ode2 13

Numerical approximation of ODE II:

boundary value problems

Grado en Ing. Civil y Territorial

(Univ. Polit´ecnica de Madrid)

April 9, 2013

Outline

1 _{Boundary value problems}

2 _{Finite differences}

3 Convergence, stability and

consistency

4 _{Other boundary conditions}

5 Eigenvalue problems

Modeling: Bending

EId 2_z

dx2 =M(x)

or

d2 dx2(EI

d2z

dx2) =q(x), dM

dx =V(x), dV

dx =−q(x)

z(x) deflection of the neural axes

E(x) is the Young’s modulus

I(x) area moment of inertia

M(x) bending moment

q(x) load

V(x) shear force

Beam equation: Equilibrium of bending moment

Strain below neutral axis ε

ε= (ρ+y)d_ρdθ_θ−ρdθ = y_ρ

Stress σ (Hook’s law)

σ =Eε=Ey_ρ

Moment equilibrium

M(x) =R_Ayσdy =R_AEy_ρ2 = EI_ρ

Curvature approximation

1 ρ =

z00_(x)

√

1+|z0_(x)|3/2 ∼z 00_(x)

Beam equation

M(x) = EI_ρ =EIz00(x)

Modeling: Boundary conditions

Cantilever

EId_dx2z2 =M(x)

z(0) =z0(0) = 0.

Simpel support

EId_dx2z2 =M(x)

z(0) =z(L) = 0.

Boundary value problem

Consider the ODE

−y00+a(x)y=f(x), y(a) =ya, y(b) =yb.

Two methods:

1 Fourier method

2 _{finite differences}

Fourier method (main idea)

Consider the boundary value problem

−y00−y =f(x), y(0) =y(1) = 0.

Introduce the associated eigenvalue problem

−ϕ00−λϕ= 0, ϕ(0) =ϕ(1) = 0.

λk =k2π2, ϕk = sin(kπx), k = 1,2,3, ...

RemarkAll the eigenfunctions are solutions of

−ϕ00_k −ϕk =(λk−1)ϕk(x),

ϕk(0) =ϕk(1) = 0.

Main idea: writef(x) as a linear combination of eigenfunctions

Fourier method in practice

Boundary value problem

−y00−y=f(x), y(0) =y(1) = 0. 1 Writey(x) andf(x) in Fourier Series

y(x) =

∞ X

k=1

aksin(kπx), f(x) = ∞ X

k=1

cksin(kπx)

2 Substitute in the differential equation ∞

X

k=1

(akk2π2−ak) sin(kπx) = ∞ X

k=1

cksin(kπx)

3 Identify coefficients

akk2π2−ak =ck, → ak =

ck

k2_π2₋₁, k = 1,2, ...

4 Approximate the solution by truncating the series

y(x)∼ K

X

k=1

aksin(kπx).

Fourier method: example

−y00−y =(1−26e5x)(1−x) + 10e5x,

y(0) =y(1) = 0.

Approximate solution

f =

K

X

k=1

cksin(kπx), → yap(x) =

K

X

k=1

aksin(kπx)

ak = _k2_πck2₋₁, k= 1,2, .... Green: exact, blue: K=3, red K=5

Matlab code

f =

K

X

k=1

cksin(kπx), → yap(x) =

K

X

k=1

aksin(kπx)

K=5;% number of Four. coef.

N=1000; h=1/N; x=0:h:1;% space discret.

f=(1-26*exp(5*x)).*(1-x)+10*exp(5*x);% f(x)

sol=0;% aproximaci´on

for k=1:K

p= sin(k∗pi∗x);% autofunci´on

c=trapz(x,f.*p)/trapz(x,p.*p); a=c/(k∧2*pi∧2-1);

sol=sol+a*p; end

plot(x,sol)

Boundary value problem: finite differerences

Consider the ODE

−y00+a(x)y=f(x), y(a) =ya, y(b) =yb.

Main idea: Approximatey00(x) by afinite difference of neighbors

Finite difference method

Take a partition of [a,b]

y0(xn) =

y(xn+1)−y(xn)

h +O(h), y

0_(x

n−1) =

y(xn)−y(xn−1)

h +O(h)

y00(xn) =

y(xn−1)−2y(xn) +y(xn+1)

h2 +O(h

2₎

whereh=tn+1−tn is the mesh size.

Finite difference approximation

Continuous

−y00+a(x)y=f(x), y(a) =ya, y(b) =yb.

Discrete

yn∼y(xn), an=a(xn), fn=f(xn),

−yn−1+2yn−yn+1

h2 +anyn=fn, n= 1, ...,N−1

y0 =ya, yN =yb.

Matrix formulation

( ˜K + ˜A) ˜Y = ˜F,

where

˜

K = 1

h2      

h2 0 0 0 0 ... 0

−1 2 −1 0 0 ... 0

0 −1 2 −1 0 ... 0

... ... ... ... ... ... ...

0 0 0 0 0 ... h2

     

, Y˜ =

      y0 y1 ... yN−1 yN       ˜ A=      

0 0 0 0

0 a1 0 0

... ... ... ...

0 0 aN 0

0 0 0 0

     

, F˜=

      ya f1 ... fN−1 yb      

Linear system to be solved

(K +A)Y =F,

where

K = 1

h2 

  

2 −1 0 0 ... 0 0

−1 2 −1 0 ... 0 0

... ... ... ... ... ... ...

0 0 0 0 ... −1 2



  

, Y =

    y1 y2 ... yN−1

    A=    

a1 0 0 0

0 a2 0 0

... ... ... ...

