APPLIED NUMERICAL METHODS USING MATLAB phần 9 docx

Replacing the first derivative on the left-side boundary x = x0 by its three-pointcentral difference approximation 5.1.8 If the boundary condition on the lower side boundary y = y0 is al

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x y

Figure 9.2 Temperature distribution over a plate— Example 9.1.

Replacing the first derivative on the left-side boundary (x = x0) by its three-pointcentral difference approximation (5.1.8)

If the boundary condition on the lower side boundary (y = y0) is also of

Neumann type, then we need to write similar equations for j = 1, 2, , M x− 1

u 0,j = r y (u 0,j+1 + u0,j−1 ) + 2r x u 1,j + r xy (g 0,j u 0,j − f0,j − 2b

y0(x j )/y) ( 9.1.10) and additionally for the left-lower corner point (x0, y0),

u 0,0 = 2(r y u 0,1 + r x u 1,0 ) + r xy (g 0,0 u 0,0 − f0,0 − 2(b

x0(y0)/x + 2b

y0(x0)/y))

(9.1.11)

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406 PARTIAL DIFFERENTIAL EQUATIONS

An example of a parabolic PDE is a one-dimensional heat equation describing

the temperature distribution u(x, t) (x is position, t is time) as

A ∂

2u(x, t)

∂x2 = ∂u(x, t)

∂t for 0≤ x ≤ x f , 0≤ t ≤ T ( 9.2.1)

In order for this equation to be solvable, the boundary conditions u(0, t)=

b0(t) & u(x f , t) = b xf (t) as well as the initial condition u(x, 0) = i0 (x) should

be provided

To apply the finite difference method, we divide the spatial domain [0, x f] into

M sections, each of length x = x f /M , and divide the time domain [0, T ] into

N segments, each of duration t = T /N, and then replace the second partial

derivative on the left-hand side and the first partial derivative on the right-handside of the above equation (9.2.1) by the central difference approximation (5.3.1)and the forward difference approximation (5.1.4), respectively, so that we have

e j iπ/P (P is any nonzero integer) ( 9.2.4)

into Eq (9.2.3) to get

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PARABOLIC PDE 407

function [u,x,t] = heat_exp(a,xf,T,it0,bx0,bxf,M,N)

%solve a u_xx = u_t for 0 <= x <= xf, 0 <= t <= T

% Initial Condition: u(x,0) = it0(x)

% Boundary Condition: u(0,t) = bx0(t), u(xf,t) = bxf(t)

% M = # of subintervals along x axis

% N = # of subintervals along t axis

dx = xf/M; x = [0:M]’*dx;

dt = T/N; t = [0:N]*dt;

for i = 1:M + 1, u(i,1) = it0(x(i)); end

for n = 1:N + 1, u([1 M + 1],n) = [bx0(t(n)); bxf(t(n))]; end

not to lose the stability

The MATLAB routine “heat_exp()” has been composed to implement thisalgorithm

In this section, we consider another algorithm called the implicit backward Eulermethod, which comes out from substituting the backward difference approxima-tion (5.1.6) for the first partial derivative on the right-hand side of Eq (9.2.1) as

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How about the case where the values of ∂u/∂x|x=0= b

0(t) at one end aregiven? In that case, we approximate this Neumann type of boundary condition by

function [u,x,t] = heat_imp(a,xf,T,it0,bx0,bxf,M,N)

dx = xf/M; x = [0:M]’*dx;

dt = T/N; t = [0:N]*dt;

for i = 1:M + 1, u(i,1) = it0(x(i)); end

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Here, let us go back to see Eq (9.2.7) and try to improve the implicit backwardEuler method The difference approximation on the left-hand side is taken at

time point k, while the difference approximation on the right-hand side is taken

at the midpoint between time k and k− 1, if we regard it as the central

differ-ence approximation with time step t/2 Doesn’t this seem to be inconsistent?

How about taking the difference approximation of both sides at the same time

point—say, the midpoint between k + 1 and k—for balance? In order to do so,

we take the average of the central difference approximations of the left-hand side

at the two points k + 1 and k, yielding

u k2+1

u k+1 3

· 0

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function [u,x,t] = heat_CN(a,xf,T,it0,bx0,bxf,M,N)

dx = xf/M; x = [0:M]’*dx;

dt = T/N; t = [0:N]*dt;

for i = 1:M + 1, u(i,1) = it0(x(i)); end

b = [r*u(1,k); zeros(M - 3,1); r*u(M + 1,k)]

+ r*(u(1:M - 1,k - 1) + u(3:M + 1,k - 1)) + r2*u(2:M,k - 1); u(2:M,k) = trid(A,b); %Eq.(9.2.17)

end

Example 9.2 One-Dimensional Parabolic PDE: Heat Flow Equation.

