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Poisson"s Equation in 2DAnalytic SolutionsA Finite Difference...A Linear System of...Direct Solution of the LSEClassification of PDEPage1of16Introduction to Scientific ComputingPoisson"s Equation in 2DMichael Bader1.Poisson"s Equation in 2DWe will now examine the general heat conduction equation,

T t=κΔT+qρc. in the 2-dimensional case, assuming a steady state problem (Tt= 0).

We get Poisson"s equation:

-uxx(x,y)-uyy(x,y) =f(x,y),(x,y)?Ω = (0,1)×(0,1), where we used the unit square as computational domain. Again, we will only deal with Dirichlet boundary conditions: u(x,y) =g(x,y)forx?∂Ω

Poisson"s Equation in 2DAnalytic SolutionsA Finite Difference...A Linear System of...Direct Solution of the LSEClassification of PDEPage2of16Introduction to Scientific ComputingPoisson"s Equation in 2DMichael Bader2.Analytic SolutionsWe will concentrate on the homogenous equation,

-uxx(x,y)-uyy(x,y) = 0,(x,y)?Ω = (0,1)×(0,1), and consider the boundary conditions

u(x,1) =g(x) 0< x <1.2.1.Separation of Variables - revisitedSimilar to the 1D heat equation, we use the ansatz

u(x,y) =X(x)Y(y) to get

Xxx(x)X(x)=Yyy(y)Y(y)

Consequently, left hand side and right hand side have to be equal to a constantλ.

Poisson"s Equation in 2DAnalytic SolutionsA Finite Difference...A Linear System of...Direct Solution of the LSEClassification of PDEPage3of16Introduction to Scientific ComputingPoisson"s Equation in 2DMichael Bader2.2.Particular solutionsFor the functionX(x), we get the eigenvalue problem

-Xxx(x) =λX(x),0< x <1,

X(0) =X(1) = 0.

We already know that the eigenvalues are

k= (kπ)2, and the respective eigenfunctions are X k(x) = sin(kπx).

The functionY(0)has to satisfy similar equations,

(Yk)yy(y) =λYk(y),0< y <1, Y k(0) = 0, however, note the absence of the minus sign! Thus,Yk(y)is a linear combination ofe⎷λkyande-⎷λky.

WithYk(0) = 0, we get the solution

Y k(y) = sinh(kπy), which satisfies the boundary conditionYk(1) = sinh(kπ).

Poisson"s Equation in 2DAnalytic SolutionsA Finite Difference...A Linear System of...Direct Solution of the LSEClassification of PDEPage4of16Introduction to Scientific ComputingPoisson"s Equation in 2DMichael Bader2.3.Fourier"s methodWe have therefore computed particular solutions

u k(x,y) = sin(kπx)sinh(kπy) that solve the boundary value problems -uxx(x)-uyy(y) = 0 0< x,y <1, u(x,1) = sinh(kπ),0< x <1. Thus, if we can represent the boundary functiong(x)by g(x) =∞? k=1g ksin(kπx), then the function u(x,y) =∞? k=1c ksin(kπx)sinh(kπy), ck=gksinh(kπ) solves the given boundary problem for u(x,1) =g(x),0< x <1.

Poisson"s Equation in 2DAnalytic SolutionsA Finite Difference...A Linear System of...Direct Solution of the LSEClassification of PDEPage5of16Introduction to Scientific ComputingPoisson"s Equation in 2DMichael Bader3.A Finite Difference Scheme3.1.Discretization of the computational domainWe use a rectangular, equidistant grid ofn×ngrid points with mesh

sizeh=1n: h:={(ih,jh):i,j= 1,...,n-1}x i,j x i-1,j x i+1,j x i,j+1 x i,j-1 h x h y•We have chosenh=hx=hy•functionuwill be computed at grid pointsxi,j, i.e. we have unknownsui,j≈u(xi,j).

Poisson"s Equation in 2DAnalytic SolutionsA Finite Difference...A Linear System of...Direct Solution of the LSEClassification of PDEPage6of16Introduction to Scientific ComputingPoisson"s Equation in 2DMichael Bader3.2.Discretization of the derivatives - Difference QuotientsReplace derivatives by difference quotients:•first derivatives: forward, backward, or central differences

∂u∂x(xk)≈?

