Poisson equation in 1D - TU Dortmund · Poisson equation in 1D Dirichlet problem (−∂2u ∂x2 =...

40
Poisson equation in 1D Dirichlet problem - 2 u ∂x 2 = f in Ω = (0, 1) u(0) = 0, u(1) = 0 Central difference discretization - ui-1-2ui+ui+1 x) 2 = f i , i =1,...,N - 1 u 0 = u N =0 Au = f, A = 1 x) 2 2 -1 -1 2 -1 -1 2 -1 ... -12

Transcript of Poisson equation in 1D - TU Dortmund · Poisson equation in 1D Dirichlet problem (−∂2u ∂x2 =...

Page 1: Poisson equation in 1D - TU Dortmund · Poisson equation in 1D Dirichlet problem (−∂2u ∂x2 = f in Ω = (0,1) u(0) = 0, u(1) = 0 Central difference discretization (−u i−1−2u

Poisson equation in 1D

Dirichlet problem −∂2u

∂x2 = f in Ω = (0, 1)

u(0) = 0, u(1) = 0

Central difference discretization−ui−1−2ui+ui+1

(∆x)2 = fi, i = 1, . . . , N − 1u0 = uN = 0

Au = f, A = 1(∆x)2

2 −1−1 2 −1−1 2 −1

. . .−1 2

Page 2: Poisson equation in 1D - TU Dortmund · Poisson equation in 1D Dirichlet problem (−∂2u ∂x2 = f in Ω = (0,1) u(0) = 0, u(1) = 0 Central difference discretization (−u i−1−2u

Error analysis

Let u(xi) be the exact solution at the mesh point xi. The localtruncation error of the CD approximation is O(∆x)2, i.e.,

−u(xi−1)− 2u(xi) + u(xi+1)(∆x)2 = fi + (∆x)2pi

The solution vectors u = ui and uex = u(xi) satisfy

Au = f, Auex = f + (∆x)2p

The error e = uex − u satisfies the linear system

Ae = (∆x)2p, ei = u(xi)− ui

Page 3: Poisson equation in 1D - TU Dortmund · Poisson equation in 1D Dirichlet problem (−∂2u ∂x2 = f in Ω = (0,1) u(0) = 0, u(1) = 0 Central difference discretization (−u i−1−2u

Error analysis

The global error is measured using a norm, e.g., the Euclidean norm

‖v‖ =

√√√√ n∑j=1

v2j , v ∈ Rn

The CD discretization of the 1D Poisson equation is consistent

∆x→ 0 ⇒ ‖Auex − f‖ = (∆x)2‖p‖ → 0

The discretization is stable if there exists a constant C > 0 s.t.

‖uex − u‖ ≤ C‖Auex − f‖

Page 4: Poisson equation in 1D - TU Dortmund · Poisson equation in 1D Dirichlet problem (−∂2u ∂x2 = f in Ω = (0,1) u(0) = 0, u(1) = 0 Central difference discretization (−u i−1−2u

Error analysis

The matrix A is non-singular and symmetric. The natural norminduced by the Euclidean vector norm ‖ · ‖ is the spectral norm

∥∥A−1∥∥ = 1λmin

, λmin = min|λ1|, . . . , |λN−1|

Ae = (∆x)2p ⇒ ‖e‖ = (∆x)2 ∥∥A−1p∥∥

≤ (∆x)2 ∥∥A−1∥∥ ‖p‖ =∥∥A−1∥∥ ‖Auex − f‖

The finite difference discretization Au = f is stable if

∃C > 0 s.t. ‖A−1‖ ≤ C

Page 5: Poisson equation in 1D - TU Dortmund · Poisson equation in 1D Dirichlet problem (−∂2u ∂x2 = f in Ω = (0,1) u(0) = 0, u(1) = 0 Central difference discretization (−u i−1−2u

