Conservation Law - Numerical Methods for Partial Differential Equations - Lecture Slides, Slides of Mathematical Methods for Numerical Analysis and Optimization

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Numerical Methods for Partial

Differential Equations

Overview

• Our final goal is to be able to solve PDE’s of the form:
• This is a conservation law with some form of dissipation (under assumptions on A )
• We will discuss boundary conditions, solution domain Ω, and suitable solution spaces for this equation later.

u u

t

u u x y t

u x y t

u x y t

x y

f A g

f f

g g

A A

0

[ , ]

x y

t t T

Divide and Conquer

• It is highly non-trivial to solve these equations analytically (i.e. with smarts, pen and paper).
• We can forget the idea of writing down closed form solutions for the general case.
• We will consider the component parts of the equations and discuss techniques to solve the reduced equations.
• Some very reduced models admit exact solutions which allow us to check how well we are doing.
• Finally we will put different methods together and aim for the big prize.

Simplification

• Let’s choose a simple example, namely the 1D

advection diffusion equation.

• This PDE is first order in time and second

order in space.

2 2

u (^) c u (^) d u t x x

∂ (^) − ∂ (^) = ∂ ∂ ∂ ∂

Necessary Information to Solve The IBVP

• The I nitial, B oundary, V alue P roblem

represented by the PDE

requires some extra information in order to

to be solvable.

• What do we need?.

∂∂ u (^) t − c ∂∂ ux = 0

Answer

In this case, because of the hyperbolic nature of the PDE (solution travels from right to left with increasing time), we need to supply: a) Extent of solution domain b) What is the solution at start of the solution process: u(x,0) c) Boundary data: u(b,t) d) Final integration time.

x

t

Need to specify the solution at t=

As we just saw we also need to specify inflow data

x=a x=b

Periodic Case

• Suppose we remove the inflow and imagine that the interval [a,b) is periodic.
• Further suppose we wish to solve for the solution at some non-negative time T.
• We can indicate this by the following specification:

( ) [ ) [ ]

( ) [ ) ( ) ( ) ( ) [ )

1. Find , such that , , 0, 0 given ( ,0) , , , [0, ]
2. Evalute , ,

o

u x t x a b t T u (^) c u t x

u x u x x a b u a t u b t t T u x T x a b

∀ ∈ Docsity.com

Analytical Solution

• Volunteer:
• For this PDE to make sense we should discuss

something about u 0 , what?

( ) [ ) [ ]

( ) [ ) ( ) ( ) ( ) [ )

1. Find , such that , , 0, 0 given ( ,0) , , , [0, ]
2. Evalute , ,

o

u x t x a b t T u (^) c u t x

u x u x x a b u a t u b t t T u x T x a b

Returning to the Advection Equation

• We wills start with a specific Fourier mode as the initial condition:
• We try to find a solution of the same type:

( ) [ ) [ ]

( ) ( ) [ )

1. Find 2 -periodic , such that 0, 2 , 0, 0 given ( ,0) = 1 ˆ 0, 2 2 where is a smooth 2 -periodic function of one frequency

i x

u x t x t T u (^) c u t x

u x f x e f x f

ω

u x t ( , (^) ) =^12 π e^ i^ ω^^ xu ˆ (^) ( ω, t ) Docsity.com

cont

• Substituting in this type of solution the PDE:
• Becomes an ODE:
• With initial condition

u (^) c u 0 t x

∂ (^) − ∂ = ∂ ∂

u c u c e i x u t ei x du i cu

t x t x dt

du i cu

dt

ω ω ω ω π π ω

∂ ∂ ^ ∂ ∂ ^ ^  ^ 

u ˆ ( ω,0) = f ˆ( ω) Docsity.com

Note on Fourier Modes

• Note that since the function should be 2pi

periodic we are able to deduce:

• We can also use the superposition principle

for the more general case when the initial

condition contains multiple Fourier modes:

ω ∈ 

( ) (^ )^ ( ) ( )

(^1) ˆ 2

, 1 ˆ 2

i x

i x ct

f x e f

u x t e f f x t

ω (^) ω ω ω (^) ω ω

ω π

ω π

=∞ =−∞ =∞ (^) + =−∞

=

⇒ = = +

∑

∑

cont

• Let’s back up a minute – the crucial part was

when we reduced the PDE to an ODE:

• The advantage is: we know how to solve ODE’s

both analytically and numerically (more about

this later on).

u (^) c u 0 t x

∂ (^) − ∂ = ∂ ∂

du ˆ (^) i cu ˆ 0 dt

− ω =

cont

• Again, we can solve this trivial ODE: dudt ˆ (^) = (^) ( i ω c − d ω (^2) ) u ˆ

( ) (^ )

2 u ˆ = u ˆ ω ,0 e ic^ ω^^ − d^ ω t

( ) (^ )^

2 u = u ˆ ω ,0 ei^ ω ct^ +^ x^ e − d^ ω t

cont

• The solution tells a story:
• The original profile travels in the direction of decreasing x (first exponential term)
• As the profile travels it decreases in amplitude (second exponential term)

( ) (^ )^

2

u = u ˆ ω ,0 e i^ ω^ ct^ +^ x^ e − d^ ω t