# Preface

This section presents basic material about numerical solutions of parabolic equations.

Introduction to Linear Algebra with Mathematica

# Numerical Heat Solutions

We demonstrate application of finite difference schemes for numerical solution of the one-dimensional heat equation

$\frac{\partial u}{\partial t} = \alpha^2 \frac{\partial^2 u}{\partial x^2} , \qquad u(x,0) = f(x) ,$
where f(x) is a given function on the interval $$(0 , \ell ) .$$ Usually, the boundary conditions are imposed on spacial variable x. For simplicity, we consider the boundary conditions of type I (called Dirichlet):
$u(0,t) = g_0 (t) , \qquad u(\ell ,t) = g_l (t) ,$
where g0 and gl are specified temperatures at end points. As a rule, these functions are just constants. The heat equation models the temperature distribution in an insulated rod with ends held at constant temperatures g0 and gl when the initial temperature along the rod is known f. Although analytic solutions to the heat conduction equation can be obtained with Fourier series, we use the problem as a prototype of a parabolic equation for numerical solution.

Assume that the rectangle $$R= \{\,(x,t)\,:\, 0 \le x \le \ell , \ 0 \le t \le b \,\}$$ is subdivided into n-1 and m-1 rectangles with sides Δx = h and Δt = k, as shown in the figure below.

l0 = Graphics[{Thick, Line[{{0, 0}, {8, 0}}]}];
l1 = Graphics[{Thick, Line[{{0, 1}, {8, 1}}]}];
l2 = Graphics[{Thick, Line[{{0, 2}, {8, 2}}]}];
l3 = Graphics[{Thick, Line[{{0, 3}, {8, 3}}]}];
l4 = Graphics[{Thick, Line[{{0, 4}, {8, 4}}]}];
l5 = Graphics[{Thick, Line[{{0, 5}, {8, 5}}]}];
la0 = Graphics[{Thick, Line[{{0, 0}, {0, 6}}]}];
la1 = Graphics[{Thick, Line[{{1, 0}, {1, 6}}]}];
la2 = Graphics[{Thick, Line[{{2, 0}, {2, 6}}]}];
la3 = Graphics[{Thick, Line[{{3, 0}, {3, 6}}]}];
la4 = Graphics[{Thick, Line[{{4, 0}, {4, 6}}]}];
la5 = Graphics[{Thick, Line[{{5, 0}, {5, 6}}]}];
la6 = Graphics[{Thick, Line[{{6, 0}, {6, 6}}]}];
la7 = Graphics[{Thick, Line[{{7, 0}, {7, 6}}]}];
la8 = Graphics[{Thick, Line[{{8, 0}, {8, 6}}]}];
d0 = Graphics[{Orange, Disk[{4, 3}, 0.15]}];
d1 = Graphics[{Orange, Disk[{4, 4}, 0.15]}];
la9 = Graphics[{Thick, Line[{{0, 6}, {8, 6}}]}];
d3 = Graphics[{Orange, Disk[{3, 3}, 0.15]}];
d4 = Graphics[{Orange, Disk[{5, 3}, 0.15]}];
r1 = Graphics[ Text[Style["\!$$\*SubscriptBox[\(x$$, $$i$$]\)", Medium], {4, -0.3}]];
r2 = Graphics[Text[ Style["\!$$\*SubscriptBox[\(x$$, $$i-1$$]\)", Medium], {3, -0.3}]];
r3 = Graphics[Text[ Style["\!$$\*SubscriptBox[\(x$$, $$i+1$$]\)", Medium], {5, -0.3}]];
r4 = Graphics[Text[ Style["\!$$\*SubscriptBox[\(x$$, $$n-1$$]\)", Medium], {7, -0.3}]];
r5 = Graphics[Text[ Style["\!$$\*SubscriptBox[\(x$$, $$n$$]\)", Medium], {8, -0.3}]];
r6 = Graphics[Text[ Style["\!$$\*SubscriptBox[\(x$$, $$2$$]\)", Medium], {1, -0.3}]];
r7 = Graphics[Text[ Style["\!$$\*SubscriptBox[\(x$$, $$1$$]\)", Medium], {0, -0.3}]];
t0 = Graphics[Text[ Style["\!$$\*SubscriptBox[\(t$$, $$1$$]\)", Medium], {-0.3, 0}]];
t1 = Graphics[Text[ Style["\!$$\*SubscriptBox[\(t$$, $$j-1$$]\)", Medium], {-0.4, 2}]];
t2 = Graphics[Text[ Style["\!$$\*SubscriptBox[\(t$$, $$j$$]\)", Medium], {-0.3, 3}]];
t3 = Graphics[Text[ Style["\!$$\*SubscriptBox[\(t$$, $$j+1$$]\)", Medium], {-0.4, 4}]];
t4 = Graphics[Text[ Style["\!$$\*SubscriptBox[\(t$$, $$m$$]\)", Medium], {-0.3, 6}]];
Show[l0, l1, l2, l3, l4, l5, la0, la1, la2, la3, la4, la5, la6, la7, la8, la9, d0, d1, d3, d4, r1, r2, r3, r4, r5, r6, r7, t0, t1, t2, t3, t4]