Error analysis

The CD approximation of ∂xx yields

λj = 4(∆x)2 sin2

2N − jN

), λmin = λN−1

For large N (on a fine mesh) we have

λN−1 = 4(∆x)2 sin2

( π

2N

)≈ 4

(∆x)2

( π

2N

)2= π2

This proves stability and convergence

‖uex − u‖ ≤1π2 ‖Auex − f‖ = 1

π2 (∆x)2‖p‖

Page 6: Poisson equation in 1D - TU Dortmund · Poisson equation in 1D Dirichlet problem (−∂2u ∂x2 = f in Ω = (0,1) u(0) = 0, u(1) = 0 Central difference discretization (−u i−1−2u

Error analysis

Assume that there is an error δf in the right-hand side f

The resulting solution u+ δu satisfies the linear system

A(u+ δu) = f + δf, Au = f

Similarly to the analysis of the approximation error, we have

Aδu = δf, ‖δu‖ ≤ 1π2 ‖δf‖

The system is well-conditioned w.r.t. the absolute error

Page 7: Poisson equation in 1D - TU Dortmund · Poisson equation in 1D Dirichlet problem (−∂2u ∂x2 = f in Ω = (0,1) u(0) = 0, u(1) = 0 Central difference discretization (−u i−1−2u

Error analysis

To estimate the relative error, divide by ‖u‖

‖δu‖‖u‖

≤ 1π2‖δf‖‖u‖

= 1π2

(‖f‖‖u‖

)‖δf‖‖f‖

The ratio of ‖f‖ and ‖u‖ determines the sensitivity of the relativeerror to relative changes in f

It is possible that the problem is well-conditioned w.r.t. absoluteerror but ill-conditioned w.r.t. relative error

Page 8: Poisson equation in 1D - TU Dortmund · Poisson equation in 1D Dirichlet problem (−∂2u ∂x2 = f in Ω = (0,1) u(0) = 0, u(1) = 0 Central difference discretization (−u i−1−2u

Other boundary conditions

Dirichlet-Neumann problem−∂2u

∂x2 = f in Ω = (0, 1)

u(0) = 0, ∂u∂x (1) = 0

Central difference discretization of the Poisson equation

−ui−1 − 2ui + ui+1

(∆x)2 = fi, i = 1, . . . , N

Discretization of the Neumann boundary condition

uN+1 − uN−1

2∆x = 0 ⇒ −uN−1 − uN

(∆x)2 = 12fN

Page 9: Poisson equation in 1D - TU Dortmund · Poisson equation in 1D Dirichlet problem (−∂2u ∂x2 = f in Ω = (0,1) u(0) = 0, u(1) = 0 Central difference discretization (−u i−1−2u

Other boundary conditions

The matrix form of the linear algebraic system becomes

Au = f, A ∈ RN×N , u, f ∈ RN

The matrix A is tridiagonal, symmetric positive-definite

A = 1(∆x)2

2 −1−1 2 −1−1 2 −1

. . .−1 2 −1−1 1

The smallest eigenvalue λmin is again O(1)

Page 10: Poisson equation in 1D - TU Dortmund · Poisson equation in 1D Dirichlet problem (−∂2u ∂x2 = f in Ω = (0,1) u(0) = 0, u(1) = 0 Central difference discretization (−u i−1−2u

Local error analysis

Consider the Taylor expansion about the point x = 1

u(x) = u(1) + (x− 1)∂u∂x

(1) + (x− 1)2

2∂2u

∂x2 (1)

+ (x− 1)3

6∂3u

∂x3 (1) + . . .

The local truncation error is given by O(∆x)

∂2u

∂x2 (1) = −2u(xN−1)− u(xN )(∆x)2 +O(∆x)

How does it influence the global error?

