Chapter 13 - Fundamentals of Numerical Computation

Functions¶

Create a tensor-product grid

tensorgrid.m

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
function [mtx, X, Y, vec, unvec, is_boundary] = tensorgrid(x, y)
% TENSORGRID Tensor-product grid.
% Input:
%   x, y           1d projections of the grid nodes (lengths m and n)
% Output:
%   mtx            evaluate a function on the grid (function)
%   X, Y           mtx applied to the coordinate functions (m×n)
%   vec            convert grid shape to vector shape (function)
%   unvec          convert vector shape to grid shape (function)
%   is_boundary    indicator of boundary nodes (logical m×n)
    m = length(x) - 1;
    n = length(y) - 1;
    vec = @(U) U(:);
    unvec = @(u) reshape(u, m+1, n+1);
    [X, Y] = ndgrid(x, y);
    function F = grideval(f)
        F = zeros(size(X));
        for k = 1:numel(X)
            F(k) = f(X(k), Y(k));
        end
    end
    mtx = @grideval;
    
    % Identify boundary points.
    is_boundary = true(m+1, n+1);
    is_boundary(2:m, 2:n) = false;
end

Solution of Poisson’s equation by finite differences

poissonfd.m

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
function [X, Y, U] = poissonfd(f,g,m,xspan,n,yspan)
%POISSONFD   Solve Poisson's equation by finite differences.
% Input:
%   f            forcing function (function of x,y)
%   g            boundary condition (function of x,y)
%   m            number of grid points in x (integer)
%   xspan        endpoints of the domain of x (2-vector)
%   n            number of grid points in y (integer)
%   yspan        endpoints of the domain of y (2-vector)
%
% Output:
%   U            solution (m+1 by n+1)
%   X,Y          grid matrices (m+1 by n+1)

% Discretize the domain.
[x, Dx, Dxx] = diffmat2(m, xspan);
[y, Dy, Dyy] = diffmat2(n, yspan);
[mtx, X, Y, vec, unvec, is_boundary] = tensorgrid(x, y);

% Form the collocated PDE as a linear system. 
Ix = speye(m+1);  Iy = speye(n+1);
A = kron(Iy, sparse(Dxx)) + kron(sparse(Dyy), Ix);  % Laplacian matrix
b = vec(mtx(f));

% Replace collocation equations on the boundary.
scale = max(abs(A(n+2, :)));
I = speye(size(A));
idx = vec(is_boundary);
A(idx, :) = scale * I(idx, :);           % Dirichet assignment
b(idx) = scale * g( X(idx),Y(idx) );     % assigned values

% Solve the linear sytem and reshape the output.
u = A \ b;
U = unvec(u);

Solution of elliptic PDE by Chebyshev collocation

elliptic.m

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
function u = elliptic(phi, g, m, xspan, n, yspan)
%ELLIPTIC   Solve an elliptic PDE in 2d.
% Input:
%   phi          defines phi)(x,y,u,u_x,u_xx,u_y,u_yy) = 0 (function)
%   g            boundary condition (function)
%   m, xspan     size and interval of x discretization (integer, 2-vector)
%   n, yspan     size and interval of y discretization (integer, 2-vector)
% Output:
%   U            solution (n+1 by n+1)
%   X,Y          coordinate matrices (n+1 by n+1)

    % Discretize the domain.
    [x, Dx, Dxx] = diffcheb(m, xspan);
    [y, Dy, Dyy] = diffcheb(n, yspan);
    [mtx, X, Y, vec, unvec, is_boundary] = tensorgrid(x, y);

    % Identify boundary locations and evaluate the boundary condition.
    idx = vec(is_boundary);
    gb = g(X(idx), Y(idx));

    % Evaluate the PDE+BC residual.
    function r = residual(u)
        U = unvec(u);
        R = phi(X, Y, U, Dx * U, Dxx * U, U * Dy', U * Dyy');  % PDE
        R(idx) = u(idx) - gb;                                  % boundary
        r = vec(R);
    end