We are going to approximate the solution u(x,t) at grid points in successive rows $$\{ \,u(x_i , t)j )\,:\, i=1,2,\ldots , n \,\}$$ for j=2,3,...,m. First, we approximate partial derivatives ut and uxx:

$\frac{\partial u(x,t)}{\partial t} = \frac{u(x,t+k) - u(x,t)}{k} + O(k)$
and
$\frac{\partial^2 u(x,t)}{\partial x^2} = \frac{u(x-h,t) - 2\, u(x,t) + u(x+h,t)}{h^2} + O\left( h^2 \right) .$
The grid spacing is uniform in every row: xi+1 = xi + h and xi-1 = xi - h, and it is uniform in every column tj+1 = tj + k. Next, we drop the terms O(k) and O(h²) and use the approximation ui,j for u(xi, tj) in the heat equation:
$\frac{u_{i,j+1} - u_{i,j}}{k} = \alpha^2 \frac{u_{i-1,j} - 2\,u_{i,j} + u_{i+1,j}}{h^2} ,$
which approximates the solution to the heat transfer equation. For convenience, the substitution
$r = \alpha^2 \frac{k}{h^2}$
is introduced, which leads to the explicit forward-difference equation
$u_{i,j+1} = \left( 1- 2r \right) u_{i,j} + r \left( u_{i-1,j} + u_{i=1,j} \right) .$
This equation is used to create the (j+1)-th row across the grid, assuming that approximations in the j-th row are known. Notice that this formula explicitly gives the value ui,j+1 in terms of ui-1,j, ui,j, and ui+1,j. The corresponding computational stencil is represented below.
l0 = Graphics[{Thick, Line[{{-4, 0}, {4, 0}}]}];
l1 = Graphics[{Thick, Line[{{0, 0}, {0, 2}}]}];
d0 = Graphics[{Red, Disk[{0, 0}, 0.15]}];
d1 = Graphics[{Red, Disk[{4, 0}, 0.15]}];
d2 = Graphics[{Red, Disk[{-4, 0}, 0.15]}];
d3 = Graphics[{Red, Disk[{0, 2}, 0.15]}];
r1 = Graphics[ Text[Style["(1-2r)\!$$\*SubscriptBox[\(u$$, $$i,j$$]\)", Medium], {0, -0.5}]];
r2 = Graphics[Text[ Style["r \!$$\*SubscriptBox[\(u$$, $$i+1,j$$]\)", Medium], {4.4, -0.3}]];
r3 = Graphics[Text[ Style["r \!$$\*SubscriptBox[\(u$$, $$i-1,j$$]\)", Medium], {-4.5, -0.3}]];
r4 = Graphics[Text[ Style["\!$$\*SubscriptBox[\(u$$, $$i,j+1$$]\)", Medium], {0, 2.4}]];
Show[l0, l1, d0, d1, d2, d3, r1, r2, r3, r4]

The simplicity of the above formula makes it appealing to use. However, it is important to utilize numerical techniques that are stable. If any error made at one stage of the calculations is eventually dampened out, the method is called stable. The above explicit forward-difference scheme is stable if and only if r is restricted to the interval $$0 \le r \le \frac{1}{2} .$$ This means that the step size k must satisfy so called the Courant--Friedrichs--Lewy or CFL condition: $$k \le h^2 /(2 \alpha^2 ) .$$ The condition is named after Richard Courant (1888--1972), Kurt Friedrichs (1901--1982), and Hans Lewy (1904--1988) who described it in their 1928 paper. If the CFL condition is not fulfilled, errors committed in one line { ui,j } might be magnified in subsequent lines { ui,p } for some p > j. The next example illustrates this point.