Page 11: Poisson equation in 1D - TU Dortmund · Poisson equation in 1D Dirichlet problem (−∂2u ∂x2 = f in Ω = (0,1) u(0) = 0, u(1) = 0 Central difference discretization (−u i−1−2u

Global error analysis

Let u(xi) be the exact solution at xi. We have

−u(xi−1)− 2u(xi) + u(xi+1)(∆x)2 = fi + (∆x)2pi

for i = 1, . . . , N − 1. The last equation is

−u(xN−1)− u(xN )(∆x)2 = 1

2fN + ∆xqN

Let u = ui and uex = u(xi) be solutions of

Au = f, Auex = f + (∆x)2p+ ∆xq

Page 12: Poisson equation in 1D - TU Dortmund · Poisson equation in 1D Dirichlet problem (−∂2u ∂x2 = f in Ω = (0,1) u(0) = 0, u(1) = 0 Central difference discretization (−u i−1−2u

Global error analysis

The approximation error e = uex − u satisfies

Ae = (∆x)2p+ ∆xq

Let e(p) and e(q) be solutions of

Ae(p) = (∆x)2p

Ae(q) = ∆xq

By linearity, we have Ae = A(e(p) + e(q)) and therefore

e = e(p) + e(q)

Page 13: Poisson equation in 1D - TU Dortmund · Poisson equation in 1D Dirichlet problem (−∂2u ∂x2 = f in Ω = (0,1) u(0) = 0, u(1) = 0 Central difference discretization (−u i−1−2u

Global error analysis

The approximation error e(p) arises at inner points

‖e(p)‖ ≤ C(∆x)2‖p‖, C = 1λmin

= O(1)

The approximation error e(q) arises at the point xN = 1

− e(q)i−1−2e

(q)i

+e(q)i+1

(∆x)2 = 0, i = 1, . . . , N − 1

− e(q)N−1−e

(q)N

(∆x)2 = ∆xqN

The exact solution of the linear system is

e(q)i = i(∆x)3qN , i = 1, . . . , N

Page 14: Poisson equation in 1D - TU Dortmund · Poisson equation in 1D Dirichlet problem (−∂2u ∂x2 = f in Ω = (0,1) u(0) = 0, u(1) = 0 Central difference discretization (−u i−1−2u

Global error analysis

Substituting e(q)i = i(∆x)3qN into the linear system, we obtain

−e

(q)i−1 − 2e(q)

i + e(q)i+1

(∆x)2 = −(i− 1− 2i+ i+ 1)(∆x)qN = 0

−e

(q)N−1 − e

(q)N

(∆x)2 = −(N − 1−N)(∆x)qN = ∆xqN

Since xi = i∆x, we have e(q)i = (∆x)2xiqN . It follows that

‖e‖ = ‖e(p) + e(q)‖ ≤ (∆x)2(C‖p‖+ qN‖x‖)

Page 15: Poisson equation in 1D - TU Dortmund · Poisson equation in 1D Dirichlet problem (−∂2u ∂x2 = f in Ω = (0,1) u(0) = 0, u(1) = 0 Central difference discretization (−u i−1−2u

Inhomogeneous boundary conditions

Consider the Dirichlet boundary condition

u(0) = g1

Substitution into the first equation yields−u0−2u1+u2

(∆x)2 = f1

u0 = g1⇒ 2u1 − u2

(∆x)2 = f1 + g1

(∆x)2

The first component of the right-hand side f changes

The matrix A is the same as in the case g1 = 0

Page 16: Poisson equation in 1D - TU Dortmund · Poisson equation in 1D Dirichlet problem (−∂2u ∂x2 = f in Ω = (0,1) u(0) = 0, u(1) = 0 Central difference discretization (−u i−1−2u

