Numerical Methods for Differential · PDF fileNumerical Methods for Differential Equations ......
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Numerical Methods for Differential Equations
Chapter 5: Partial differential equations – elliptic and pa rabolic
Gustaf Soderlind and Carmen Arevalo
Numerical Analysis, Lund University
Textbooks: A First Course in the Numerical Analysis of Differential Equations, by Arieh Iserles
and Introduction to Mathematical Modelling with Differential Equations, by Lennart Edsberg
c© Gustaf Soderlind, Numerical Analysis, Mathematical Sciences, Lund University, 2008-09
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1. Brief overview of PDE problems
Classification: Three basic types, four prototype equations
Elliptic
∆u = 0 + BC
Parabolic
ut = ∆u + BC & IC
Hyperbolic
utt = ∆u + BC & IC
ut + a(u)ux = 0 + BC & IC
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Classification of PDEs
Linear PDE with two independent variables
Auxx + 2Buxy + Cuyy + L(ux, uy, u, x, y) = 0
with L linear in ux, uy, u. Study
δ := det
A B
B C
= AC − B2
Elliptic δ > 0 Parabolic δ = 0 Hyperbolic δ < 0
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Standard PDEs. Prototypical problems
δ > 0 Elliptic PDE Laplace equation
uxx + uyy = 0 A = C = 1; B = 0
δ = 0 Parabolic PDE Diffusion equation
ut = uxx A = 1; B = C = 0
δ < 0 Hyperbolic PDEs Wave equation
utt = uxx A = 1; B = 0; C = −1
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PDE method types
FDM Finite difference methods
FEM Finite element methods
FVM Finite volume methods
BEM Boundary element methods
We will mostly study FDM to cover basic theory
Industrial relevance: FEM
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PDE methods for elliptic problems
Simple geometry FDM or Fourier methods
Complex geometry FEM
Special problems FVM or BEM
Large sparse systemsCombine with iterative solvers such as multigrid methods
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PDE methods for parabolic problems
Simple geometry FDM or Fourier methods
Complex geometry FEM
StiffnessAlways use A–stable time-stepping methods
Need Newton-type solvers for large sparse systems
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PDE methods for hyperbolic problems
FDM, FVM. Sometimes FEM
ShocksSolutions may be discontinuous – example: “sonic boom”
TurbulenceMultiscale phenomena
Hyperbolic problems have several complications and manyhighly specialized techniques are often needed
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2. Elliptic problems. FDM
Laplacian ∆ =∂2
∂x2+
∂2
∂y2+
∂2
∂z2
Laplace equation ∆u = 0
with boundary conditions u = u0(x, y, z) x, y, z ∈ ∂Ω
Poisson equation −∆u = f
with boundary conditions u = u0(x, y, z) x, y, z ∈ ∂Ω
Other boundary conditions also of interest (∂Ω = bdry of Ω)
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Elliptic problems. Some applications
Equilibrium problems
Structural analysis (strength of materials)
Heat distribution
Potential problems
Potential flow (inviscid, subsonic flow)
Electromagnetics (fields, radiation)
Eigenvalue problems
Acoustics
Microphysics
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Poisson equation – an elliptic model problem
∂2u
∂x2+
∂2u
∂y2= f(x, y)
Computational domain Ω = [0, 1] × [0, 1] (unit square)Dirichlet conditions u(x, y) = 0 on boundary
Uniform grid xi, yjN,Mi,j=1 with equidistant mesh widths
∆x = 1/(N + 1) and ∆y = 1/(M + 1)
Discretization Finite differences with ui,j ≈ u(xi, yj)
ui−1,j − 2ui,j + ui+1,j
∆x2+
ui,j−1 − 2ui,j + ui,j+1
∆y2= f(xi, yj)
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Equidistant mesh ∆x = ∆y
ui−1,j + ui,j−1 − 4ui,j + ui,j+1 + ui+1,j
∆x2= f(xi, yj)
Participating approximations and mesh points
xi−1 xi xi+1
yj
yj+1
yj−1
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Computational “stencil” for ∆x = ∆y
ui−1,j + ui,j−1 − 4ui,j + ui,j+1 + ui+1,j
∆x2= f(xi, yj)
“Five-point operator”1
1 −4 1
1
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The FDM linear system of equations
Lexicographic ordering of unknowns ⇒ partitioned system
1
∆x2
T I 0 . . .