Consider the parabolic PDE

∂2u(x, t)

∂x2 = ∂u(x, t)

∂t for 0≤ x ≤ 1, 0 ≤ t ≤ 0.1 ( E9.2.1)

with the initial condition and the boundary conditions

u(x, 0) = sin πx, u( 0, t) = 0, u( 1, t)= 0 ( E9.2.2)

We made the MATLAB program “solve_heat.m” in order to use the routines

“heat_exp()”, “heat_imp()”, and “heat_CN()” in solving this equation and ranthis program to obtain the results shown in Fig 9.3 Note that with the spatial

interval x = x f /M = 1/20 and the time step t = T /N = 0.1/100 = 0.001,

we have

r = A t

x2 = 1 0.001

( 1/20)2 = 0.4 ( E9.2.3) which satisfies the stability condition (r ≤ 1/2) (9.2.6) and all of the three meth-

ods lead to reasonably fair results with a relative error of about 0.013 But,

if we decrease the spatial interval to x = 1/25 for better resolution, we have

r = 0.625, violating the stability condition and the explicit forward Euler method

(“heat_exp()”) blows up because of instability as shown in Fig 9.3a, whilethe implicit backward Euler method (“heat_imp()”) and the Crank–Nicholsonmethod (“heat_CN()”) work quite well as shown in Figs 9.3b,c Now, with the

spatial interval x = 1/25 and the time step t = 0.1/120, the explicit method

as well as the other ones works well with a relative error less than 0.001 in return

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1 0.5

0.5 x

0 0

0.05t0.1 1

(c) The Crank-Nicholson method

5

0

x

0.05t0.1 1

1 0.5

0.5 x

0 0

0.05t0.1 1

(a) The explicit method (b) The implicit method

0 0

Figure 9.3 Results of various algorithms for a one-dimensional parabolic PDE: heat equation.

for somewhat (30%) more computations, despite that r = 0.5208 doesn’t strictly

satisfy the stability condition

This implies that the condition (r ≤ 1/2) for stability of the explicit forward

Euler method is not a necessary one, but only a sufficient one Besides, if itconverges, its accuracy may be better than that of the implicit backward Eulermethod, but generally no better than that of the Crank–Nicholson method

%solve_heat

a = 1; %the parameter of (E9.2.1)

it0 = inline(’sin(pi*x)’,’x’); %initial condition

bx0 = inline(’0’); bxf = inline(’0’); %boundary condition

Uo = uo(x,t); aUo = abs(Uo)+eps; %values of true analytical solution

%How far from the analytical solution?

err1 = norm((u1-Uo)./aUo)/MN

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Another example of a parabolic PDE is a two-dimensional heat equation

describ-ing the temperature distribution u(x, y, t)((x, y) is position, t is time) as

u(x0, y, t) = b x (y, t), u(x f , y, t) = b xf (y, t),

u(x, y0, t) = b y (x, t), and u(x, y f , t) = b yf (x, t)

as well as the initial condition u(x, y, 0) = i0 (x, y)

We replace the first-order time derivative on the right-hand side by the

three-point central difference at the midthree-point (t k+1+ t k )/2 just as with the Crank–

Nicholson method We also replace one of the second-order derivatives, u xx and

u yy , by the three-point central difference approximation (5.3.1) at time t kand the

other at time t k+1, yielding

k

i,j+1− 2u k

i,j + u k i,j−1

t ( 9.2.20)

which seems to be attractive, since it can be formulated into a tridiagonal system

of equations with respect to u k i +1,j+1 , u k i,j+1, and u k i −1,j+1 But, why do we treat u xx and u yy with discrimination—that is, evaluate one at time t kand the other at time

t k+1in a fixed manner? In an alternate manner, we write the difference equation

for the next time point t k+1 as

This formulation, proposed by Peaceman and Rachford [P-1], is referred to as thealternating direction implicit (ADI) method and can be cast into the followingalgorithm:

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imple-function [u,x,y,t] = heat2_ADI(a,D,T,ixy0,bxyt,Mx,My,N)