Inhomogeneous boundary conditions

Consider the Neumann boundary condition

∂u

∂x(1) = g2,

uN+1 − uN−1

2∆x = g2

Substitution into the first equation yields−uN−1−2uN +uN+1

(∆x)2 = fN

uN+1 = uN−1 + 2∆xg2⇒ −uN−1 − uN

(∆x)2 = 12fN + g2

∆x

The first component of the right-hand side f changes

The matrix A is the same as in the case g2 = 0

Page 17: Poisson equation in 1D - TU Dortmund · Poisson equation in 1D Dirichlet problem (−∂2u ∂x2 = f in Ω = (0,1) u(0) = 0, u(1) = 0 Central difference discretization (−u i−1−2u

Inhomogeneous boundary conditions

Consider the Robin boundary condition

∂u

∂x(1) + αu(1) = g3

uN+1 − uN−1

2∆x + αuN = g3

Substitution into the last equation yields

−uN−1 − (1 + α∆x)uN

(∆x)2 = 12fN + g3

∆x

The matrix A and the right-hand side f change

Page 18: Poisson equation in 1D - TU Dortmund · Poisson equation in 1D Dirichlet problem (−∂2u ∂x2 = f in Ω = (0,1) u(0) = 0, u(1) = 0 Central difference discretization (−u i−1−2u

Convection-diffusion in 1D

Dirichlet problem−∂2u

∂x2 = −Pe∂u∂x in (0, 1)

u(0) = 0, u(1) = 1

For the time being let Pe ≥ 00 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

Pe=0Pe=10Pe=100

Exact solution and its derivative

u(x) = eP e x − 1eP e − 1 ,

dudx = Pe eP e x

eP e − 1

Page 19: Poisson equation in 1D - TU Dortmund · Poisson equation in 1D Dirichlet problem (−∂2u ∂x2 = f in Ω = (0,1) u(0) = 0, u(1) = 0 Central difference discretization (−u i−1−2u

Convection-diffusion in 1D

The exact solution satisfies the maximum principle

0 ≤ u(x) ≤ 1 ∀x ∈ (0, 1)

A boundary layer forms near x = 1 at large Peclet numbers

Pe 1 ⇒ dudx (1) = Pe eP e

eP e − 1 ≈ Pe

This is an example of a singularly perturbed problem

Page 20: Poisson equation in 1D - TU Dortmund · Poisson equation in 1D Dirichlet problem (−∂2u ∂x2 = f in Ω = (0,1) u(0) = 0, u(1) = 0 Central difference discretization (−u i−1−2u

Convection-diffusion in 1D

First-order BD-CD discretization

Peui − ui−1

∆x − ui−1 − 2ui + ui+1

(∆x)2 = 0

Second-order CD-CD discretization

Peui+1 − ui−1

2∆x − ui−1 − 2ui + ui+1

(∆x)2 = 0

Dirichlet boundary conditions

u0 = 0, uN = 1

Page 21: Poisson equation in 1D - TU Dortmund · Poisson equation in 1D Dirichlet problem (−∂2u ∂x2 = f in Ω = (0,1) u(0) = 0, u(1) = 0 Central difference discretization (−u i−1−2u

Convection-diffusion in 1D

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1

1.2

Pe=100, exact

h=10−1, 1st−order

h=10−1, 2nd−order

h=10−2, 1st−order

h=10−2, 2nd−order

Page 22: Poisson equation in 1D - TU Dortmund · Poisson equation in 1D Dirichlet problem (−∂2u ∂x2 = f in Ω = (0,1) u(0) = 0, u(1) = 0 Central difference discretization (−u i−1−2u

Convection-diffusion in 1D

The CD solution oscillates and the sign changes at each node

The BD gradient is not as steep as that of the exact solution

As the mesh is refined, oscillations in the CD solution disappearand the slopes of the BD solution become steeper

The error of the CD approximation decreases quadratically

The error of the BD approximation decreases linearly

Page 23: Poisson equation in 1D - TU Dortmund · Poisson equation in 1D Dirichlet problem (−∂2u ∂x2 = f in Ω = (0,1) u(0) = 0, u(1) = 0 Central difference discretization (−u i−1−2u