I T I
I T I
. . . I
. . . 0 I T
u·,1
u·,2
u·,3
...
u·,N
=
f(x·, y1)
f(x·, y2)
f(x·, y3)...
f(x·, yN )
with Toeplitz matrix T = tridiag(1 −4 1)
The system is N 2 × N 2, hence large and very sparse
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3. Elliptic problems. FEM
Finite Element Method
PDE Lu = 0
Ansatz u =∑
ciϕi ⇒ Lu =∑
ciLϕi
Requirement 〈ϕi, Lu〉 = 0 gives coefficients ci
FEM is a least squares approximation, fitting a linearcombination of basis functions ϕi to the differentialequation using orthogonality
Simplest case Piecewise linear basis functions
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Strong and weak forms
Strong form
−∆u = f ; u = 0 on ∂Ω
Take v with v = 0 on ∂Ω∫
Ω
−∆u · v =
∫
Ω
f · v
Integrate by parts to get weak form∫
Ω
∇u · ∇v =
∫
Ω
f · v
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Strong and weak forms. 1D case
Recall integration by parts in 1D
∫ 1
0
−u′′v = [−u′v]10 +
∫ 1
0
u′v′
or in terms of an inner product
−〈u′′, v〉 = 〈u′, v′〉
Generalization to 2D, 3D uses vector calculus
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Weak form of −∆u = f
Define inner product
〈v, u〉 =
∫
Ω
vu dΩ
and energy norm (note scalar product!)
a(v, u) =
∫
Ω
∇v · ∇u dΩ
to get the weak form of −∆u = f as
a(v, u) = 〈v, f〉
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Galerkin method (Finite Element Method)
1. Basis functions ϕi2. Approximate u =
∑
cjϕj
3. Determine cj from∑
cj
∫
∇ϕi · ∇ϕj =∫
fϕi
The cj are determined by the linear system
Kc = F
The matrix K is called stiffness matrix
Stiffness matrix elements kij =∫
∇ϕi · ∇ϕj = a(ϕi, ϕj)
Right-hand side Fi =∫
ϕif = 〈ϕi, f〉
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The FEM mesh. Domain triangulation
Piecewise linear basis ϕj require domain triangulation
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
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4. Parabolic problems
The prototypical equation is the
Diffusion equation ut = ∆u
Nonlinear diffusion
ut = div (k(u)gradu)
Boundary and initial conditions are needed
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Parabolic problems. Some applications
Diffusive processes
Heat conduction ut = d · uxx
Chemical reactions
Reaction–diffusion ut = d · uxx + f(u)
Convection–diffusion ut = ux + 1Pe
uxx
Irreversibility ut = −∆u is not well-posed!
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Diffusion – a parabolic model problem
Equation ut = uxx
Initial values u(x, 0) = g(x)
Boundary values u(0, t) = u(1, t) = 0
Separation of variables u(x, t) := X(x)T (t) ⇒
ut = XT , uxx = X ′′T ⇒ T
T=
X ′′
X=: λ
T = Ceλt X = A sin√−λx + B cos
√−λx
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Parabolic model problem. . .
Boundary values
X(0) = X(1) = 0 ⇒ λk = −(kπ)2, therefore
Xk(x) =√
2 sin kπx Tk(t) = e−(kπ)2t
Initial values
Fourier expansion g(x) =∑
∞
1 ck
√2 sin kπx ⇒
u(x, t) =√
2
∞∑
k=1
cke−(kπ)2tsin kπx
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5. Method of lines (MOL) discretization
In ut = uxx, discretize ∂2/∂x2 by
uxx ≈ ui−1 − 2ui + ui+1
∆x2
System of ODEs (semidiscretization) u = T∆xu
u =1
∆x2
−2 1
1 −2 1. . .