    % Solve the equation.
    u = levenberg(@residual, vec(zeros(size(X))));
    U = unvec(u(:, end));

    function u = evaluate(xi, eta)
        v = zeros(1, n+1);
        for j = 1:n+1
            v(j) = chebinterp(x, U(:, j), xi);
        end
        u = chebinterp(y, v, eta);
    end

    u = @evaluate;
end

function f = chebinterp(x, v, xi)
    n = length(x) - 1;
    w = (-1.0) .^ (0:n)';
    w([1, n+1]) = w([1, n+1]) / 2;

    terms = w ./ (xi - x(:));
    if any(isinf(terms))     % exactly at a node
        f = v(xi == x);
    else
        f = sum(v(:) .* terms) / sum(terms);
    end
end

Examples¶

13.1 Tensor-product discretizations¶

Example 13.1.2

Here is the grid from Example 13.1.1.

m = 4;   
x = linspace(0, 2, m+1);
n = 2;   
y = linspace(1, 3, n+1);

For a given $f(x,y)$ we can find $\operatorname{mtx}(f)$ by using a comprehension syntax.

[mtx, X, Y] = tensorgrid(x, y);
f = @(x, y) cos(pi * x .* y - y);
F = mtx(f)

We can make a nice plot of the function by first choosing a much finer grid. However, the contour and surface plotting functions expect the transpose of mtx( $f$ ).

m = 80;  x = linspace(0, 2, m+1);
n = 60;  y = linspace(1, 3, n+1);
[mtx, X, Y] = tensorgrid(x, y);
F = mtx(f);

pcolor(X', Y', F')
shading interp
colormap(redsblues),  colorbar
axis equal
xlabel("x"),  ylabel("y")

Example 13.1.3

For a function given in polar form, such as $f(r,\theta)=1-r^4$ , construction of a function over the unit disk is straightforward using a grid in $(r,\theta)$ space.

r = linspace(0, 1, 41);
theta = linspace(0, 2*pi, 121);
[mtx, R, Theta] = tensorgrid(r, theta);
F = mtx(@(r, theta) 1 - r.^4);
clf,  colormap(parula)
contourf(R', Theta', F', 20)
shading flat
xlabel("r"),  ylabel("\theta"), 
title("A polar function")

Of course, we are used to seeing such plots over the $(x,y)$ plane, not the $(r,\theta)$ plane. For this we create matrices for the coordinate functions $x$ and $y$ .

X = R .* cos(Theta);  Y = R .* sin(Theta);
contourf(X', Y', F', 20)
axis equal,  shading interp  
xlabel('x'),  ylabel('y')
title('Function over the unit disk')

In such functions the values along the line $r=0$ must be identical, and the values on the line $\theta=0$ should be identical to those on $\theta=2\pi$ . Otherwise the interpretation of the domain as the unit disk is nonsensical. If the function is defined in terms of $x$ and $y$ , then those can be defined in terms of $r$ and θ using (13.1.6).

Example 13.1.4

We define a function and, for reference, its two exact partial derivatives.

u = @(x, y) sin(pi * x .* y - y);
du_dx = @(x, y) pi * y .* cos(pi * x .* y - y);
du_dy = @(x, y) (pi * x - 1) .* cos(pi * x .* y - y);

We will use an equispaced grid and second-order finite differences as implemented by diffmat2. First, we have a look at a plots of the exact partial derivatives.

m = 80;  [x, Dx] = diffmat2(m, [0, 2]);
n = 60;  [y, Dy] = diffmat2(n, [1, 4]);
[mtx, X, Y] = tensorgrid(x, y);
U = mtx(u);
dU_dX = mtx(du_dx);
dU_dY = mtx(du_dy);

clf,  subplot(1, 2, 1)
pcolor(X', Y', dU_dX')
axis equal,  shading interp
title('∂u/∂x')
subplot(1, 2, 2)
pcolor(X', Y', dU_dY')
axis equal,  shading interp
title('∂u/∂y')

Now we compare the exact partial derivatives with their finite-difference approximations. Since these are signed errors, we use a colormap that is symmetric around zero.

err = dU_dX - Dx * U;
subplot(1, 2, 1)
pcolor(X', Y', err')
colorbar,  clim([-.4, .4])
axis equal,  shading interp
title('error in ∂u/∂x')

err = dU_dY - U * Dy';
subplot(1,2,2)
pcolor(X', Y', err')
colorbar,  clim([-.1, .1])
axis equal,  shading interp
colormap(redsblues)
title('error in ∂u/∂y')

Error using -
Arrays have incompatible sizes for this operation.

