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Dr. Noemi Friedman, 28.10.2015.
Introduction to PDEs and Numerical Methods
Lecture 1:
Introduction
28. 10. 2015. | PDE lecture | Seite 2
Basic information on the course
Course Title:
Introduction to PDEs and Numerical Methods
Lecturer:
Noémi Friedman
Mühlenpfordtstr. 23, 8th floor
Room: 819
Assistant (exercises):
Jaroslav Vondřejc
Mühlenpfordtstr. 23, 8th floor
Room: 822
Assistant2 (small tutorials):
Stephan Lenz (CSE student)
28. 10. 2015. | PDE lecture | Seite 3
Basic information on the course
Credits and work load:
5 credits: 6-7 hours/week
Pre-requisits:
Differential operators,
elementary knowledge of PDEs,
basics of linear algebra,
basic MATLAB coding skills
Requisits:
Weekly assignments in group of two or three (min 50%)
Written exam: 29.2.2016., 10:30 - 12:00, room ZI 24.1
Script, recommended literature:
See webpage: https://www.tu-braunschweig.de/wire/lehre/ws15/pde1
Software used:
MATLAB, FEniCS (Python interface)
28. 10. 2015. | PDE lecture | Seite 4
Information about the assignments
Homework assignments in groups (max. group of three)
Submission of homework
Written homework
Submit on the beginning of the tutorial
(include cover sheet with subject name (PDE1), names and matriculation
number of students, assignment number)
For program codes:
e-mail: [email protected]
subject: assignment# NAMES
(#: number of the assignment, NAMES: names of students)
(e.g.: assignment1 J. Smith, K. Park)
Homework is due to the beginning of the tutorials
Consultation:
Noemi Friedman (after the lecture, office hours by arrangement, please, take
appointment first by e-mail: [email protected])
Jaroslav Vondřejc (will be assigned on the tutorial
28. 10. 2015. | PDE lecture | Seite 5
Definition: ODEs PDEs
Partial differential equation:
Equation specifying a relation between the partial derivative(s) of an unkown
multivariable function and maybe the function itself:
𝐹 𝑢 𝑥, 𝑦, 𝑧, 𝑡 ,𝜕𝑢 𝑥, 𝑦, 𝑧, 𝑡
𝜕𝑥,𝜕𝑢 𝑥, 𝑦, 𝑧, 𝑡
𝜕𝑦, …
𝜕2𝑢 𝑥, 𝑦, 𝑧, 𝑡
𝜕𝑥𝑦, … = 𝑓(𝑥, 𝑦, 𝑧, 𝑡)