1 −2
u
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Full FDM discretization
Note that ui(t) ≈ u(x, t) along the line x = xi in (x, t) plane
Use time-stepping to solve the IVP; Explicit Euler withui,j ≈ u(xi, tj) implies
ui,j+1 − ui,j
∆t=
ui−1,j − 2ui,j + ui+1,j
∆x2
With the Courant number µ = ∆t/∆x2 we obtain recursion
ui,j+1 = ui,j + µ · (ui−1,j − 2ui,j + ui+1,j)
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Method of lines. The grid
Rectangular grid (i · ∆x, j · ∆t), i = 0 : N + 1, j ≥ 0 with∆x = 1/(N + 1)
0 1 2 3 4 5 6 70
1
2
3
4
5
6
7
8
t
x
ui,j ≈ u(i · ∆x, j · ∆t)
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Method of lines. Computational stencil
Explicit Euler time stepping. Participating grid points
xi−1 xi xi+1
tj
tj+1
Courant number µ = ∆t/∆x2
1
−µ 2µ−1 −µ
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Method of lines. Stability and the CFL condition
Explicit Euler with ∆x = 1/(N + 1) implies recursion
u·,j+1 = u·,j + ∆t · T∆xu·,j
Recall λk[T∆x] = −4(N + 1)2 sin2 kπ
2(N + 1)for k = 1 : N
Stability requires ∆t · λk ∈ S for all eigenvalues
−2.5 −2 −1.5 −1 −0.5 0 0.5−1.5
−1
−0.5
0
0.5
1
1.5
Re
Im
∆t · λk ∈[
− 4∆t
∆x2,−π2∆t
]
S
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The CFL condition
For stability we need 4∆t/∆x2 ≤ 2
CFL condition (Courant, Friedrichs, Lewy 1928)
∆t
∆x2≤ 1
2
The CFL condition is a severe restriction on time step ∆t
Stiffness The CFL condition can be avoided by usingA-stable methods, e.g. Trapezoidal Rule or Implicit Euler
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Experimental stability investigation
N = 30 internal pts in [0, 1], M = 187 time steps on [0, 0.1]
Stable solution at CFL = .514
00.02
0.040.06
0.080.1
0
0.2
0.4
0.6
0.8
10
0.2
0.4
0.6
0.8
1
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Violating the CFL condition. Instability
N = 30 internal pts in [0, 1], M = 184 time steps on [0, 0.1]
Unstable solution at CFL = .522
00.02
0.040.06
0.080.1
0
0.2
0.4
0.6
0.8
1−0.4
−0.2
0
0.2
0.4
0.6
0.8
1
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Crank–Nicolson method (1947)
Crank–Nicolson method ⇔ Trapezoidal Rule for PDEs
The trapezoidal rule is
implicit ⇒ more work/step
A–stable ⇒ no restriction on ∆t
Theorem Crank–Nicolson is unconditionally stable
There is no CFL condition on the time-step ∆t
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Crank–Nicolson method. . .
Courant number µ = ∆t/∆x2 ⇒ recursion
(I − µ
2T )u·,j+1 = (I +
µ
2T )u·,j
with Toeplitz matrix T = tridiag(1 − 2 1)
Tridiagonal structure ⇒ low complexity
Refactorize only if Courant number µ = ∆t/∆x2 changes
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6. Error analysis. Convergence
MOL with explicit Euler for ut = uxx
Global error ei,j = ui,j − u(xi, tj)
Local error Insert exact solution to get
u(xi, tj+1) − u(xi, tj)
∆t=
=u(xi−1, tj) − 2u(xi, tj) + u(xi+1, tj)
∆x2− li,j
Expand local error li,j in Taylor series
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Local error
Taylor expansion
−li,j = ut − uxx +∆t
2utt +
∆x2
12uxxxx + O(∆t2,∆x4)
Therefore
−li,j =∆t
2utt +
∆x2
12uxxxx + O(∆t2,∆x4)
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The Lax Principle
Conclusion
Consistency li,j → 0 as ∆t,∆x → 0
Stability CFL condition ∆t/∆x2 ≤ 1/2
Convergence ei,j → 0 as ∆t,∆x → 0
Theorem (Lax Principle)
Consistency + Stability ⇒ Convergence
Note Choice of norm is very important
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The order of the method
With local error
−li,j =∆t
2utt +
∆x2
12uxxxx = O(∆t,∆x2)
and stability in terms of CFL condition µ = ∆t/∆x2 ≤ 1/2
we have global error ei,j = O(∆t,∆x2)
For fixed µ we have ∆t ∼ ∆x2 and it follows that
Global error ei,j = O(∆t,∆x2) = O(∆x2) ⇒
Theorem The order of convergence is p = 2
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Order of PDE discretizations
Discretization process from PDE to SD to FD
ut = uxx → v∆x = P∆xv∆x + h∆x(t) → uj+1µ = Aµu
jµ + kj
µ
Suppose
order of SD scheme for spatial variables is p1
order of ODE time discretization is p2
