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Numerical Methods for Partial Differential Equations. CAAM 452 Spring 2005 Instructor: Tim Warburton. 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 ) - PowerPoint PPT Presentation
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Numerical Methods for Partial Differential Equations
CAAM 452
Spring 2005
Instructor: Tim Warburton
CAAM 452 Spring 2005
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 , and suitable solution spaces for this equation later.
, ,
, , ,
, , ,
,
uu
tu 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
CAAM 452 Spring 2005
Physical Examples
• These and similar equations and vector analogs are pervasive:– Fluid mechanics (Euler equations, compressible Navier-Stokes
equations, magnetohydrodynamics).– Electromagnetics (Maxwell’s equations)– Heat equation– Shallow water equations– Atmospheric models– Ocean models– Bio-population models (morphogenesis, predator prey, epidemiology)– …..
uu
t
f A g
CAAM 452 Spring 2005
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.
CAAM 452 Spring 2005
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 u uc d
t x x
CAAM 452 Spring 2005
Further Simplification
• We can simplify even further by dropping the second order diffusion or dissipation term:
• This PDE is first order in time and first order in space.
• Volunteer to solve this equation analytically?.
0u uc
t x
CAAM 452 Spring 2005
Necessary Information to Solve The IBVP
• The Initial, Boundary, Value Problem represented by the PDErequires some extra information in order to to be solvable.
• What do we need?.
0u uc
t x
CAAM 452 Spring 2005
Answer
In this case, because of the hyperbolic nature of the PDE(solution travels from right to left with increasing time), weneed to supply:a) Extent of solution domainb) 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=0
As we just sawwe also need tospecify inflowdata
x=a x=b
CAAM 452 Spring 2005
Brief Summary
• There is a checklist of conditions we will need to consider to obtain a hopefully unique solution of a PDE
1) The PDE (duh)
2) Boundary values (also known as boundary conditions)
3) Initial values (if there is a time-like variable)
4) Solution domain
CAAM 452 Spring 2005
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 uc
t x
u x u x x a b
u a t u b t t T
u x T x a b
CAAM 452 Spring 2005
Analytical Solution
• Volunteer:
• For this PDE to make sense we should discuss something about u0, what?
1) Find , such that , , 0,
0
given ( ,0) ,
, , [0, ]
2) Evalute , ,
o
u x t x a b t T
u uc
t x
u x u x x a b
u a t u b t t T
u x T x a b
CAAM 452 Spring 2005
Fourier Series Representation (p4 GKO)
1
Assume that C , is 2 periodic.Then
has a Fourier series representation:
1 ˆ 2
ˆwhere the Fourier coefficients are given by
1ˆ
Theorem:
2
i x
i
f
f
f x f e
f
f e
2
0
Finally, the series converges uniformly to
xf x dx
f x
In other words, we can express a sufficiently smooth function in terms of an infinitetrigonometric polynomial. The fhats are the Fourier coefficients of the polynomial.
CAAM 452 Spring 2005
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
1 ˆ( ,0) = 0,22
where is a smooth 2 -periodic function of one frequency
i x
u x t x t T
u uc
t x
u x f x e f x
f
1, ,ˆ
2i xu x t e u t
CAAM 452 Spring 2005
cont
• Substituting in this type of solution the PDE:
• Becomes an ODE:
• With initial condition
0u uc
t x
1 1 ˆ,ˆ ˆ
2 2ˆ
0ˆ
i x i xu u duc c e u t e i cu
t x t x dt
dui cu
dt
ˆ,0u f
CAAM 452 Spring 2005
cont
• We have Fourier transformed the PDE into an ODE.• We can solve the ODE:
• And it follows that the PDE solution is:
ˆ0ˆ ˆ, ,0ˆ ˆ
ˆ,0ˆ
i ct i ct
dui cu
u t e u e fdtu f
1: , ,ˆ
21ˆ ˆsolution : , ,ˆ2
1 ˆinitial condition: 2
i x
i x cti ct
i x
ansatz u x t e u t
u t e f u x t e f f x ct
f x e f
CAAM 452 Spring 2005
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:
1 ˆ2
1 ˆ,2
i x
i x ct
f x e f
u x t e f f x t
CAAM 452 Spring 2005
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).
0u uc
t x
ˆ0ˆ
dui cu
dt
CAAM 452 Spring 2005
Add Diffusion Back In
• So we have a good handle on the advection equation, let’s reintroduce the diffusion term:
• Again, let’s assume 2-pi periodicity and assume the same ansatz:
• This time:
2
2
u u uc d
t x x
1ˆ
2i xu e u t
2
2
u u uc d
t x x
2ˆˆ ˆ
dui cu d u
dt 2ˆ
ˆdu
ic d udt
CAAM 452 Spring 2005
cont
• Again, we can solve this trivial ODE:
2ˆˆ
dui c d u
dt
2
,0ˆ ˆic d t
u u e
2
,0ˆ i ct x d tu u e e
CAAM 452 Spring 2005
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
,0ˆ i ct x d tu u e e
CAAM 452 Spring 2005
What Did Diffusion Do??
• Advection:
• Diffusion:
• Adding the diffusion term shifted the multiplier on the right hand side of the Fourier transformed PDE (i.e. the ODE) into the left half plane.
• We summarize the role of the multiplier…
0u uc
t x
2
2
u u uc d
t x x
ˆ0ˆ
dui cu
dt
2ˆˆ
dui c d u
dt
CAAM 452 Spring 2005
Categorizing a Linear ODE
Re
Im
Exponential growth Exponential decay
Incr
easi
ngly
osc
illat
ory
In
crea
sing
ly
osc
illat
ory
Here we plot the behavior of the solution to the top right ODE for mu in the complex plane
ˆ 0ˆ ˆ ˆ tduu u u e
dt
CAAM 452 Spring 2005
Solving the Scalar ODE Numerically
• We know the solution to the scalar ODE
• However, it is also reasonable to ask if we can solve it approximately.
• We have now simplified as far as possible.
• Once we can solve this model problem numerically, we will apply this technique using the method of lines to approximate the solution of the PDE.
ˆˆ
duu
dt
CAAM 452 Spring 2005
ODE Prototype
• We will consider ODE’s of the kind
0
0
duf u
dtu u
CAAM 452 Spring 2005
ODE Time Stepping Topics
We will cover the following details on time stepping
the ODE.– Stability of time stepping– Stability regions in the complex plane– Accuracy of time stepping– Convergence of time stepping method with decreasing
time step– Examples of explicit time stepping methods:
• Euler forward• Leap-frog• Adams-Bashford • Runge-Kutta
CAAM 452 Spring 2005
Reading for Next Week
• Study: Gustaffson-Kreiss-Oliger (GKO) p3-17 and p38-39
• Brush up your programming and PDE knowledge – there will be frequent implementation exercises.