Not surprisingly, the errors are largest where the derivatives themselves are largest.

13.2 Two-dimensional diffusion and advection¶

Example 13.2.1

m = 4;  n = 3;
x = linspace(0, 2, m+1);
y = linspace(-3, 0, n+1);

f = @(x, y) cos(0.75*pi * x .* y - 0.5*pi * y);
[mtx, X, Y, vec, unvec] = tensorgrid(x, y);
F = mtx(f);
disp("function on a 4x3 grid:")
disp(F)

function on a 4x3 grid:

   -0.0000   -1.0000    0.0000    1.0000
    0.3827    0.7071    0.9239    1.0000
   -0.7071    0.0000    0.7071    1.0000
    0.9239   -0.7071   -0.3827    1.0000
   -1.0000    1.0000   -1.0000    1.0000

disp("vec(F):")
disp(vec(F))

vec(F):

  -1.8370e-16
   3.8268e-01
  -7.0711e-01
   9.2388e-01
  -1.0000e+00
  -1.0000e+00
   7.0711e-01
   6.1232e-17
  -7.0711e-01
   1.0000e+00
   6.1232e-17
   9.2388e-01
   7.0711e-01
  -3.8268e-01
  -1.0000e+00
   1.0000e+00
   1.0000e+00
   1.0000e+00
   1.0000e+00
   1.0000e+00

The unvec operation is the inverse of vec.

disp("unvec(vec(F)):")
disp(unvec(vec(F)))

unvec(vec(F)):

  Columns 1 through 3

  -0.000000000000000  -1.000000000000000   0.000000000000000
   0.382683432365090   0.707106781186548   0.923879532511287
  -0.707106781186547   0.000000000000000   0.707106781186548
   0.923879532511287  -0.707106781186548  -0.382683432365090
  -1.000000000000000   1.000000000000000  -1.000000000000000

  Column 4

   1.000000000000000
   1.000000000000000
   1.000000000000000
   1.000000000000000
   1.000000000000000

Example 13.2.2

m = 60;  n = 40;
[x, Dx, Dxx] = diffper(m, [-1, 1]);
[y, Dy, Dyy] = diffper(n, [-1, 1]);
[mtx, X, Y, vec, unvec] = tensorgrid(x, y);

Note that the initial condition should also be periodic on the domain.

U0 = sin(4*pi*X) .* exp( cos(pi*Y) );
clf,  surf(X', Y', U0')
mx = max(abs(vec(U0)));
clim([-mx, mx]),  shading interp
colormap(redsblues)
xlabel('x'),  ylabel('y')  
title('Initial condition')

This function computes the time derivative for the unknowns. The actual calculations take place using the matrix shape.

f13_2_heat.m

function du_dt = timederiv(t, u, p)
    [alpha, Dxx, Dyy, vec, unvec] = p{:};
    U = unvec(u);
    Uxx = Dxx * U;  Uyy = U * Dyy';     % 2nd partials
    dU_dt = alpha * (Uxx + Uyy);  % PDE
    du_dt = vec(dU_dt);
end

Since this problem is parabolic, a stiff integrator is appropriate.

ivp = ode(ODEFcn=@f13_2_heat);
ivp.InitialTime = 0;
ivp.InitialValue = vec(U0);
ivp.Parameters = {0.1, Dxx, Dyy, vec, unvec};
ivp.Solver = "stiff";
sol = solutionFcn(ivp, 0, 0.2);
U = @(t) unvec(sol(t));

Error using *
Incorrect dimensions for matrix multiplication. Check that the number of columns in the first matrix matches the number of rows in the second matrix. To operate on each element of the matrix individually, use TIMES (.*) for elementwise multiplication.