Convection-diffusion in 1D

Finite difference discretization

aui−1 + bui + cui+1

(∆x)2 = 0, i = 1, . . . , N − 1

uN = 1 ⇒ auN−2 + buN−1

(∆x)2 = − c

(∆x)2

Coefficients of the linear system

a b cBD-CD −1− Pe∆x 2 + Pe∆x −1CD-CD −1− P e∆x

2 2 −1 + P e∆x2

Page 24: Poisson equation in 1D - TU Dortmund · Poisson equation in 1D Dirichlet problem (−∂2u ∂x2 = f in Ω = (0,1) u(0) = 0, u(1) = 0 Central difference discretization (−u i−1−2u

Convection-diffusion in 1D

Matrix form of the linear system Au = f

A = 1(∆x)2

b ca b ca b c

. . .a b

, f =

000·

− c(∆x)2

The entries of the coefficient matrix A satisfy

a < 0, b > 0, a+ b+ c = 0

A is tridiagonal but not symmetric for Pe > 0

Page 25: Poisson equation in 1D - TU Dortmund · Poisson equation in 1D Dirichlet problem (−∂2u ∂x2 = f in Ω = (0,1) u(0) = 0, u(1) = 0 Central difference discretization (−u i−1−2u

Analysis of the discrete problem

Consider a discretization leading to the linear system

aui−1 + bui + cui+1 = 0, i = 1, . . . , N − 1

The solution of this discrete problem is a polynomial of the form

ui = α+ βri, i = 0, 1, . . . , N

The constants are determined using the boundary conditions

u0 = 0, uN = 1

Page 26: Poisson equation in 1D - TU Dortmund · Poisson equation in 1D Dirichlet problem (−∂2u ∂x2 = f in Ω = (0,1) u(0) = 0, u(1) = 0 Central difference discretization (−u i−1−2u

Analysis of the discrete problem

Substitution of ui = α+ βri into the i-th equation yields

α(a+ b+ c) + β(ari−1 + bri + cri+1) = 0

wherea+ b+ c = 0 ⇒ cr2 − (a+ c)r + a = 0

The roots of this quadratic equation are

r1,2 =(a+ c)±

√(a+ c)2 − 4ac2c = (a+ c)± (a− c)

2c

Page 27: Poisson equation in 1D - TU Dortmund · Poisson equation in 1D Dirichlet problem (−∂2u ∂x2 = f in Ω = (0,1) u(0) = 0, u(1) = 0 Central difference discretization (−u i−1−2u

Analysis of the discrete problem

The equation cr2 − (a+ c)r + a = 0 has two roots

r1 = a

c, r2 = 1

The root r = r2 does not satisfy the boundary conditions for

ui = α+ βri

Substituting r = r1 into this formula, we obtain

ui = α+ β(ac

)i

Page 28: Poisson equation in 1D - TU Dortmund · Poisson equation in 1D Dirichlet problem (−∂2u ∂x2 = f in Ω = (0,1) u(0) = 0, u(1) = 0 Central difference discretization (−u i−1−2u

Analysis of the discrete problem

The Dirichlet boundary conditions are given by

u0 = α+ β = 0, uN = α+ β(ac

)N

= 1

The corresponding values of α and β are

α = − 11−

(ac

)N, β = 1

1−(

ac

)N

The solution of the discrete problem is

ui =1−

(ac

)i

1−(

ac

)N, i = 1, . . . , N

Page 29: Poisson equation in 1D - TU Dortmund · Poisson equation in 1D Dirichlet problem (−∂2u ∂x2 = f in Ω = (0,1) u(0) = 0, u(1) = 0 Central difference discretization (−u i−1−2u

