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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

[email protected]

Mühlenpfordtstr. 23, 8th floor

Room: 819

Assistant (exercises):

Jaroslav Vondřejc

[email protected]

Mühlenpfordtstr. 23, 8th floor

Room: 822

Assistant2 (small tutorials):

Stephan Lenz (CSE student)

mailto:[email protected]



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)


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




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


Motivation – simulation of planets


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


Motivation – structural analysis

Source: ANSYS

http://wildeanalysis.co.uk/system/photos/838/preview/ansys_ex

plicit_str.png?1273430962


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


Source: https://33463d8ba37cd0a930f1-

eb07ed6f28ab61e35047cec42359baf1.ssl.cf5.rackcdn.com/ugc/entry/5

44867a78ef07-133981577813387476428_0129_fast15.jpg

Source: TU Braunschweig SFB 880


Motivation – highly coupled systems


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


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


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:


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:


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:


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


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:


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:


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:


Differential operators - Laplace

Example:

Laplace operator:


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


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


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



Linearity:



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



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



∆𝑥

𝜕

𝜕𝑡 𝑥

𝑥+∆𝑥

𝐴𝜌𝑐𝑢 𝑠, 𝑡 𝑑𝑠 = 𝑥

𝑥+∆𝑥


𝜕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


𝜕𝑥𝑑𝑠 = −

𝑥

𝑥+∆𝑥


𝜕𝑡𝑑𝑠

𝑥

𝑥+∆𝑥


𝜕𝑥+ 𝐴𝜌𝑐

𝜕𝑢 𝑠, 𝑡

𝜕𝑡𝑑𝑠 = 0



∆𝑥

𝑞(𝑥, 𝑡) 𝑞(𝑥 + ∆𝑥, 𝑡) 𝑥

𝑥+∆𝑥



𝜕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)



∆𝑥

𝑞(𝑥, 𝑡) 𝑞(𝑥 + ∆𝑥, 𝑡) 𝑥

𝑥+∆𝑥



𝜕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


𝜕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)



0 𝑙

𝑞(0, 𝑡) 𝑞(𝑙, 𝑡)

𝐴


𝜕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

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