%solve u_t = c(u_xx + u_yy) for D(1) <= x <= D(2), D(3) <= y <= D(4), 0 <= t <= T

% Initial Condition: u(x,y,0) = ixy0(x,y)

% Boundary Condition: u(x,y,t) = bxyt(x,y,t) for (x,y)cB

% Mx/My = # of subintervals along x/y axis

bx = [ry*u(i,1) zeros(1,Mx - 3) ry*u(i,My + 1)]

+rx*(u_1(i-1,jj)+ u_1(i + 1,jj)) + rx2*u_1(i,jj);

u(i,jj) = trid(Ay,bx’)’; %Eq.(9.2.22a)

end

else

for j = 2:Mx

ii = 2:My;

by = [rx*u(1,j); zeros(My-3,1); rx*u(Mx + 1,j)]

+ ry*(u_1(ii,j-1) + u_1(ii,j + 1)) + ry2*u_1(ii,j);

u(ii,j) = trid(Ax,by); %Eq.(9.2.22b)

end

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2 0

0 2

y 4

Example 9.3 A Parabolic PDE: Two-Dimensional Temperature Diffusion.

Consider a two-dimensional parabolic PDE

with the initial conditions and boundary conditions

u(x, y, t) = e y cos x − e x cos y for x = 0, x = 4, y = 0, y = 4 ( E9.3.2b)

We made the following MATLAB program “solve_heat2.m” in order to usethe routine “heat2_ADI()” to solve this equation and ran this program to get theresult shown in Fig 9.4 at the final time

%solve_heat2

clear, clf

a = 1e-4;

it0 = inline(’0’,’x’,’y’); %(E9.3.2a)

bxyt = inline(’exp(y)*cos(x)-exp(x)*cos(y)’,’x’,’y’,’t’); %(E9.3.2b)

D = [0 4 0 4]; T = 5000; Mx = 40; My = 40; N = 50;

[u,x,y,t] = heat2_ADI(a,D,T,it0,bxyt,Mx,My,N);

mesh(x,y,u)

An example of a hyperbolic PDE is a one-dimensional wave equation for the

amplitude function u(x, t)(x is position, t is time) as

A ∂

2u(x, t)

∂x2 = ∂2u(x, t)

∂t2 for 0≤ x ≤ x f , 0≤ t ≤ T ( 9.3.1)

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HYPERBOLIC PDE 415

In order for this equation to be solvable, the boundary conditions u(0, t)=

b0(t) and u(x f , t) = b xf (t) as well as the initial conditions u(x, 0) = i0 (x)and

∂u/∂t|t=0(x, 0) = i

0(x)should be provided

In the same way as with the parabolic PDEs, we replace the second derivatives

on both sides of Eq (9.3.1) by their three-point central difference approximation(5.3.1) as

i directly from this

We use Eq (9.3.6) together with the initial conditions to get u1i and then go

on with Eq (9.3.3) for k = 1, 2, Note the following facts:

ž We must have r≤ 1 to guarantee the stability

ž The accuracy of the solution gets better as r becomes larger so that x

decreases

It is therefore reasonable to select r= 1

The stability condition can be obtained by substituting Eq (9.2.4) into

Eq (9.3.3) and applying the Jury test [P-3]:

λ = 2r cos(π/P ) + 2(1 − r) − λ−1, λ2+ 2(r(1 − cos(π/P )) − 1)λ + 1 = 0

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We need the solution of this equation to be inside the unit circle for stability,which requires

Example 9.4 A Hyperbolic PDE: One-Dimensional Wave (Vibration) Consider

a one-dimensional hyperbolic PDE

∂2u(x, t)

∂x2 = ∂u2(x, t)

∂t2 for 0≤ x ≤ 2, 0 ≤ y ≤ 2, and 0 ≤ t ≤ 2 (E9.4.1)

with the initial conditions and boundary conditions

u(x, 0) = x(1 − x), ∂u/∂t(x, 0) = 0 for t = 0 (E9.4.2a)

u( 0, t) = 0 for x = 0, u(1, t) = 0 for x = 1 (E9.4.2b)

We made the following MATLAB program “solve_wave.m” in order to usethe routine “wave()” to solve this equation and ran this program to get the resultshown in Fig 9.5 and see a dynamic picture

function [u,x,t] = wave(a,xf,T,it0,i1t0,bx0,bxf,M,N)