Example: Consider the initial boundary value problem

$u_t (x,t) = u_{xx} (x,t) \qquad \mbox{for } 0 < x < 2, \quad 0 < t < 1,$
with the initial condition
$u(x,0) = 2x-x^2 \qquad \mbox{for } t=0 \quad\mbox{and}\quad 0 \le x \le 2,$
and the boundary conditions
$\begin{split} u(0,t) =0 \qquad\mbox{for} \quad 0 \le t \le 1, \\ u(2,t) =0 \qquad\mbox{for} \quad 0 \le t \le 1 . \end{split}$
For the first illustration, we use the step sizes Δx = h = 0.2 and Δt = k = 0.02 and α = 1, so the ratio is r = 0.5. the grid will be n=12 columns wide by m=11 rows high. In this case, we have
$u_{i,j+1} = \frac{1}{2} \left( u_{i-1,j} + u_{i+1,j} \right) .$
This formula is stable for r=0. and can be used successfully to generate reasonably accurate approximation to u(x,t).

For our second illustration, we use the step sizes Δx = h = 0.2 and Δt = k = 0.05, so the ratio becomes r = 1.25. In this case, we get the forward-difference scheme:

$u_{i,j+1} = =1.25 \left( u_{i-1,j} + u_{i+1,j} \right) - 1.5\,u_{i,j} .$
Since r > = 1.25 > 0.5, the above recurrence is unstable, and errors committed at one row will be magnified in successive rows.    ■

Program (Forward-difference method for the heat equation): to approximate the solution of $$u_t = \alpha^2 u_{xx}$$ over $$R = \{\,(x,t) \,:\, 0 \le x \le \ell , \ 0 \le t \le b$$ with u(x,0) = f(x), for 0 ≤ x ≤ ℓ, and u(0,t) = c0, u(ℓ,t) = cl, for 0 ≤ tb.


function U=forward(f,c0,cl,a,b,c,n,m)
% Input  -- f=u(x,0) as a string 'f'
%        -- c0=u(0,t) and cl=u(a,t)
%        -- a and b right end points of [0,a] and [0,b]
%        -- c=alpha^2 the thermal diffusivity in heat equation
%        -- n and m number of grid points over [0,a] and [0,b]
% Output -- U solution matrix;
% Initialize parameters and U
h=a/(n-1);
k=b/(m-1);
r=c*k/h^2;
s=1-2*r;
U=zeros(n,m);
% Boundary conditions
U(1,1:m)=c0;
U(n,1:m)=cl;
% Generate the first row
U(2:n-1,1)=feval(f,h:h:(n-2)*h)';
% Generate remaining rows of U
for j=2:m
for j=2:n-1
U(i,j)=s*U(i,j-1)+r*(U(i-1,j-1)+U(i+1,j-1));
end
end
U=U';


The Crank--Nicholson Method

An implicit finite difference scheme, invented in 1947 by John Crank (1916--2006) and Phyllis Nicholson (1917--1968), is based on numerical approximations for solutions of heat equation at the point (x,t+k/2) and that lies between the rows in the grid. Specifically, the approximation used for ut(x,t+k/2) is obtained from the central-difference formula:

$u_{t} \left( x, t + \frac{k}{2} \right) = \frac{u(x,t+k)-u(x,t)}{k} + O\left( k^2 \right) .$
The approximation used for the second derivative uxx(x,t+k/2) is the average of the approximations uxx(x,t) and uxx(x,t+k), which has an accuracy of the order O(h²):
$u_{xx} \left( x, t + \frac{k}{2} \right) = \frac{1}{2\,h^2} \left[ u(x-h,t+k) -2\,u(x,t+k) + u(x+h,t+k) + u(x-h,t) -2\,u(x,t) + u(x+h,t) \right] + O\left( k^2 \right) .$
In a fashion similar to the previous derivation, we neglect the error terms O(k²) and O(h²) and substitute the remaining terms into the heat equation. Then employing the notation $$u_{i,j} = u(x_i , t_j )$$ will produce the finite difference scheme
$\frac{u_{i,j+1} - u_{i,j}}{k} = \alpha^2 \frac{u_{i-1,j+1} -2\,u_{i,j+1} + u_{i+1,j+1} + u_{i-1,j} -2\,u_{i,j} + u_{i+1,j}}{h^2} .$
We also substitute $$r=\alpha^2 k/h^2$$ into the above formula. However, this time we must solve for the three "et to be computed" values ui-1,j+1, ui,j+1, and ui+1,j+1. This is accomplished by placing them all on the left side of the equation. then rearrangement of the terms in the above difference equation results in the following Crank--Nicholson algorithm
$-r\,u_{i-1,j+1} + \left( 2 + 2\,r \right) u_{i,j+1} -r\,u_{i+1,j+1} = \left( 2 - 2\,r \right) u_{i,j} + r \left( u_{i-1,j} + u_{i+1,j} \right) ,$
for i = 2,3,...,n-1. The terms on the right-hand side are all known. Hence the above equations form a tridiagonal linear system A X = B.
$\begin{bmatrix} 2+ 2r &-r& 0& \cdots &0& 0 \\ -r& 2+2r&-r& 0& \cdots & 0 \\ &&\ddots&&& \\ &&&&-r&2+2r& \end{bmatrix} \begin{bmatrix} u_{2,j+1} \\ u_{3,j+1} \\ \vdots \\ u_{n-1,j+1} \end{bmatrix} = \begin{bmatrix} (2-2r)\,u_{2,j} + r u_{3,j} \\ \end{bmatrix} .$
Solving such tridiagonal systems was presented previously.
l0 = Graphics[{Thick, Line[{{-4, 0}, {4, 0}}]}];
l1 = Graphics[{Thick, Line[{{-4, 2}, {4, 2}}]}];
l2 = Graphics[{Thick, Line[{{0, 0}, {0, 2}}]}];
d0 = Graphics[{Red, Disk[{0, 0}, 0.15]}];
d1 = Graphics[{Red, Disk[{4, 0}, 0.15]}];
d2 = Graphics[{Red, Disk[{-4, 0}, 0.15]}];
d3 = Graphics[{Red, Disk[{0, 2}, 0.15]}];
d4 = Graphics[{Red, Disk[{4, 2}, 0.15]}];
d5 = Graphics[{Red, Disk[{-4, 2}, 0.15]}];
r1 = Graphics[ Text[Style["(2-2r)\!$$\*SubscriptBox[\(u$$, $$i,j$$]\)", Medium], {0, -0.5}]];
r2 = Graphics[Text[ Style["r \!$$\*SubscriptBox[\(u$$, $$i+1,j$$]\)", Medium], {4.4, -0.3}]];
r3 = Graphics[Text[ Style["r \!$$\*SubscriptBox[\(u$$, $$i-1,j$$]\)", Medium], {-4.5, -0.3}]];
r4 = Graphics[Text[ Style["(2+2r)\!$$\*SubscriptBox[\(u$$, $$i,j+1$$]\)", Medium], {0, 2.4}]];
r5 = Graphics[Text[ Style["r \!$$\*SubscriptBox[\(u$$, $$i-1,j+1$$]\)", Medium], {-4.5, 2.5}]];
r6 = Graphics[Text[ Style["r \!$$\*SubscriptBox[\(u$$, $$i+1,j+1$$]\)", Medium], {4.5, 2.5}]];
Show[l0, l1, l2, d0, d1, d2, d3, d4, d5, r1, r2, r3, r4, r5, r6]

Program (Crank--Nicholson method for heat equation): to approximate the solution of $$u_t = \alpha^2 u_{xx}$$ over $$R = \{\,(x,t) \,:\, 0 \le x \le \ell , \ 0 \le t \le b$$ with u(x,0) = f(x), for 0 ≤ x ≤ ℓ, and u(0,t) = c0, u(ℓ,t) = cl, for 0 ≤ tb.


function U=crnich(f,c0,cl,a,b,c,n,m)
% Input  -- f=u(x,0) as a string 'f'
%        -- c0=u(0,t) and cl=u(a,t)
%        -- a and b right end points of [0,a] and [0,b]
%        -- c=alpha^2 the thermal diffusivity in heat equation
%        -- n and m number of grid points over [0,a] and [0,b]
% Output -- U solution matrix;
% Initialize parameters and U
h=a/(n-1);
k=b/(m-1);
r=c*k/h^2;
s1=2+2/r;
s2=2/r-2;
U=zeros(n,m);
% Boundary conditions
U(1,1:m)=c0;
U(n,1:m)=cl;
% Generate the first row
U(2:n-1,1)=feval(f,h:h:(n-2)*h)';
% Form the diagonal and off-diagonal elements of A and
% the constant vector B and solve trigiagonal system AX=B
Vd(1,1:n)=s1*ones(1,n);
Vd(n)=1;
Vd(1)=1;
Va=-ones(1,n-1);
Va(n-1)=0;
Vc=-ones(1,n-1);
Vc(1)=0;
Vb(1)=c1;
Vb(n)=c2;
for j=2:m
for i=2:n-1
Vb(i)=U(i-1,j-1)+U(i+1,j-1)+s2*U(i,j-1);
end
X=trisys(Va,Vd,Vc,Vb);
U(1:n,j)=X';
end
U=U'