Ordinary differential equation:
Equation specifying a relation between the derivative(s) of an unkown univariable
function and maybe the function itself:
𝐹 𝑢 𝑡 ,𝑑𝑢 𝑡
𝜕𝑡,
𝑑2𝑢 𝑡
𝑑𝑡2 , … = 𝑓(𝑡)
Boundary Value Problem (BVP), Initial Boundary Value Problem (IBVP):
PDE with initial/boundary conditions
28. 10. 2015. | PDE lecture | Seite 6
Motivation – simulation of planets
28. 10. 2015. | PDE lecture | Seite 7
Motivation – heat convection
Source:
ANSYS http://www.ansys.com/staticassets/ANSYS/staticassets/product/16-highlights/underhood-simulation-surface-temps-heat-
transfer-manifold-2.jpg
computed surface temperatures due to
convective and radiative heat transfer from
the exhaust manifold to surrounding
objects
28. 10. 2015. | PDE lecture | Seite 8
Motivation – structural analysis
Source: ANSYS
http://wildeanalysis.co.uk/system/photos/838/preview/ansys_ex
plicit_str.png?1273430962
28. 10. 2015. | PDE lecture | Seite 9
Motivation – flow problems
The Stokes equation
Source:
FENICS documentation:
http://fenicsproject.org/documentation/dolfin/1.6.0/python/demo/documented/stokes-taylor-hood/python/documentation.html
−𝛻 ⋅ 𝛻 𝑢 + 𝑝 𝐼 = 𝑓 in Ω∇⋅ 𝑢 = 0 in Ω
𝑝(x)𝑢(x)
Motivation – flow problems
28. 10. 2015. | PDE lecture | Seite 10
Source: https://33463d8ba37cd0a930f1-
eb07ed6f28ab61e35047cec42359baf1.ssl.cf5.rackcdn.com/ugc/entry/5
44867a78ef07-133981577813387476428_0129_fast15.jpg
Source: TU Braunschweig SFB 880
28. 10. 2015. | PDE lecture | Seite 11
Motivation – highly coupled systems
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Overview of the course
Introduction (definition of PDEs, classification, basic math,
introductory examples of PDEs)
Analitical solution of elementary PDEs (Fourier series/transform,
seperation of variables, Green’s function)
Numerical solutions of PDEs:
Finite difference method
Finite element method
28. 10. 2015. | PDE lecture | Seite 13
Overview of this lecture
Basic definitions, motivation
Differential operators: basic notations, divergence, Laplace, curl, grad