Error in f13_2_heat (line 4)
    Uxx = Dxx * U;  Uyy = U * Dyy';     % 2nd partials

Error in odearguments (line 93)
f0 = ode(t0,y0,args{:});

Error in ode15s (line 148)
    odearguments(odeIsFuncHandle, odeTreatAsMFile, solver_name, ode, tspan, y0, options, varargin);

Error in matlab.ode.internal.LegacySolver/extend (line 56)
                obj.sol = obj.SolverFcn(args{:});

Error in matlab.ode.internal.SolverPair/extend (line 162)
                [obj.rSolver,tmax] = obj.rSolver.extend(t2);

Error in ode/solutionFcn (line 382)
            [ODESOL,L,R] = ODESOL.extend(L,R);

surf(X', Y', U(0.05)')
clim([-mx, mx]),  shading interp
colormap(redsblues)
xlabel('x'),  ylabel('y')  
title('Solution at t = 0.05')

Array indices must be positive integers or logical values.

Here is an animation of the solution.

title('Heat equation on a periodic domain')
vid = VideoWriter("figures/2d-heat.mp4","MPEG-4");
vid.Quality = 85;
open(vid);
for t = linspace(0, 0.2, 61)
    cla, surf(X', Y', U(t)')
    zlim([-3, 3]),  clim([-mx, mx])
    shading interp
    str = sprintf("t = %.2f", t);
    text(-0.9, 0.75, 2, str, fontsize=14);
    writeVideo(vid, frame2im(getframe(gcf)));
end
close(vid)

Example 13.2.3

The first step is to define a discretization of the domain and the initial state.

m = 50;  n = 40;
[x, Dx, Dxx] = diffcheb(m, [-1, 1]);
[y, Dy, Dyy] = diffcheb(n, [-1, 1]);
[mtx, X, Y] = tensorgrid(x, y);
u_init = @(x, y) (1+y) .* (1-x).^4 .* (1+x).^2 .* (1-y.^4);

There are really two grids now: the full grid and the subset grid of interior points. Since the IVP unknowns are on the interior grid, that is the one we need to change shapes on. We also need the functions extend and chop to add and remove boundary values.

[~, ~, ~, vec, unvec] = tensorgrid(x(2:m), y(2:n));
chop = @(U) U(2:m, 2:n);
z = zeros(1, n-1);
extend = @(U) [ zeros(m+1, 1) [z; U; z] zeros(m+1, 1)];
pack = @(U) vec(chop(U));
unpack = @(u) extend(unvec(u));

Index exceeds the number of array elements. Index must not exceed 15.

Now we can define and solve the IVP using a stiff solver.

f13_2_advdiff.m

function du_dt = timederiv(t, u, p)
    [ep, Dx, Dxx, Dy, Dyy, pack, unpack] = p{:};
    U = unpack(u);
    Uxx = Dxx * U;  Uyy = U * Dyy'; 
    dU_dt = 1 - Dx * U + ep * (Uxx + Uyy);  % PDE
    du_dt = pack(dU_dt);
end

ivp = ode(ODEFcn=@f13_2_advdiff);
ivp.InitialTime = 0;
ivp.InitialValue = pack(mtx(u_init));
ivp.Parameters = {0.05, Dx, Dxx, Dy, Dyy, pack, unpack};
ivp.Solver = "stiff";
sol = solutionFcn(ivp, 0, 2);

Index in position 1 is invalid. Array indices must be positive integers or logical values.

Error in tensorgrid/grideval (line 19)
            F(k) = f(X(k), Y(k));

When we evaluate the solution at a particular value of $t$ , we get a vector of the interior grid values. The same unpack function above converts this to a complete matrix of grid values.

