MAE502 Partial Differential Equations in Engineering (Spring 2009)


Syllabus (updated 22 Jan 2009)
New time for office hours on Wednesdays: After class from 6:45-8:00 PM Detail here


Projected schedule 

Wed 2/4  HW1 posted
Wed 2/18 HW1 due, HW2 posted 
Wed 3/4  HW2 due, HW3 posted 
--- 3/9 - 3/13 spring break ---
Wed 3/18 HW3 due, HW4 posted
Mon 3/23 Midterm 5:15-7:15 PM, open book, no laptop (calculator OK)
Wed 4/1  HW4 due, HW5 posted
Wed 4/15 HW5 due, HW6 posted
Wed 4/29 HW6 due
Mon 5/4  Last lecture before final
Final Exam: Monday 5/11/2009, 4:40-6:50 PM (at SCOB201), open book, no laptop (calculator OK)  
(University designated time is 4:50-6:40. We will start early.) Six homework assignments = 50% of total score One midterm = 20% Final exam = 30% ========================== We have finished: Ch. 1, whole chapter except pp. 27-32 (polar coordinate) Ch. 2, whole chapter except pp. 76-85 (Sec. 2.5.2-2.5.4) Ch. 3, whole chapter except Sec. 3.6 Ch. 4, whole chapter except Sec. 4.6 Ch. 7, Sec. 7.1-7.3 only ------ midterm -------- Ch. 8, Sec. 8.1-8.3 only Ch. 10, pp. 445-452, 459-465, and 471-483 PDEs in non-cartesian coordinates, Sec. 1.5, 2.5.2, 7.7 (7.8, 7.9) Ch. 5 (quick survey only - see summary in slides #17) Ch. 6, Sec. 6.1, 6.2, 6.6 (ignore detailed discussion on the iteration schemes),
6.3-6.5 (sketch only), slides #19 and #20 Ch. 12, Sec. 12.1, 12.2, 12.6 (pp. 561-568 & 12.6.5 only) Reading guide for finals (slightly revised) Reading guide for midterm


Homework #1 Due Wed 2/18 before class.
Solutions for HW1

Homework #2 Due Wed 3/4 before class.
Solutions for HW2

Homework #3 Due Wed 3/18 before class.
Solutions for HW3

Homework #4 Due Wed 4/1 before class.
Solutions for HW4

Homework #5 Due Wed 4/15 before class.
Solutions for HW5

Homework #6 Due Wed 4/29 before class.
Solutions for HW6


Slides are "rough cut" at this point. Beware of possible typos and glitches.

Slides #1 (1/21 2009)

Slides #2 (general remarks 1/26)

Slides #3 (Placeholder only)

Slides #4 (Placeholder only)

Slides #5 (boundary condition - ODE 1/26, 1/28)

Slides #6 (boundary condition - eigenvalue prob. 2/2)

Slides #7 (boundary condition - heat equation 1/28, 2/2)

Slides #8 (separation of variables, heat equation 2/2, 2/4)

Slides #9 (some properties of heat equation (II) 2/4)

Slides #10 (Addendum to Slides #8; orthogonality relationship 2/9)

Slides #11 (Placeholder only)

Slides #12 (equilibrium solution; intro to Laplace equation 2/4, 2/9)

Slides #13 (solution to Laplace equation 2/11)

Slides #14 (Fourier transform for PDE (I), 3/18-4/1)

Slides #15 (Fourier transform for PDE (II), 3/18-4/1)

Slides #16 (Placeholder only)

Slides #17 (Sturm-Liouville problem and orthogonality)

Slides #18 (Quick review of numerical method for ODE - BVP)

Slides #19 (Two examples of numerical methods for PDE)

Slides #20 (Two additional examples for numerical solutions of Laplace equation)

Slides are "rough cut" at this point. Beware of possible errors and glitches.



Matlab sample programs

Example_1 - codes
Example_1 - result
This program plots the two functions, u(x) = sin(6πx)*exp(-x) 
and v(x) = cos(6πx)exp(-x), for x ∈ [0, 1]. The discretization 
interval (resolution of plot) is 0.01.

Example_2 - codes
Example_2 - result
This program generates the color+contour map for the function,  
u(x,y) = sin(2πx)sin(2πy)exp(-(x2+y2)), for x ∈ [0,1], y ∈ [0,1]. 
The discretization interval (resolution of plot) is 0.01 for both 
x and y. Contour interval is 0.1 (contour levels are -0.9, -0.8, ...,
-0.2, -0.1, 0.1, 0.2, ... , 0.8, 0.9). Contours for negative 
values are dashed. 
Note: With a given 2-D array, u(q,p), Matlab plots the contours
of u as a "map" of the matrix u(q,p), i.e., q goes up and down
and p goes left and right.  The index q would then correspond to our y,
and p to x.  This is somewhat counterintuitive so beware.  

Example_3 - codes
Example_3 - result
This program plots the solution to the Laplace equation in Slides #12.
Discretization interval (resolution of plot) is 0.01 for both x and y.
Contour interval is 0.05 (contour levels are 0.05, 0.1, 0.15, ..., 0.45).
The infinite series in the solution is truncated at n = 20 for this plot.

Example_4 - codes
Example_4 - result
This program plots the Fourier Sine series representation of the 
function defined on x ∈ [0,1]: F(x) = 1 for 0 ≤ x ≤ 1/2,  
F(x) = 0 for 1/2 < x ≤ 1.  Black is the original function. Magenta,
red, and blue are the Fourier series representation truncated at n = 5,
25, and 100.

Example_5 - codes
Example_5 - result
This program plots the solution to a 2-D wave equation to be 
discussed in class. The expression of the solution is 
u(x,y,t) = sin(2πx)sin(3πy)cos(sqrt(13)πt)
The 4 panels, top-left, top-right, bottom-left, and bottom-right,
correspond to the solutions at t = 0, 0.45/sqrt(13), 0.55/sqrt(13),
and 1/sqrt(13). 

Example_6 - codes
Example_6 - result
This program demonstrates how to make a contour/color map for a 
function defined in polar coordinate. The result is a map for 
u(r,θ) = r*θ, where (r,θ) are the usual polar coordinate
and u is defined within a 60-degree wedge with radius = 1. More 
precisely, the domain is 0 ≤ r ≤ 1, 0 ≤ θ ≤ π/3.

Example_7 - codes
Example_7 - result
Just another example for plotting a function in polar coordinate.
The result is a contour/color map of u(r,θ) = r*cos(3*θ) for   
the annular domain, 1/2 ≤ r ≤ 1.  This program is not needed for
homework but might help you understand Matlab Example 6.

Example_8 - codes
Example_8 - result
This code plots Bessel function of 1st kind and of 0th, 1st, and 
2nd order, J0(z), J1(z), and J2(z), and Bessel function of 2nd kind and
of 0th, 1st, and 2nd order, Y0(z), Y1(z), and Y2(z), for 0 ≤ z ≤ 20.

Example_9 - codes
Example_9 - result
This code solves the nonlinear (quasilinear) PDE, the "shock equation",
∂u/∂t + u∂u/∂x = 0, with the b.c., u(x,0) = sin(x).
Shown are the solution u(x,t) at t = 0.3 and 0.7, and the initial state.
The solution is periodic in x. Only one full period (x = 0 to 2π) is shown.
The method of characteristics is used to obtain an analytic expression of
the solution. (This is done by hand, not part of the code.) Numerical evaluation
of u(x,t) requires root-finding of a nonlinear equation, which is dealt with
by using the bisection method. This code will work only up to a certain value
of t (what is it?), beyond which the solution becomes a multiple-valued
function of x.