%solve a u_xx = u_tt for 0<=x<=xf, 0<=t<=T

% Initial Condition: u(x,0) = it0(x), u_t(x,0) = i1t0(x)

% Boundary Condition: u(0,t)= bx0(t), u(xf,t) = bxf(t)

u(2:M,k) = r*u(1:M - 1,k - 1) + r2*u(2:M,k-1) + r*u(3:M + 1,k - 1)

- u(2:M,k - 2); %Eq.(9.3.3) end

%solve_wave

a = 1;

it0 = inline(’x.*(1-x)’,’x’); i1t0 = inline(’0’); %(E9.4.2a)

bx0t = inline(’0’); bxft = inline(’0’); %(E9.4.2b)

for n = 1:N %dynamic picture

plot(x,u(:,n)), axis([0 xf -0.3 0.3]), pause(0.2)

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Figure 9.5 A solution for a 1-D hyperbolic PDE obtained by using ‘‘wave()’’ (Example 9.4).

In this section, we consider a two-dimensional wave equation for the amplitude

function u(x, y, t) ((x, y) is position, t is time) as

u( 0, y, t) = b x (y, t), u(x f , y, t) = b xf (y, t),

u(x, 0, t) = b y (x, t), and u(x, y f , t) = b yf (x, t)

as well as the initial condition u(x, y, 0) = i0 (x, y) and ∂u/∂t|t=0(x, y, 0)=

i0(x, y)

In the same way as with the one-dimensional case, we replace the ond derivatives on both sides by their three-point central difference approxi-mation (5.3.1) as

k

i,j+1− 2u k

i,j + u k i,j−1

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Since u−1i,j = u(x j , y i , −t) is not given, we cannot get u1

i,j directly from this

0(x j , y i ) ( 9.3.12) and make use of this to remove u−1i,j from Eq (9.3.11) to have

We use this Eq (9.3.13) together with the initial conditions to get u1

i,j and then

go on using Eq (9.3.10) for k = 1, 2, A sufficient condition for stability [S-1,

algo-Example 9.5 A Hyperbolic PDE: Two-Dimensional Wave (Vibration) Over a

Square Membrane Consider a two-dimensional hyperbolic PDE

u( 0, y, t) = 0, u(2, y, t) = 0, u(x, 0, t) = 0, u(x, 2, t) = 0 (E9.5.2)

u(x, y, 0) = 0.1 sin(πx) sin(πy/2), ∂u/∂t(x, y, 0) = 0 for t = 0 (E9.5.3)

We made the following MATLAB program “solve_wave2.m” in order to usethe routine “wave2()” for solving this equation and ran this program to get theresult shown in Fig 9.6 and see a dynamic picture Note that we can be sure ofstability, since we have

x2+ y2 = 4(1/4)(2/20)2

( 2/20)2+ (2/20)2 = 1

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HYPERBOLIC PDE 419

function [u,x,y,t] = wave2(a,D,T,it0,i1t0,bxyt,Mx,My,N)

%solve a(u_xx + u_yy) = u_tt for D(1) <= x <= D(2), D(3) <= y <= D(4), 0 <= t <= T

% Initial Condition: u(x,y,0) = it0(x,y), u_t(x,y,0) = i1t0(x,y)

% Boundary Condition: u(x,y,t) = bxyt(x,y,t) for (x,y) on Boundary

% Mx/My = # of subintervals along x/y axis

adt2 = a*dt*dt; rx = adt2/(dx*dx); ry = adt2/(dy*dy);

rxy1 = 1- rx - ry; rxy2 = rxy1*2;

u_1 = u;

for k = 0:N

t = k*dt;

for i = 1:My + 1 %Boundary condition

u(i,[1 Mx + 1]) = [bxyt(x(1),y(i),t) bxyt(x(Mx + 1),y(i),t)];

u(i,j) = 0.5*(rx*(u_1(i,j - 1) + u_1(i,j + 1))

+ ry*(u_1(i - 1,j)+u_1(i + 1,j))) + rxy1*u(i,j) + dt*ut(i,j);

u(i,j) = rx*(u_1(i,j - 1)+ u_1(i,j + 1))

+ ry*(u_1(i - 1,j) + u_1(i + 1,j)) + rxy2*u(i,j) -u_2(i,j);

end

u_2 = u_1; u_1 = u; %update the buffer memory

mesh(x,y,u), axis([0 2 0 2 -.1 1]), pause

end

%solve_wave2

it0 = inline(’0.1*sin(pi*x)*sin(pi*y/2)’,’x’,’y’); %(E9.5.3)

i1t0 = inline(’0’,’x’,’y’); bxyt = inline(’0’,’x’,’y’,’t’); %(E9.5.2)

(a) At t = 0.1

−0.1 0.1

0 0

0

2 2

(b) At t = 1.8

Figure 9.6 The solution of a two-dimensional hyperbolic PDE: vibration of a square membrane (Example 9.5).