Classification of PDEs
Introductory example: heat flow in a bar
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Differential operators – partial derivative
Partial derivative:
𝜑(𝑥, 𝑦, 𝑧, 𝑡)
𝜕𝜑
𝜕𝑥, 𝜕𝑥𝜑, 𝜑𝑥 ,
𝜕
𝜕𝑥𝜑, 𝜑,𝑥 , (𝜑′)
(Image source: Wikipedia)
𝑧 = 𝑓(𝑥, 𝑦) = 𝑥2 + 𝑥𝑦 + 𝑦2𝜕𝜑
𝜕𝑥= 2𝑥 + 𝑦
𝜕𝜑
𝜕𝑥𝑥=1,𝑦=1
= 3
𝜕𝜑
𝜕𝑡= 𝜑
Example:
28. 10. 2015. | PDE lecture | Seite 15
Differential operators – mixed derivative
Mixed derivative:
𝜑(𝑥, 𝑦)
𝜕2𝜑
𝜕𝑥𝜕𝑦=
𝜕2𝜑
𝜕𝑦𝜕𝑥
𝑓(𝑥, 𝑦, 𝑧) = 𝑥𝑦2cos(𝑧)
𝜕𝑓
𝜕𝑥= 𝑦2cos(𝑧)
𝜕𝑓
𝜕𝑦= 2𝑥𝑦 cos(𝑧)
𝜕2𝜑
𝜕𝑥𝜕𝑦= 2𝑦cos(𝑧)
𝜕2𝜑
𝜕𝑦𝜕𝑥= 2𝑦cos(𝑧)
Example:
28. 10. 2015. | PDE lecture | Seite 16
Differential operators – total derivative
Total derivative:
Total differential (differential change of f):
𝑑𝜑
𝑑𝑡=
𝜕𝜑
𝜕𝑡
𝑑𝑡
𝑑𝑡+
𝜕𝜑
𝜕𝑥
𝑑𝑥
𝑑𝑡+
𝜕𝜑
𝜕𝑦
𝑑𝑦
𝑑𝑡+
𝜕𝜑
𝜕𝑧
𝑑𝑧
𝑑𝑡
𝜑 𝒓 = 𝜑 𝑥, 𝑦, 𝑧, 𝑡
𝑑𝜑 =𝑑𝜑
𝑑𝑡+
𝜕𝜑
𝜕𝑥𝑑𝑥 +
𝜕𝜑
𝜕𝑦𝑑𝑦 +
𝜕𝜑
𝜕𝑧𝑑𝑧
𝜑 𝑥, 𝑦 = 𝑥2 + 2𝑦
𝑑𝜑
𝑑𝑥=
𝜕𝜑
𝜕𝑥
𝑑𝑥
𝑑𝑥+
𝜕𝜑
𝜕𝑦
𝑑𝑦
𝑑𝑥= 2𝑥 + 2𝑦 𝑥 = 𝑥
𝜕𝜑
𝜕𝑥= 2𝑥
total derivative
partial derivative
Example:
28. 10. 2015. | PDE lecture | Seite 17
Differential operators – gradient
Nabla operator:
Gradient:
𝜑 𝒓 = 𝜑 𝑥, 𝑦, 𝑧 : ℝ3 → ℝ (vector-scalar function)
𝛻 =
𝜕
𝜕𝑥𝜕
𝜕𝑦𝜕
𝜕𝑧
Example:
• direction: greatest rate of increase of
the function
• magnitude: the slope of the function in
that direction
28. 10. 2015. | PDE lecture | Seite 18
Differential operators – directional derivative
𝜑 𝒓 = 𝜑 𝑥, 𝑦, 𝑧 : ℝ3 → ℝ (vector-scalar function)
𝐷𝒗𝑓 𝒓 = limℎ→0
𝜑 𝒓 + ℎ𝒗 − 𝜑 𝒓
ℎ= 𝒗 ∙ 𝛻 𝑓 𝒓 = 𝒗𝑇 𝛻 𝑓 𝒓 = 𝑣𝑥 𝑣𝑦 𝑣𝑧
𝜕𝜑 𝑥, 𝑦, 𝑧
𝜕𝑥𝜕𝜑 𝑥, 𝑦, 𝑧
𝜕𝑦
𝜕𝜑 𝑥, 𝑦, 𝑧
𝜕𝑧
Directional derivative:
(normalised):
𝐷𝒗𝑓 𝒓 = limℎ→0
𝑓 𝒓 + ℎ𝒗 − 𝑓 𝒓
ℎ 𝒗=
𝒗
𝒗∙ 𝛻 𝑓 𝒓
𝑡
𝑥
𝑢What is the differential equation to define a wave
traveling with speed 𝑐?
In the direction 𝑥 − 𝑐𝑡 𝑢 is constant→ directional
derivative is zero:
𝐷𝒗𝑢 𝑥, 𝑡 =𝒗
𝒗∙ 𝛻 𝑢 𝑥, 𝑡 = 0 (𝒗 ∙ 𝛻 𝑢 𝑥, 𝑡 = 0)
Example 1:
28. 10. 2015. | PDE lecture | Seite 19
Differential operators – directional derivative
𝒗 ∙ 𝛻 𝑢 𝑥, 𝑡 = 𝒗𝑇 𝛻 𝑢 𝑥, 𝑡 = 𝑐 1
𝜕𝑢
𝜕𝑥𝜕𝑢
𝜕𝑡
= 0
𝑡
𝑥
𝑢 𝒗 =𝑐1 𝒗 ∙ 𝛻 𝑢 𝑥, 𝑡 = 0
Let’s suppose 𝑢 = sin(𝑥 − 𝑐𝑡) is a solution of the transport equation.