?????u(xk+1)-u(xk)hxu(xk)-u(xk-1)hxu(xk+1)-u(xk-1)2hx•second derivatives: standard second-order discretization (3-

point-stencil)

2u∂x2(xk)≈u(xk+1)-2u(xk) +u(xk-1)h2x

Instencil notation:•[0-1 1],[-1 1 0], or[-1 0 1]for the first derivatives•[1-2 1]for the second derivatives

Poisson"s Equation in 2DAnalytic SolutionsA Finite Difference...A Linear System of...Direct Solution of the LSEClassification of PDEPage7of16Introduction to Scientific ComputingPoisson"s Equation in 2DMichael Bader3.3.2D stencilsUsing standard second order discretization, we get

Thus, discretization of the PDE

-uxx(x,y)-uyy(x,y) =f(x,y) leads to the linear system of equations

1h2(ui+1,j+ui,j+1-4ui,j+ui,j-1+ui-1,j) =f(xi,j)

fori,j= 1,...,n-1.

This is often represented using a 2D stencil:

?-1 -1 4-1 -1? ?4-1 -1 -1-1

Poisson"s Equation in 2DAnalytic SolutionsA Finite Difference...A Linear System of...Direct Solution of the LSEClassification of PDEPage8of16Introduction to Scientific ComputingPoisson"s Equation in 2DMichael Bader4.A Linear System of Equations4.1.Homogeneous Dirichlet boundary conditionsTo account for homogeneous Dirichlet boundary conditions, we set

u

0,j=un,j=ui,0=ui,n= 0

fori,j= 0,...,n. Altogether, we get the following linear system of equations: u

0,j=un,j=ui,0=ui,n= 0fori,j= 0,...,n

(ui+1,j+ui,j+1-4ui,j+ui,j-1+ui-1,j)h2=f(xi,j)fori,j= 1,...,n-1

This system has•N= (n+ 1)×(n+ 1)unknownsui,j, (i,j= 0,...,n);•N= (n+ 1)×(n+ 1)equations.

Question:Is there a unique solution?

Poisson"s Equation in 2DAnalytic SolutionsA Finite Difference...A Linear System of...Direct Solution of the LSEClassification of PDEPage9of16Introduction to Scientific ComputingPoisson"s Equation in 2DMichael Bader4.2.LSE in matrix-vector notationWe can write the previous system of equation in matrix-vector nota-

tion, A huh=fh, where•uhis the vector u

h:= (u1,1,...,u1,n,u2,1,...un-2,n-1,un-1,1,...,un-1,n-1)T,•fh, similar touh, is the vector of the right hand sidesfi,j=

f(xi,j);•Ahis the following sparse matrix 1h2( (((((((((((4-1-1 -1......... -1.........-1 .........-1 -1-1 4)

Poisson"s Equation in 2DAnalytic SolutionsA Finite Difference...A Linear System of...Direct Solution of the LSEClassification of PDEPage10of16Introduction to Scientific ComputingPoisson"s Equation in 2DMichael BaderBlock-Tridiagonal Matrix:

A his more precisely described by the following block-tridiagonal ma- trix: A h=1h2( ((((((B h-I0···0 -I Bh-I...... 0 .........0......-I Bh-I

0···0-I Bh)

whereBhandIare(n-1)×(n-1)matrices.

WhileIis the unity matrix,Bhis given by

B h=( ((((((4-1 0···0 -1 4-1...... 0 .........0......-1 4-1

0···0-1 4)

Poisson"s Equation in 2DAnalytic SolutionsA Finite Difference...A Linear System of...Direct Solution of the LSEClassification of PDEPage11of16Introduction to Scientific ComputingPoisson"s Equation in 2DMichael Bader4.3.Properties of the system matrixAhResume:•Ahis a sparse(n-1)2×(n-1)2-matrix, and has a so-calledband structure;•Ahis block-tridiagonal•all diagonal elements ofAhare 4;•in each row, between 2 and 4 elements are non-zeroes (-1),

depending on whether the respective unknown is close to the boundary;

Conclusions:•Ahis thereforediagonal dominant•AsAhis diagonal dominant, it is also positive definite.•Thus, the system has aunique solution!