Analysis of the discrete problem

Solution of the difference equations

ui = P

Q, P = 1−

(ac

)i

, Q = 1−(ac

)N

a < 0, b > 0, a+ b+ c = 0

In the case c < 0, we have ac > 0

ac > 1 ⇒ P < 0, Q < 0, 0 < ui < 1

ac < 1 ⇒ P > 0, Q > 0, 0 < ui < 1

The solution satisfies the maximum principle

Page 30: Poisson equation in 1D - TU Dortmund · Poisson equation in 1D Dirichlet problem (−∂2u ∂x2 = f in Ω = (0,1) u(0) = 0, u(1) = 0 Central difference discretization (−u i−1−2u

Analysis of the discrete problem

Solution of the difference equations

ui = P

Q, P = 1−

(ac

)i

, Q = 1−(ac

)N

a < 0, b > 0, a+ b+ c = 0

In the case c > 0, we have ac < 0

a+ c = −b < 0 ⇒ c < −a ⇒ ac < −1

Alternating signs: if ui > 0 then ui±1 < 0

The solution violates the maximum principle

Page 31: Poisson equation in 1D - TU Dortmund · Poisson equation in 1D Dirichlet problem (−∂2u ∂x2 = f in Ω = (0,1) u(0) = 0, u(1) = 0 Central difference discretization (−u i−1−2u

Analysis of the discrete problem

In the case c = 0, the difference equations reduce to

ui = −abui−1 = 0, i = 1, . . . , N − 1

Invoking the Dirichlet boundary conditions, we find that

u0 = u1 = · · · = uN−1 = 0, uN = 1

The maximum principle holds but the solution is inaccurate

Page 32: Poisson equation in 1D - TU Dortmund · Poisson equation in 1D Dirichlet problem (−∂2u ∂x2 = f in Ω = (0,1) u(0) = 0, u(1) = 0 Central difference discretization (−u i−1−2u

Summary

In the case c > 0, violations of the maximum principle are possible;the matrix A is not diagonally dominant

a < 0, b > 0, b = −a− c ⇒ |b| < |a|+ |c|

In the case c ≤ 0, violations of the maximum principle are ruled out;the matrix A is diagonally dominant

a < 0, b = −(a+ c) ⇒ |b| = |a|+ |c|

The case c = 0 should be avoided for accuracy reasons

Page 33: Poisson equation in 1D - TU Dortmund · Poisson equation in 1D Dirichlet problem (−∂2u ∂x2 = f in Ω = (0,1) u(0) = 0, u(1) = 0 Central difference discretization (−u i−1−2u

Analysis of the CD-CD scheme

Equations of the difference scheme

Peui+1 − ui−1

2∆x − ui−1 − 2ui + ui+1

(∆x)2 = 0

This discretization corresponds to c = −1 + P e∆x2 . The sufficient

(not necessary) condition of the maximum principle is

c ≤ 0 ⇔ Pe∆x ≤ 2

If Pe is very large, the mesh size ∆x must be very small

It is worthwhile to refine the mesh just in subdomains whereviolations of the maximum principle would occur otherwise

Page 34: Poisson equation in 1D - TU Dortmund · Poisson equation in 1D Dirichlet problem (−∂2u ∂x2 = f in Ω = (0,1) u(0) = 0, u(1) = 0 Central difference discretization (−u i−1−2u

Analysis of the BD-CD scheme

Equations of the difference scheme

Peui − ui−1

∆x − ui−1 − 2ui + ui+1

(∆x)2 = 0

For this discretization c = −1 regardless of the mesh size

The maximum principle is satisfied but the error is O(∆x)

For accuracy reasons ∆x must be chosen very small; otherwisethe solution will be strongly smeared by numerical diffusion

In contrast to CD-CD local mesh refinement is insufficient

Page 35: Poisson equation in 1D - TU Dortmund · Poisson equation in 1D Dirichlet problem (−∂2u ∂x2 = f in Ω = (0,1) u(0) = 0, u(1) = 0 Central difference discretization (−u i−1−2u

Analysis of the BD-CD scheme

Taylor expansions for ui±1 = u(xi±1)

ui±1 = ui ±∆x(∂u

∂x

)i

+ (∆x)2

2

(∂2u

∂x2

)i

± (∆x)3

6

(∂3u

∂x3

)i

+ . . .