What is its directional derivative in the direction:
𝑢𝑡 + 𝑐𝑢𝑥 = 0Transport equation:
𝒗 =𝑐1
𝒗 ∙ 𝛻 𝑢 𝑥, 𝑡 = 𝒗𝑇 𝛻 𝑢 𝑥, 𝑡 = 𝑐 1cos(𝑥 − 𝑐𝑡)
−𝑐 cos(𝑥 − 𝑐𝑡= 0
Example 2:
28. 10. 2015. | PDE lecture | Seite 20
Differential operators - divergence
of 𝒈(𝑥, 𝑦, 𝑧): ℝ3 → ℝ3((of a vector field):
𝒈 𝑥, 𝑦, 𝑧 =
𝑔𝑥(𝑥, 𝑦, 𝑧)𝑔𝑦(𝑥, 𝑦, 𝑧)
𝑔𝑧(𝑥, 𝑦, 𝑧)
=
𝑔𝑥
𝑔𝑦
𝑔𝑧
𝑑𝑖𝑣𝒈 𝑥, 𝑦, 𝑧 = 𝛻 ∙ 𝒈 𝑥, 𝑦, 𝑧 =𝜕
𝜕𝑥
𝜕
𝜕𝑦
𝜕
𝜕𝑧
𝑔𝑥
𝑔𝑦
𝑔𝑧
=𝜕𝑔𝑥
𝜕𝑥+
𝜕𝑔𝑦
𝜕𝑦+
𝜕𝑔𝑧
𝜕𝑧
Example:
Divergence:
28. 10. 2015. | PDE lecture | Seite 21
Differential operators - Laplace
Example:
Laplace operator:
28. 10. 2015. | PDE lecture | Seite 22
Differential operators – rotation (curl)
of 𝒈(𝑥, 𝑦, 𝑧): ℝ3 → ℝ3((of a vector field):
𝒈 𝑥, 𝑦, 𝑧 =
𝑔𝑥(𝑥, 𝑦, 𝑧)𝑔𝑦(𝑥, 𝑦, 𝑧)
𝑔𝑧(𝑥, 𝑦, 𝑧)
=
𝑔𝑥
𝑔𝑦
𝑔𝑧
𝑟𝑜𝑡𝒈 𝑥, 𝑦, 𝑧 = 𝛻 × 𝒈 𝑥, 𝑦, 𝑧 = 𝑑𝑒𝑡
𝒊 𝒋 𝒌𝜕
𝜕𝑥
𝜕
𝜕𝑦
𝜕
𝜕𝑧𝑔𝑥 𝑔𝑦 𝑔𝑧
=
𝜕𝑔𝑧
𝜕𝑦−
𝜕𝑔𝑦
𝜕𝑧𝜕𝑔𝑥
𝜕𝑧−
𝜕𝑔𝑧
𝜕𝑥𝜕𝑔𝑦
𝜕𝑥−
𝜕𝑔𝑥
𝜕𝑦
Example:
Rotation (curl):
• direction: axis of rotation
• magnitude: magnitude of rotation
28. 10. 2015. | PDE lecture | Seite 23
Classification of PDEs
Constant/variable coefficients
Stationary/instationary (not time dependent/time dependent)
Linear/nonlinear
linearity condition:
order
order of the highest derivative
homogeneous/inhomogenous
inhomogeneous: additive terms which do not depend on unknown function
homogenous: 𝑢 = 0 is a solution of the equation
elliptic/parabolic/hyperbolic (only for second order PDEs)
𝐴𝑢𝑥𝑥 + 2𝐵𝑢𝑥𝑦 + 𝐶𝑢𝑦𝑦 + lower order derivatives = 0
• 𝐴𝐶 − 𝐵2 = 0 parabolic
• 𝐴𝐶 − 𝐵2 < 0 hyperbolic
• 𝐴𝐶 − 𝐵2 > 0 elliptic
28. 10. 2015. | PDE lecture | Seite 24
Classification of PDEs, examples of PDEs
Wave equation
𝑢𝑡𝑡 − 𝒄𝟐𝑢𝑥𝑥 = 0Laplace equation
𝑢𝑥𝑥 + 𝑢𝑦𝑦 = 0Heat equation
𝑢𝑡 − 𝑢𝑥𝑥 = 0
Order 2 2 2
Constant
coefficient?