Poisson"s Equation in 2DAnalytic SolutionsA Finite Difference...A Linear System of...Direct Solution of the LSEClassification of PDEPage12of16Introduction to Scientific ComputingPoisson"s Equation in 2DMichael Bader5.Direct Solution of the LSE5.1.Gaussian elimination for band matrices•Ahis aband-matrix(band-width isn-1)•Gaussian elimination on a LSE ofNunknowns usually requires

O(N3)operations.•for band matrices this reduces to

O(N×bandwidth2)operations.

Solving our LSE using Gaussian elimination therefore requires

O(N2) =O(n4)operations.

The numerical error of our method is onlyO(n-2).

Thus, to reduce an error to half of its size, we have to•use twice as many unknowns;•wait four times as long!

?Faster methods to solve the LSE are needed!

Poisson"s Equation in 2DAnalytic SolutionsA Finite Difference...A Linear System of...Direct Solution of the LSEClassification of PDEPage13of16Introduction to Scientific ComputingPoisson"s Equation in 2DMichael Bader5.2.Faster methodsDirect methods:•use a clever numbering of the unknowns (not line by line but

"divide and conquer") ?nested dissection,O(n3)•use eigenvectors of the matrix, and Fast Sine Transform ?Fast Poisson Solvers,O(n2logn) Iterative methods:•solve system line by line, but do this again and again ?JacobiorGauss-Seidel relaxation,O(n4)•clever weghting of corrections ?SOR(successive over-relaxation),O(n3)•reformulate LSE as minimization problem ?Krylow methods,Conjugate Gradients,O(n3)•use different mesh sizes and combine their solutions ?Multigrid methods,O(n2)

Poisson"s Equation in 2DAnalytic SolutionsA Finite Difference...A Linear System of...Direct Solution of the LSEClassification of PDEPage14of16Introduction to Scientific ComputingPoisson"s Equation in 2DMichael Bader6.Classification of PDEClassification similar to ODE:•first/second/...order according to order of derivatives•linear/non-linear

Example: General linear PDE of second order:

d i=1d j=1a i,j(?x)·uxi,xj(?x) +d? i=1a i(?x)·uxi(?x) +a(?x)·u(?x) =f(?x)

Some linear PDE of second order:•1D heat equation:x1?=t,x2?=x•2D Poisson equation:x1?=x,x2?=y•1D wave equation (utt=uxx):x1?=t,x2?=x•etc.

Poisson"s Equation in 2DAnalytic SolutionsA Finite Difference...A Linear System of...Direct Solution of the LSEClassification of PDEPage15of16Introduction to Scientific ComputingPoisson"s Equation in 2DMichael Bader6.1.Boundary ConditionsAmong other things, elliptic, parabolic, and hyperbolic PDE differ in

the type of boundary or initial conditions they require. Examples:•Laplace equation, Poisson equation (elliptic): -uxx-uyy=f ?boundary conditions•heat equation (parabolic): u t= Δu ?initial condition fort, boundary conditions forx•wave equation (hyperbolic): u tt= Δu ?initial conditions foruandut

Poisson"s Equation in 2DAnalytic SolutionsA Finite Difference...A Linear System of...Direct Solution of the LSEClassification of PDEPage16of16Introduction to Scientific ComputingPoisson"s Equation in 2DMichael Bader6.2.Elliptic, Parabolic, and Hyperbolic PDEGeneral linear PDE of second order:

d? i=1d j=1a i,j(?x)·uxi,xj(?x) +d? i=1a i(?x)·uxi(?x) +a(?x)·u(?x) =f(?x)

Three types of PDE:•ellipticPDE:

the matrixAof theai,jis positive or negative definite•parabolicPDE: one eigenvalue ofAis zero, the others have the same sign, and the rank ofAtogether with the vector of theaiis full (d)•hyperbolicPDE:

Ahas 1 pos. andd-1neg. eigenvalues or vice versa.

If theai,jare functions of?x:•elliptic in?x:

the matrixAof theai,j(?x)is positive or negative definite for a certain?x•elliptic: the matrixAof theai,j(?x)is positive or negative definitefor all?xquotesdbs_dbs16.pdfusesText_22

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