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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.
∂∂ 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:
( ) [ ) [ ]
( ) [ ) ( ) ( ) ( ) [ )
- Find , such that , , 0, 0 given ( ,0) , , , [0, ]
- 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?
( ) [ ) [ ]
( ) [ ) ( ) ( ) ( ) [ )
- Find , such that , , 0, 0 given ( ,0) , , , [0, ]
- 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:
( ) [ ) [ ]
( ) ( ) [ )
- 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