0 0 0 aN−1



  

, F =



  

f1+ya/h2

f1 ...

fN−1+yb/h2



  

Solving the linear system

K+Amust be invertible. Sufficient conditions:

Maximum rank condition: Rank (K +A) =N−1.

K +Ais definite positive (K +A)ξ·ξ >0 for all nonzero

ξ ∈_RN−1

K +Ais symmetric with positive eigenvalues

K +Ais diagonal dominant and strictly in at least one row.

Ex:

K = 1

h2 

  

2 −1 0 0 ... 0 0

−1 2 −1 0 ... 0 0

... ... ... ... ... ... ...

0 0 0 0 ... −1 2



   ,

Numerical methods for linear system

The size of the linear system depends on the number of equations

we solve and the number of interior nodesN.

For relevant 3-d problems the size can be of the order of 107 and

larger. Efficient numerical methods are essential

1 Direct methods (small scale problems): LU decomposition,

Cholesky, etc.

2 Iterative methods (large scale problems): Jacobi,

Gauss-Seidel, etc.

Convergence

Definition

A numerical scheme is said to be convergent if the solutionyn

satisfies

lim

h→0+_n=1max_,...,Nen,h= lim_h→0+_n=1max_,...,N|y(xn)−yn|= 0.

Two main ingredients for convergence:

1 _{Small local error: consistency}

−y(xn−1) + 2y(xn)−y(xn+1)

h2 −a(xn)y(xn)−f(xn)

=y00(xn)−a(xn)y(xn)−f(xn) +O(h2) =O(h2)

Second order accurate.

2 Global estimate of the discrete solution ash →0 (via

maximum principle): Stability

Matlab code

Consider the following ODE inx ∈[0,1],

−y00= 1,

y(0) =y(1) = 0.

a=0; b=1; ya=0; yb=0; N=40; h=(b-a)/N;

K=2*diag(ones(1,N-1))-diag(ones(1,N-2),1)-diag(ones(1,N-2),-1); K=1/h2*K;

F=ones(N-1,1); Y =K\F; y=[ya;Y;yb]; x=a:h:b; plot(x,y,’x’);

Remark

Numerical solutions of ODE:

1 _{Initial value problems are reduced to algebraic systems that}

are solved by time advancing methods: Euler, Trapezoidal, Runge-Kutta,...

2 Boundary value problems are reduced to a (big) linear system

of equations. This is usually more expensive in terms of computational time.

Finite difference approximation of a mixed problem

Continuous

−y00+a(x)y =f(x), y(a) =ya, y0(b) =yb.

Discrete

yn∼y(xn), an=a(xn), fn=f(xn),

( _−y

n−1+2yn−yn+1

h2 +anyn=fn, n = 1, ...,N−1

y0 =ya, yN

−yN−1 h =yb.

Matrix formulation

( ˜K + ˜A) ˜Y = ˜F,

where

˜

K = 1

h2      

h2 0 0 0 0 ... 0

−1 2 −1 0 0 ... 0

0 −1 2 −1 0 ... 0

... ... ... ... ... ... ...

0 0 0 0 0 −h h

     

, Y˜ =

      y0 y1 ... yN−1 yN       ˜ A=      

0 0 0 0

0 a1 0 0

... ... ... ...

0 0 aN 0

0 0 0 0

     

, F˜=

      ya f1 ... fN−1 yb      

Linear system to be solved

(K +A)Y =F,

where

K = 1

h2      

2 −1 0 0 ... 0 0 0

−1 2 −1 0 ... 0 0 0

... ... ... ... ... ... ... ...

0 0 0 0 ... −1 2 −1

0 0 0 0 ... ... −h h

     

, Y =

    y1 y2 ... yN     A=      

a1 0 0 0

0 a2 0 0

... ... ... ...

0 0 aN−1 0

0 0 0 0

     

, F =

     

f1+ya/h2

f1 ... fN−1 yb      

Example

Consider the following ODE inx ∈[0, π/4],

−y00−4y = 0,

y(0) = 2, y0(π/4) =−2.

There is no solution of the continuous system. Numerical approximation:

Remarks:

1 The linear system associated to the approximate system has a

solution even if the matrix is not definite positive.

2 The numerical method is consistent. It is 1st order accurate.

3 However, there is no stability. The solution becomes larger as

h →0 and it does not converge to anything. This is due to

the fact that the continuous problem has no solution.

Concluding remark

Two methods to solve boundary value problems:

1 _{Finite differences. Reduce the problem to a linear system of}

equations. Can be applied to any boundary value problem.

2 Fourier method. Reduce the problem to compute Fourier

coefficients. Few Fourier coefficients provide a good approximation. But it requires to know the eigenfunctions!

Modeling: Buckling

Consider a column with a loadP. The load provokes a momentum

which depends on the deflectionM(x) =−Py(x). Substituting in the beam equation,

(

EId_dx2y2 =−Py

y(0) =y(L) = 0

The column is stable while the only possible deflection isy(x) = 0 everywhere. This corresponds to

P < the first eigenvalue =Pcr

If the column is homogeneous (E =cte) and of constant section,

then

Pcr

EI =

π2

L , Pcr =

π2EI

L =

π2EI

L ,(Euler formula)

Eigenvalue problems

Continuous problem. Findλsuch that there exists a solution

of

−y00 =λy, y(0) =y(1) = 0.

Discrete problem. Find λh such that there exists a vector

Y ∈_RN−1 _{solution of}

KY =λY.