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The FEM method is another procedure used in finding approximate numericalsolutions to BVPs/PDEs It can handle irregular boundaries in the same way asregular boundaries [R-1, S-2, Z-1] It consists of the following steps to solve theelliptic PDE:

∂2u(x, y)

∂x2 +∂2u(x, y)

∂y2 + g(x, y)u(x, y) = f (x, y) ( 9.4.1) for the domain D enclosed by the boundary B on which the boundary condition

is given as

1 Discretize the (two-dimensional) domain D into, say, N s subregions

{S1 , S2, , S N s} such as triangular elements, neither necessarily of thesame size nor necessarily covering the entire domain completely andexactly

2 Specify the positions of N n nodes and number them starting from the

boundary nodes, say, n = 1, , N b , and then the interior nodes, say, n=

N b + 1, , N n

3 Define the basis/shape/interpolation functions

φ n (x, y) = {φ n,s , for s = 1, , N s } ∀ (x, y) ∈ D (9.4.3a)

φ n,s (x, y) = p n,s ( 1) + p n,s ( 2)x + p n,s ( 3)y

for each subregion S s (9.4.3b)

collectively for all subregions s = 1 : N s and for each node n = 1 : N n, so

that φ n is 1 only at node n, and 0 at all other nodes Then, the

approxi-mate solution of the PDE is a linear combination of basis functions

ϕ1= [ φ1 φ2 · φ N b]T , c1= [ c1 c2 · c N b]T (9.4.5a)

ϕ2= [ φ N b+1 φ N b+2 · φ N n]T , c2= [ c N b+1 c N b+2 · c N n]T

(9.4.5b)

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FINITE ELEMENT METHOD (FEM) FOR SOLVING PDE 421

For each subregion s = 1, , N s, this solution can be written as

(x s , y s ): the centroid (gravity center) of the sth subregion S s

The FEM is based on the variational principle that a solution to Eq (9.4.1)can be obtained by minimizing the functional

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which, with u(x, y)= cT ϕ(x, y), can be written as

See [R-1] for details

The objectives of the MATLAB routines “fem_basis_ftn()” and

“fem_coef()” are to construct the basis function φ n,s (x, y)’s for each node

n = 1, , N n and each subregion s = 1, , N s and to get the coefficient vector

c of the solution (9.4.4) via Eq (9.4.7) and the solution polynomial φ s (x, y)’s

via Eq (9.4.6) for each subregion s = 1, , N s, respectively

Before going into a specific example of applying the FEM method to solve

a PDE, let us take a look at the basis (shape) function φ n (x, y) for each node

n = 1, , N n, which is defined collectively for all of the (triangular) subregions

so that φ n is 1 only at node n, and 0 at all other nodes and can be generated by

the routine “fem_basis_ftn()”

function p = fem_basis_ftn(N,S)

%p(i,s,1:3): coefficients of each basis ftn phi_i

%N(n,1:2) : x & y coordinates of the n-th node

%S(s,1:3) : the node #s of the s-th subregion(triangle)

N_n = size(N,1); % the total number of nodes

N_s = size(S,1); % the total number of subregions(triangles)

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function [U,c] = fem_coef(f,g,p,c,N,S,N_i)

%p(i,s,1:3): coefficients of basis ftn phi_i for the s-th subregion

%c = [ 1 1 0 0 ] with value for boundary and 0 for interior nodes

%N(n,1:2) : x & y coordinates of the n-th node

%S(s,1:3) : the node #s of the s-th subregion(triangle)

%N_i : the number of the interior nodes

%U(s,1:3) : the coefficients of p1 + p2(s)x + p3(s)y for each subregion N_n = size(N,1); % the total number of nodes = N_b + N_i

N_s = size(S,1); % the total number of subregions(triangles)