Theorem If ∆t = µ∆x2 the FD order of convergence isp = minp1, 2p2
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Example. The Crank–Nicolson method
From Trapezoidal rule yn+1 = yn + h2[f(yn) + f(yn+1)]
uj+1i = uj
i +∆t
2∆x2[(uj
i−1 − 2uji + uj
i+1) + (uj+1i−1 − 2uj+1
i + uj+1i+1 )]
−µ
2uj+1
i−1 + (1 + µ)uj+1i − µ
2uj+1
i+1 =µ
2uj
i−1 + (1 − µ)uji +
µ
2uj
i+1
Same order p = min2, 4 = 2 as with Explicit Euler
xi−1 xi xi+1
tj
tj+1
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Crank–Nicolson. Stability
Aµ = (I − µ
2T )−1(I +
µ
2T ) with usual Toeplitz matrix T
Theorem The eigenvalues are λ[Aµ] =1 + µ
2λ[T ]
1 − µ
2λ[T ]
Note λ[T ] ∈ (−4, 0) ⇒ −1 < λ[Aµ] < 1 This implies thatthere is no CFL stability condition on the Courant ratio µ!The method is stable for all ∆t > 0
Theorem Crank–Nicolson is unconditionally stable
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Experimental stability investigation
N = 30 internal pts in [0, 1], M = 30 time steps on [0, 0.1]
Stable solution at CFL = 3.2
00.02
0.040.06
0.080.1
0
0.2
0.4
0.6
0.8
10
0.2
0.4
0.6
0.8
1
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7. Parabolic problems. FEM
Consider diffusion problem in strong form ut − uxx = 0 withDirichlet boundary conditions
Multiply by test function v and integrate by parts
∫ 1
0
vut dx +
∫ 1
0
v′u′ dx = 0
In terms of inner product and energy norm –
Weak form 〈v, ut〉+a(v, u) = 0 for all v with v(0) = v(1) = 0
Numerical Methods for Differential Equations – p. 43/50
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Galerkin method (Finite Element Method)
1. Basis functions ϕi2. Approximate u(t, x) =
∑
cj(t)ϕj(x)
3. Determine cj from 〈ϕi, ut〉 + a(ϕi, u) = 0
Note 〈ϕi, ut〉 =∑
cj〈ϕi, ϕj〉 and a(ϕi, u) =∑
cj〈ϕ′
i, ϕ′
j〉
We get an initial value problem
B∆xc + K∆xc = 0
for the determination of the coefficients cj(t) with c(0)
determined by the initial condition
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Galerkin method
Simplest casePiecewise linear basis functions on equidistant grid
Stiffness matrix elements kij = 〈ϕ′
i, ϕ′
j〉
K∆x =1
∆xtridiag(−1 2 −1)
and mass matrix elements bij = 〈ϕi, ϕj〉
B∆x =∆x
6tridiag(1 4 1)
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Simplest Galerkin FEM. . .
Note that in the initial value problem
B∆xc + K∆xc = 0
the matrix B∆x is tridiagonal ⇒ no advantage fromexplicit time stepping methods
Explicit Euler
B∆x(cn+1 − cn) = −∆t · K∆xcn
requires the solution of a tridiagonal system on every step
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Simplest Galerkin FEM. . .
As the system is stiff, consider implicit A-stable method
B∆x(cn+1 − cn) = − ∆t
2· K∆x(cn + cn+1)
and solve tridiagonal system
(B∆x +∆t
2K∆x)cn+1 = (B∆x −
∆t
2K∆x)cn
on every step
Trapezoidal rule has same cost, but better stability
Numerical Methods for Differential Equations – p. 47/50
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8. Well-posedness
Linear partial differential equationut = Lu + f, 0 ≤ x ≤ 1, t ≥ 0, u(x, 0) = h(x),
u(0, t) = φ0(t), u(1, t) = φ1(t)
Suppose
wt = Lw + f, w(x, 0) = h(x)
vt = Lv + f, v(x, 0) = h(x) + g(x)
Subtract to get homogeneous Dirichlet problem
ut = Lu, u(x, 0) = g(x), φ0(t) ≡ 0, φ1(t) ≡ 0
Numerical Methods for Differential Equations – p. 48/50
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Well-posedness. Time evolution
Suppose time evolution u(x, t) = E(t)g(x)
Definition The equation is well-posed if for every t∗ > 0
there is a constant 0 < C(t∗) < ∞ such that‖E(t)‖ ≤ C(t∗) for all 0 ≤ t ≤ t∗
Theorem A well-posed equation has a solution that
depends continuously on the initial value (the “data”)
is uniformly bounded in any compact interval
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ut = uxx is well posed
Fourier series expansion g(x) =√
2∑
∞
1 ck sin kπx implies
u(x, t) =√
2∞
∑
k=1
cke−(kπ)2t sin kπx
‖E(t)g‖22 =
∫ 1
0
|u(x, t)|2 dx
= 2∞
∑
k=1
∞∑
j=1
ckcje−(k2+j2)π2t
∫ 1
0
sin kπx sin jπx dx
=∞
∑
k=1
c2ke
−2(kπ)2t ≤∞
∑
k=1
c2k = ‖g‖2
2
Hence ‖E(t)‖2 ≤ 1 for every t ≥ 0Numerical Methods for Differential Equations – p. 50/50