Example: Consider the initial boundary value problem

$u_t (x,t) = u_{xx} (x,t) \qquad \mbox{for } 0 < x < 2, \quad 0 < t < 1,$
with the initial condition
$u(x,0) = f(x) = \sin (\pi x) + 3\,\sin (5\pi x) \qquad \mbox{for } t=0 \quad\mbox{and}\quad 0 \le x \le 2,$
and the boundary conditions
$\begin{split} u(0,t) =0 \qquad\mbox{for} \quad 0 \le t \le 1, \\ u(2,t) =0 \qquad\mbox{for} \quad 0 \le t \le 1 . \end{split}$
For simplicity, we use the step sizes Δx = h = 0.1 and Δt = k = 0.01 so that the ratio is r = 1. The grid will be n = 21 columns wide by m = 11 rows high.    ■

Other Heat Transfer Equations

Nonlinear Diffusion Equations

Example: Consider the coupled Burgers equations

$\begin{split} u_t - u_{xx} -2u\,u_x + \left( u\,v \right)_x &= 0 , \\ v_t - v_{xx} - 2v\,v_x + \left( u\,v \right)_x &= 0 , \end{split}$
subject to the initial condition
$u(x,0) = \sin x , \qquad v(x,0) = \sin x .$
In order to solve the given system using the power series method, the solutions u and v are considered as
\begin{align*} u(x,t) &= \sum_{i\ge 0} \sum_{j\ge 0} a_{i,j} x^i t^j , \\ v(x,t) &= \sum_{i\ge 0} \sum_{j\ge 0} b_{i,j} x^i t^j . \end{align*}
Upon substitution these series into the given system of differential equations, we get the recurrence:
\begin{align*} a_{i, j+1} &= \frac{1}{j+1} \left[ \left( i+1 \right) \left( i+2 \right) a_{i+2, j} + \sum_{r=0}^i \sum_{s=0}^j 2 \left( r+1 \right) a_{i-r, j-s} a_{r+1, j} \right. \\ & \quad + \left. \sum_{r=0}^i \sum_{s=0}^j \left( r+1 \right) a_{i-r, j-s} b_{r+1, s} - \sum_{r=0}^i \sum_{s=0}^j \left( r+1 \right) a_{r+1,s} b_{i-r, j-s} \right] , \\ b_{i, j+1} &= \frac{1}{j+1} \left[ \left( i+1 \right) \left( i+2 \right) b_{i+2, j} + \sum_{r=0}^i \sum_{s=0}^j 2 \left( r+1 \right) b_{i-r, j-s} b_{r+1, j} \right. \\ & \quad + \left. \sum_{r=0}^i \sum_{s=0}^j \left( r+1 \right) a_{i-r, j-s} b_{r+1, s} - \sum_{r=0}^i \sum_{s=0}^j - \sum_{r=0}^i \sum_{s=0}^j \left( r+1 \right) a_{r+1,s} b_{i-r, j-s} \right] . \end{align*}
Then the desired coefficients are obtained by repeated application of the recurrence relations. Using Mathematica program to solve the preceding recurrence system, we obtain the following polynomials approximations:
\begin{align*} u(x,t) &= x - xt - \frac{xt^3}{3} - \frac{x^3}{6} + \frac{x^3 t}{6} + \frac{2x^3 t^3}{9} + \cdots , \\ v(x,t) &= x - xt - \frac{xt^3}{3} - \frac{x^3}{6} + \frac{x^3 t}{6} + \frac{2x^3 t^3}{9} + \cdots . \end{align*}
■

Example: Consider the coupled Burgers equations    ■

1. Crank, J. and Nicolson, P. (1947). "A practical method for numerical evaluation of solutions of partial differential equations of the heat conduction type," Proceedings of the Cambridge Philosophical Society, Vol. 43, No. 1, pp. 50--67. doi:10.1007/BF02127704
2. Mark Davis, Finite Difference Methods, Imperial College London
3. Stability of Finite Difference Methods, http://web.mit.edu/16.90/BackUp/www/pdfs/Chapter14.pdf