U = @(t) unpack(sol(t));

clf,  pcolor(X', Y', U(0.5)')
clim([0, 2]), shading interp
axis equal,  colormap(sky), colorbar
title('Advection-diffusion at t = 0.5')  
xlabel('x'),  ylabel('y')

unpack requires one of the following:
  MATLAB Support Package for Quantum Computing
  Vehicle Network Toolbox

Error in Notebook>@(t)unpack(sol(t)) (line 1)
U = @(t) unpack(sol(t));

hold on
vid = VideoWriter("figures/2d-advdiff.mp4","MPEG-4");
vid.Quality = 85;
open(vid);
title("Advection-diffusion in 2d")
for t = linspace(0, 2, 81)
    cla, pcolor(X', Y', U(t)')
    shading interp
    str = sprintf("t = %.2f", t);
    text(-1.5, 0.75, str, fontsize=14);
    writeVideo(vid, frame2im(getframe(gcf)));
end
close(vid)

Example 13.2.4

We start with the discretization and initial condition.

m = 40; n = 42;
[x, Dx, Dxx] = diffcheb(m, [-2, 2]);
[y, Dy, Dyy] = diffcheb(n, [-2, 2]);
[mtx, X, Y] = tensorgrid(x, y);

u_init = @(x, y) (x+0.2) .* exp(-12*(x.^2 + y.^2));
U0 = mtx(u_init);
V0 = zeros(size(U0));

Note that because $u$ is known on the boundary, while $v$ is unknown over the full grid, there are two different sizes of vec/unvec operations. We also need to define functions to pack grid unknowns into a vector and to unpack them. When the unknowns for $u$ are packed, the boundary values are chopped off, and these are restored when unpacking.

[~, ~, ~, vec_v, unvec_v] = tensorgrid(x, y);
[~, ~, ~, vec_u, unvec_u] = tensorgrid(x(2:m), y(2:n));

chop = @(U) U(2:m, 2:n);
z = zeros(1, n-1);
extend = @(U) [ zeros(m+1, 1) [z; U; z] zeros(m+1, 1)];
pack = @(U, V) [vec_u(chop(U)); vec_v(V)];
N = (m-1) * (n-1);
unpack = @(u) f13_2_wave_unpack(u, N, unvec_u, unvec_v, extend);

Error using zeros
Size inputs must be integers.

f13_2_wave_unpack.m

function [U, V] = unpack(w, N, unvec_u, unvec_v, extend)
    U = extend( unvec_u(w(1:N)) );
    V = unvec_v(w(N+1:end));
end

We can now define and solve the IVP. Since this problem is hyperbolic, not parabolic, a nonstiff integrator is faster than a stiff one.

f13_2_wave.m

function dw_dt = timederiv(t, w, p)
    [Dxx, Dyy, pack, unpack] = p{:};
    [U, V] = unpack(w);
    dU_dt = V;
    dV_dt = Dxx * U + U * Dyy';
    dw_dt = pack(dU_dt, dV_dt);
end

ivp = ode(ODEFcn=@f13_2_wave);
ivp.InitialTime = 0;
ivp.InitialValue = pack(U0, V0);
ivp.Parameters = {Dxx, Dyy, pack, unpack};
ivp.Solver = "nonstiff";
sol = solutionFcn(ivp, 0, 4);

pack requires Vehicle Network Toolbox.

clf
[U, V] = unpack(sol(0.5));
pcolor(X', Y', U')
axis equal,  clim([-0.1, 0.1])
colormap(redsblues),  shading interp
xlabel("x"),  ylabel("y")
title("Wave equation at t = 0.5")

Array indices must be positive integers or logical values.

hold on
vid = VideoWriter("figures/2d-wave.mp4","MPEG-4");
vid.Quality = 85;
open(vid);
title("Wave equation in 2d")
for t = linspace(0, 4, 121)
    [U, V] = unpack(sol(t));
    cla, pcolor(X, Y, U)
    shading interp
    str = sprintf("t = %.2f", t);
    text(-3, 1.75, str, fontsize=14);
    writeVideo(vid, frame2im(getframe(gcf)));
end
close(vid)

