝐵_𝑚𝑛 sin ^𝑛𝜋𝑦

𝑏 = ²

𝑎 ׬₀^𝑎 𝑓 𝑥, 𝑦 sin ^𝑚𝜋𝑥

𝑎 𝑑𝑥 𝐵_𝑚𝑛 = ⁴

𝑎𝑏 ׬₀^𝑏 ׬₀^𝑎 𝑓 𝑥, 𝑦 sin ^𝑚𝜋𝑥

𝑎 sin ^𝑛𝜋𝑦

𝑏 𝑑𝑥𝑑𝑦

• Similarly, 𝐵_𝑚𝑛^∗ = ⁴

𝑎𝑏𝜆_𝑚𝑛 ׬₀^𝑏 ׬₀^𝑎 𝑔 𝑥, 𝑦 sin ^𝑚𝜋𝑥

𝑎 sin ^𝑛𝜋𝑦

𝑏 𝑑𝑥𝑑𝑦