yes yes yes
Linear? yes yes yes
Homogenous? yes yes yes
Class A=1, B=0, C=−𝒄𝟐
𝐴𝐶 − 𝐵2 = −𝒄𝟐
Hyperbolic
A=1, B=0, C=1𝐴𝐶 − 𝐵2 =1
Elliptic
A=-1, B=0, C=0𝐴𝐶 − 𝐵2 = 0
Parabolic
28. 10. 2015. | PDE lecture | Seite 25
Classification of PDEs, examples of PDEs
Linearity:
28. 10. 2015. | PDE lecture | Seite 26
Classification of PDEs, examples of PDEs
28. 10. 2015. | PDE lecture | Seite 27
Introductory example: heat flow in a bar
Basic assumptions:
Uniform cross-section
Temperature varies only in the longitudinal direction
Relationship between heat energy and temperature is linear
𝑐𝐽
𝑔𝐾: cpecific heat capacity Energy [𝐽]
𝑐𝐽 energy is required to raise the tempreture by 1𝐾 of 1𝑔 material
Homogenous material properties along the bar (𝜌 and 𝑐 are constants along the bar)
𝜌[𝑔
𝑚3]: density of the material of the bar
Problem description 𝑢 𝑥, 𝑡 =? (temparature)
0 𝑙𝑥 𝑥 + ∆𝑥
∆𝑥
𝑞(0, 𝑡) 𝑞(𝑙, 𝑡)
𝐴
28. 10. 2015. | PDE lecture | Seite 28
Introductory example: heat flow in a bar
0 𝑙𝑥 𝑥 + ∆𝑥
∆𝑥
𝑞(0, 𝑡) 𝑞(𝑙, 𝑡)
𝐴
∆𝑥Total energy in the bar section of length ∆𝑥 is:
𝐸𝑡𝑜𝑡 = 𝐸0 + 𝑥
𝑥+∆𝑥
𝐴𝜌𝑐(𝑢 𝑠, 𝑡 − 𝑇0))𝑑𝑠 =
𝐸0 + 𝑥
𝑥+∆𝑥
𝐴𝜌𝑐𝑢 𝑠, 𝑡 𝑑𝑠 − 𝑥
𝑥+∆𝑥
𝐴𝜌𝑐𝑇0𝑑𝑠
𝐴𝜌𝑐𝑇0𝑑𝑠But I’m only interested in:
𝜕
𝜕𝑡 𝑥
𝑥+∆𝑥
𝐴𝜌𝑐𝑢 𝑠, 𝑡 𝑑𝑠 = 𝑥
𝑥+∆𝑥
𝐴𝜌𝑐𝜕𝑢 𝑠, 𝑡
𝜕𝑡𝑑𝑠
28. 10. 2015. | PDE lecture | Seite 29
Introductory example: heat flow in a bar
∆𝑥
𝜕
𝜕𝑡 𝑥
𝑥+∆𝑥
𝐴𝜌𝑐𝑢 𝑠, 𝑡 𝑑𝑠 = 𝑥
𝑥+∆𝑥
𝐴𝜌𝑐𝜕𝑢 𝑠, 𝑡
𝜕𝑡𝑑𝑠
Change of heat energy by time in the section of length ∆𝑥
𝑞(𝑥, 𝑡) 𝑞(𝑥 + ∆𝑥, 𝑡)
Change of energy by time in the section of length ∆𝑥 from heat flux:
𝑞 𝑥, 𝑡𝐽
𝑚2𝑠: heat flux
𝐴𝑞 𝑥 + ∆𝑥, 𝑡 − 𝐴𝑞 𝑥, 𝑡 = 𝑥
𝑥+∆𝑥
𝐴𝜕𝑞 𝑠, 𝑡
𝜕𝑥𝑑𝑠
Fundamental
theorem of calculus:
Conservation of energy:
𝑥
𝑥+∆𝑥
𝐴𝜕𝑞 𝑠, 𝑡
𝜕𝑥𝑑𝑠 = −
𝑥
𝑥+∆𝑥
𝐴𝜌𝑐𝜕𝑢 𝑠, 𝑡
𝜕𝑡𝑑𝑠
𝑥
𝑥+∆𝑥
𝐴𝜕𝑞 𝑠, 𝑡
𝜕𝑥+ 𝐴𝜌𝑐
𝜕𝑢 𝑠, 𝑡
𝜕𝑡𝑑𝑠 = 0
28. 10. 2015. | PDE lecture | Seite 30
Introductory example: heat flow in a bar
∆𝑥
𝑞(𝑥, 𝑡) 𝑞(𝑥 + ∆𝑥, 𝑡) 𝑥
𝑥+∆𝑥
𝐴𝜕𝑞 𝑠, 𝑡
𝜕𝑥+ 𝐴𝜌𝑐
𝜕𝑢 𝑠, 𝑡
𝜕𝑡𝑑𝑠 = 0
𝜕𝑞 𝑥, 𝑡
𝜕𝑥+ 𝜌𝑐
𝜕𝑢 𝑥, 𝑡
𝜕𝑡= 0 0 < 𝑥 < 𝑙
𝜕𝑢 𝑥,𝑡
𝜕𝑥: change of temperature with increasing 𝑥
Assumption: Fourier’s law of heat conduction: 𝑞 𝑥, 𝑡 = −𝜅𝜕𝑢 𝑥, 𝑡
𝜕𝑥
𝜌𝑐𝜕𝑢 𝑥, 𝑡
𝜕𝑡− 𝜅
𝜕2𝑢 𝑥, 𝑡
𝜕𝑥2 = 0Heat equation
𝜌(𝑥)𝑐(𝑥)𝜕𝑢 𝑥, 𝑡
𝜕𝑡−
𝜕
𝜕𝑥𝜅(𝑥)
𝜕𝑢 𝑥, 𝑡
𝜕𝑥= 0 (for inhomogenous material
properties)
(for homogenous material properties)
28. 10. 2015. | PDE lecture | Seite 31
Introductory example: heat flow in a bar
∆𝑥
𝑞(𝑥, 𝑡) 𝑞(𝑥 + ∆𝑥, 𝑡) 𝑥
𝑥+∆𝑥
𝐴𝜕𝑞 𝑠, 𝑡
𝜕𝑥+ 𝐴𝜌𝑐
𝜕𝑢 𝑠, 𝑡
𝜕𝑡𝑑𝑠 =
𝑥
𝑥+∆𝑥
𝐴𝑓(𝑠, 𝑡) 𝑑𝑠
𝜕𝑞 𝑥, 𝑡
𝜕𝑥+ 𝜌𝑐
𝜕𝑢 𝑥, 𝑡
𝜕𝑡= 𝑓(𝑥, 𝑡) 0 < 𝑥 < 𝑙
𝜌𝑐𝜕𝑢 𝑥, 𝑡
𝜕𝑡− 𝜅
𝜕2𝑢 𝑥, 𝑡
𝜕𝑥2= 𝑓(𝑥, 𝑡)
Heat equation
𝜌(𝑥)𝑐(𝑥)𝜕𝑢 𝑥, 𝑡
𝜕𝑡−
𝜕
𝜕𝑥𝜅(𝑥)
𝜕𝑢 𝑥, 𝑡
𝜕𝑥= 𝑓(𝑥, 𝑡) (for inhomogenous material properties)
(for homogenous material properties)
The heat equation with source or sink (inhomogenous heat equation)
28. 10. 2015. | PDE lecture | Seite 32
Introductory example: heat flow in a bar
0 𝑙
𝑞(0, 𝑡) 𝑞(𝑙, 𝑡)
𝐴
𝜌𝑐𝜕𝑢 𝑥, 𝑡
𝜕𝑡− 𝜅
𝜕2𝑢 𝑥, 𝑡
𝜕𝑥2 = 𝑓(𝑥, 𝑡)
Boundary conditions:
a) Perfect isolation at the end (flux across the boundaries is zero):
−𝜅𝜕𝑢 0, 𝑡
𝜕𝑥= 0 −𝜅
𝜕𝑢 𝑙, 𝑡
𝜕𝑥= 0 ∀𝑡
b) Perfect thermal contact:
𝑢 0, 𝑡 = 0 𝑢 𝑙, 𝑡 = 0 ∀𝑡