13.3 Laplace and Poisson equations¶

Example 13.3.1

A = [1, 2; -2, 0];
B = [1, 10, 100; -5, 5, 3];
disp("A:")
disp(A)
disp("B:")
disp(B)

A:

     1     2
    -2     0

B:

     1    10   100
    -5     5     3

Applying the definition manually, we get

A_kron_B = [
    A(1,1)*B  A(1,2)*B;
    A(2,1)*B  A(2,2)*B
    ]

But it makes more sense to use kron.

kron(A, B)

Example 13.3.2

We make a crude discretization for illustrative purposes.

m = 6;  n = 5;
[x, Dx, Dxx] = diffmat2(m, [0, 3]);
[y, Dy, Dyy] = diffmat2(n, [-1, 1]);
[mtx, X, Y, vec, unvec, is_boundary] = tensorgrid(x, y);

Next, we define ϕ and evaluate it on the grid to get the forcing vector of the linear system.

phi = @(x, y) x.^2 - y + 2;
b = vec(mtx(phi));

Here are the coefficients for the PDE collocation, before any modifications are made for the boundary conditions. The combination of Kronecker products and finite differences produces a characteristic sparsity pattern.

A = kron(speye(n+1), sparse(Dxx)) + kron(sparse(Dyy), speye(m+1));
clf,  spy(A)
title("System matrix before boundary conditions")

Error using +
Arrays have incompatible sizes for this operation.

The number of equations is equal to $(m+1)(n+1)$ , which is the total number of points on the grid.

N = length(b)

We now use the Boolean array that indicates where the boundary points lie in the grid.

spy(is_boundary)
title("Boundary points")

In order to impose Dirichlet boundary conditions, we replace the boundary rows of the system by rows of the identity.

I = speye(N);
idx = vec(is_boundary);
A(idx, :) = I(idx, :);

spy(A)
title("System matrix with boundary conditions")

Index exceeds array bounds.

Finally, we must replace the rows in the vector $\mathbf{b}$ by the boundary values being assigned to the boundary points. Here, we let the boundary values be zero everywhere.

b(idx) = 0;

Now we can solve for $\mathbf{u}$ and reinterpret it as the matrix-shaped $\mathbf{U}$ , the solution on our grid.

u = A \ b;
U = unvec(u)

Error using reshape
Number of elements must not change. Use [] as one of the size inputs to automatically calculate the appropriate size for that dimension.

Error in tensorgrid>@(u)reshape(u,m+1,n+1) (line 14)
    unvec = @(u) reshape(u, m+1, n+1);

Example 13.3.3

First we define the problem on $[0,1]\times[0,2]$ .

f = @(x, y) -sin(3*x .* y - 4*y) .* (9*y.^2 + (3*x - 4).^2);
g = @(x, y) sin(3*x .* y - 4*y);
xspan = [0, 1];
yspan = [0, 2];

Here is the finite-difference solution.

[X, Y, U] = poissonfd(f, g, 40, xspan, 60, yspan);

clf, surf(X', Y', U')
colormap(parula),  shading interp
colorbar
title("Solution of Poisson's equation")
xlabel("x"),  ylabel("y"),  zlabel("u(x,y)")

Since this is an artificial problem with a known solution, we can plot the error, which is a smooth function of $x$ and $y$ . It must be zero on the boundary; otherwise, we have implemented boundary conditions incorrectly.

err = g(X, Y) - U;
mx = max(abs(vec(err)));
pcolor(X', Y', err')
colormap(redsblues),  shading interp
clim([-mx, mx]),  colorbar
axis equal,  xlabel("x"),  ylabel("y")
title("Error")

13.4 Nonlinear elliptic PDEs¶

Example 13.4.2

All we need to define are ϕ from (13.4.2) for the PDE, and a trivial zero function for the boundary condition.

lambda = 1.5;
phi = @(X, Y, U, Ux, Uxx, Uy, Uyy) Uxx + Uyy - lambda ./ (U + 1).^2;
g = @(x, y) zeros(size(x));

Here is the solution for $m=15$ , $n=8$ .

u = elliptic(phi, g, 15, [0, 2.5], 8, [0, 1]);
disp(sprintf("solution at (2, 0.6) is %.7f", u(2, 0.6)))

Error using LiveEditorEvaluationHelperE1fb1e3a3-29d1-439a-b6eb-1cf408c2605a>@(x,u,du_dx)(u.^3-u)/epsilon
Too many input arguments.