%phi_i,x*phi_n,x + phi_i,y*phi_n,y - g(x,y)*phi_i*phi_n

p_vctr = [p([i n],s,1) p([i n],s,2) p([i n],s,3)];

tmpg(s) = sum(p(i,s,2:3).*p(n,s,2:3))

-g(xy(1),xy(2))*p_vctr(1,:)*[1 xy]’*p_vctr(2,:)*[1 xy]’;

dS(s) = det([N(S(s,1),:) 1; N(S(s,2),:) 1;N(S(s,3),:) 1])/2;

%area of triangular subregion

if n == 1, tmpf(s) = -f(xy(1),xy(2))*p_vctr(1,:)*[1 xy]’; end

“trimesh()” to plot the shape functions for nodes n= 2, 3, 4, and 5 as depicted

in Figs 9.8b–e, each of which is 1 only at the corresponding node n and is

0 at all other nodes Figure 9.8f is the graph of a linear combination of basisfunctions

having the given value c n at each node n This can obtained by using the

MAT-LAB command “trimesh()” as

>>trimesh(S,N(:,1),N(:,2),c)

where the first input argument Shas the node numbers for each subregion, thesecond/third input argument N has the x/y coordinates for each node, and the

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1

0.2 0.5]

S = [1 2

2 3

3 4

4 5] 5; 5; 5;

Figure 9.7 A region (domain) divided into four triangular subregions.

fourth input argumentchas the function values at each node as follows:

The meaning of this N n ( the number of nodes:5) × N s (the number of

subre-gions:4) × 3 array p is that, say, the second rows of the three sub-arrays constitute

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Figure 9.8 The basis (shape) functions for nodes in Fig 9.7 and a composite function.

the coefficient vectors of the basis function for node 2 as

−7/10 + (1/2)x + (6/5)y for subregion S1

−7/16 + (15/16)x + (1/2)y for subregion S2

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With the program “show_basis.m” in your computer, type the following mands into the MATLAB command window and see the graphical/textual output

N = [-1 1;1 1;1 -1;-1 -1;0.2 0.5]; %the list of nodes in Fig.9.7

N_n = size(N,1); % the number of nodes

S = [1 2 5;2 3 5;3 4 5;1 4 5]; %the list of subregions in Fig.9.7

N_s = size(S,1); % the number of subregions

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Example 9.6 Laplace’s Equation: Electric Potential Over a Plate with Point

Charge Consider the following Laplace’s equation:

∇2u(x, y)= ∂2u(x, y)

∂x2 +∂2u(x, y)

∂y2 = f (x, y) (E9.6.1)for − 1 ≤ x ≤ +1, −1 ≤ y ≤ +1

23

8 9

1

28

26 16 27

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(−0.5, −0.5), since they are only two points at which the value of the right-hand side of Eq (9.6.1) is not zero, and consequently the value of the solution u(x, y)

is expected to change sensitively around them

We made the following MATLAB program “do_fem.m” in order to use theroutines “fem_basis_ftn()” and “fem_coef()” for solving this equation Forcomparison, we have added the statements to solve the same equation by using theroutine “poisson()” (Section 9.1) The results obtained by running this programare depicted in Fig 9.10a–c

−1 −1

x

(c) 16 ×15-point FDM (Finite Difference Method) solution

Figure 9.10 Results of Example 9.6.

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GUI OF MATLAB FOR SOLVING PDES: PDETOOL 429

N_n = size(N,1); %the total number of nodes

N_i = N_n - N_b; %the number of interior nodes

c = zeros(1,N_n); %boundary value or 0 for boundary/interior nodes

p = fem_basis_ftn(N,S);

[U,c] = fem_coef(f,g,p,c,N,S,N_i);

%Output through the triangular mesh-type graph

figure(1), clf, trimesh(S,N(:,1),N(:,2),c)

%Output through the rectangular mesh-type graph

N_s = size(S,1); %the total number of subregions(triangles)

by0 = inline(’0’); byf = inline(’0’);

[U,x,y] = poisson(f,g,bx0,bxf,by0,byf,[x0 xf y0 yf],Mx,My);

figure(3), clf, mesh(x,y,U)

In this section, we will see what problems can be solved by using the GUI (graphicuser interface) tool of MATLAB for PDEs and then apply the tool to solve theelliptic/parabolic/hyperbolic equations dealt with in Examples 9.1/9.3/9.5 and 9.6

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