Residual of the difference equation

Peui − ui−1

∆x − ui−1 − 2ui + ui+1

(∆x)2 = Pe

(∂u

∂x

)i

−(∂2u

∂x2

)i︸ ︷︷ ︸

=0

− Pe∆x2

(∂2u

∂x2

)i

+O(∆x)2 = O(∆x)

Page 36: Poisson equation in 1D - TU Dortmund · Poisson equation in 1D Dirichlet problem (−∂2u ∂x2 = f in Ω = (0,1) u(0) = 0, u(1) = 0 Central difference discretization (−u i−1−2u

Analysis of the BD-CD scheme

The BD-CD scheme is a first-order accurate approximation to

Pe∂u

∂x− ∂2u

∂x2 = 0

but a second-order accurate approximation to

Pe∂u

∂x−(

1 + Pe∆x2

)∂2u

∂x2 = 0

The numerical diffusion coefficient P e∆x2 is proportional to the

mesh size ∆x and tends to zero very slowly

Page 37: Poisson equation in 1D - TU Dortmund · Poisson equation in 1D Dirichlet problem (−∂2u ∂x2 = f in Ω = (0,1) u(0) = 0, u(1) = 0 Central difference discretization (−u i−1−2u

Upwind difference scheme

The analysis of the BD-CD scheme was performed for Pe > 0

If Pe < 0, then a = −1− Pe∆x may assume positive values

The maximum principle holds for any ∆x if the convective term isdiscretized using the first-order upwind difference approximation

upwind difference =

backward difference, P e > 0forward difference, P e < 0

Upwind difference approximations of second and higher order mayproduce oscillations in the vicinity of steep gradients

Page 38: Poisson equation in 1D - TU Dortmund · Poisson equation in 1D Dirichlet problem (−∂2u ∂x2 = f in Ω = (0,1) u(0) = 0, u(1) = 0 Central difference discretization (−u i−1−2u

Adaptive schemes

Equations of the difference scheme

Pe(1 + ωi−1/2)(ui − ui−1) + (1− ωi+1/2)(ui+1 − ui)

2∆x−ui−1 − 2ui + ui+1

(∆x)2 = 0, i = 1, . . . , N − 1

backward difference for ωi±1/2 = 1, central difference forωi±1/2 = 0, forward difference for ωi±1/2 = −1

optimal choice of ωi±1/2 depends on the solution

Page 39: Poisson equation in 1D - TU Dortmund · Poisson equation in 1D Dirichlet problem (−∂2u ∂x2 = f in Ω = (0,1) u(0) = 0, u(1) = 0 Central difference discretization (−u i−1−2u

Nonuniform meshes

So far we considered difference approximations on uniform meshes

For accuracy and efficiency reasons, it is worthwhile to use meshesdesigned to equidistribute local truncation errors

εi ≈ ε, i = 1, . . . , N − 1

Taylor expansion reveals that εi depends not only on ∆x but also onthe derivatives of the exact solution at xi

Hence, the use of nonuniform meshes is generally required to achievea (nearly) uniform distribution of local errors

Page 40: Poisson equation in 1D - TU Dortmund · Poisson equation in 1D Dirichlet problem (−∂2u ∂x2 = f in Ω = (0,1) u(0) = 0, u(1) = 0 Central difference discretization (−u i−1−2u

Nonuniform meshes

To balance εi, the local mesh size ∆xi = xi − xi−1 should be smallif derivatives of u are large and large if derivatives are small

Since the exact solution is unknown, the magnitude of the derivativesneeds to be estimated in terms of approximate solution values

The location of boundary layers is usually known a priori but largederivatives are also found in internal layers and moving fronts

If the mesh is nonuniform, the finite difference approximationsconsidered so far require generalization and further analysis