Error in elliptic/residual (line 24)
        R = phi(X, Y, U, Dx * U, Dxx * U, U * Dy', U * Dyy');  % PDE

Error in levenberg (line 14)
x = x1(:);     fk = f(x1);

Error in elliptic (line 30)
    u = levenberg(@residual, vec(zeros(size(X))));

x = linspace(0, 2.5, 91);
y = linspace(0, 1, 51);
[mtx, X, Y] = tensorgrid(x, y);
clf,  pcolor(x, y, mtx(u)')
colormap(flipud(sky)),  shading interp,  colorbar
axis equal
xlabel("x"),  ylabel("y")
title("Deflection of a MEMS membrane")

Index in position 1 is invalid. Array indices must be positive integers or logical values.

Error in tensorgrid/grideval (line 19)
            F(k) = f(X(k), Y(k));

In the absence of an exact solution, how can we be confident that the solution is accurate? First, the Levenberg iteration converged without issuing a warning, so we should feel confident that the discrete equations were solved. Assuming that we encoded the PDE correctly, the remaining source of error is truncation from the discretization. We can estimate that by refining the grid a bit and seeing how much the numerical solution changes.

x_test = linspace(0, 2.5, 6);
y_test = linspace(0, 1 , 5);
mtx_test = tensorgrid(x_test, y_test);
format long
mtx_test(u)

Index in position 1 is invalid. Array indices must be positive integers or logical values.

Error in tensorgrid/grideval (line 19)
            F(k) = f(X(k), Y(k));

u = elliptic(phi, g, 25, [0, 2.5], 14, [0, 1]);
mtx_test(u)

Error using LiveEditorEvaluationHelperE1fb1e3a3-29d1-439a-b6eb-1cf408c2605a>@(x,u,du_dx)(u.^3-u)/epsilon
Too many input arguments.

Error in elliptic/residual (line 24)
        R = phi(X, Y, U, Dx * U, Dxx * U, U * Dy', U * Dyy');  % PDE

Error in levenberg (line 14)
x = x1(:);     fk = f(x1);

Error in elliptic (line 30)
    u = levenberg(@residual, vec(zeros(size(X))));

The original solution seems to be accurate to about four digits.

Example 13.4.3

phi = @(X, Y, U, Ux, Uxx, Uy, Uyy) 1 - Ux - 2*Uy + 0.05*(Uxx + Uyy);
g = @(x, y) zeros(size(x));
u = elliptic(phi, g, 32, [-1, 1], 32, [-1, 1]);

x = linspace(-1, 1, 80);
y = x;
mtx = tensorgrid(x, y);
clf,  pcolor(x, y, mtx(u)')
colormap(parula),  shading interp,  colorbar
axis equal,  xlabel("x"),  ylabel("y")
title("Steady advection–diffusion")

Example 13.4.4

The following defines the PDE and a nontrivial Dirichlet boundary condition for the square $[0,1]^2$ .

phi = @(X, Y, U, Ux, Uxx, Uy, Uyy) U .* (1-U.^2) + 0.05*(Uxx + Uyy);
g = @(x, y) tanh(5*(x + 2*y - 1));

We solve the PDE and then plot the result.

u = elliptic(phi, g, 36, [0, 1], 36, [0, 1]);

x = linspace(0, 1, 80);
y = x;
mtx = tensorgrid(x, y);
clf,  pcolor(x, y, mtx(u)')
colormap(parula),  shading interp,  colorbar
axis equal,  xlabel("x"),  ylabel("y")
title("Steady Allen–Cahn")