Ver el código de «Approximation of the heat equation: Fourier method»

== Introduction ==
Let <math>{\bf u(x_1,x_2,x_3,t)}</math> be the temperature at the point <math>(x_1,x_2,x_3)\in </math> bar, and time <math>t>0</math>.

<gallery>
Bar.jpg
</gallery>

Assume that the cross section is so small that we can consider the bar as an unidimensional object in the interval <math>x\in [0,L]</math>,
<math>
u(x_1,x_2,x_3,t)=u(x_1,t)=u(x,t).
</math>

If we assume that the extreme are at zero temperature, the system of equations for u is given by
<math>
\left\{ \begin{array}{l}
u_t-u_{xx}=0, \qquad x\in(0,L), \qquad t>0, \\
u(0,t)=0, \qquad t>0, \\
u(L,t)=0, \qquad t>0, \\
u(x,0)=u^0(x), \qquad x\in(0,L).
\end{array} \right.
</math>
where <math>u_0(x)</math> is a function that describes the initial temperature of the bar.

We describe below the Fourier method to approximate the solutions of this system.

== Fourier method ==

The main point is to observe that, if <math>\varphi(x)</math> is solution of the eigenvalue problem

<math>
\left\{ \begin{array}{l}
\varphi''(x)+\lambda \varphi(x)=0, \\
\varphi(0)=0, \quad \varphi(L)=0, \end{array} \right.
</math>

for some <math>\lambda</math>, then

<math>
u(x,t)=\varphi(x) e^{-\lambda t}
</math>

is a solution of the heat equation <math>u_t-u_{xx}=0</math> and the boundary conditions <math>u(0,t)=u(L,t)=0.</math>

The eigenvalues of the above eigenvalue problem are <math>\lambda_k=\frac{k^2 \pi^2}{L^2}</math> with <math>k=1,2,3,...</math> and the associated eigenfunctions are

<math>
\varphi_k=\sin(\frac{k\pi x}{L}).
</math>

Therefore, all these functions are solutions of the heat equation and boundary conditions

<math>
u_k (x,t)= e^{-k^2\pi^2 /L^2 t}\sin(\frac{k\pi x}{L})
</math>

Thus, we only have to deal with the initial conditions. We consider two different cases:

==The initial condition is a linear combination of eigenfunctions==

We assume that we can write the initial condition <math>u_0(x)</math> as

<math>
u^0(x)=\sum_{k=1}^N c_k \sin(\frac{k\pi x}{L}),
</math>

for some coefficients <math>c_k \in \mathbb{R}</math>. Then

<math>
u(x,t)=\sum_{k=1}^N c_k e^{-k^2\pi^2/L^2t} \sin(\frac{k\pi x}{L}),
</math>

is a solution of the full system: heat equation, boundary conditions and initial conditions.

'''Example''' 

Consider the problem

<math>
\left\{   \begin{array}{lll}
u_t-u_{xx}=0, \;\; &x \in (0,1), \;\; t\in (0,3], \\ u(0, t)= 0, \hspace{0.3cm} u(1,t)=0, \;\; & t \in (0, 3], \\ u(x,0)= 2\sin(\pi x) -4\sin(3\pi x), \;\; & x \in [0,2],
\end{array}   \right.
</math>

In this case <math>c_1=2</math> and <math>c_3=-4</math>. The solution is:

<math>
u(x,t)= 2 e^{-\pi^2 t} \sin(\pi x) -4 e^{-3^2\pi^2 t} \sin(3\pi x) ,
</math>

==  MATLAB code  ==
{{matlab|codigo=
% Draw solutions of heat equation
clear all
dx=1/100; % space mesh step
x=0:dx:1; %space mesh
dt=1/100; % time mesh step
t=0:dt:2; % time mesh
% Fourier coefficients of u0
ck=[2, 0, -4];

% meshgrid
[xx,tt]=meshgrid(x,t);
% first term:
u=ck(1)*exp(-pi^2*tt).*sin(pi*xx);
for k=2:length(ck)
    % add the k-term
   u=u+ck(k)*exp(-k^2*pi^2*tt).*sin(k*pi*xx);
end
figure(2)
surf(xx,tt,u)
}}

==The initial condition is not a finite linear combination of eigenfunctions==

In this case, the main idea is to approximate the initial condition <math>u_0(x)</math> by

<math>
u^0(x) \sim \sum_{k=1}^N c_k \sin(\frac{k\pi x}{L}),
</math>

for some coefficients <math>c_k \in \mathbb{R}</math>. Then

<math>
u(x,t)\sim \sum_{k=1}^N c_k e^{-k^2\pi^2/L^2t} \sin(\frac{k\pi x}{L}),
</math>

is an approximated solution of the full system: heat equation, boundary conditions and initial conditions.

As we know, any <math>L^2(0,1)</math>-function can be approximated with a finite number of terms of the Fourier series and this allows to find approximations of heat solutions.

<gallery>
Eig4.jpg|Initial condition and approximation
Heat4.jpg|Approximate solution
</gallery>

==Example: ==

Consider the problem

<math>
\left\{   \begin{array}{lll}
u_t-u_{xx}=0, \;\; &x \in (0,2), \;\; t\in (0,3], \\ u(0, t)= 0, \hspace{0.3cm} u(2,t)=0, \;\; & t \in (0, 3], \\ u(x,0)= e^{-k(x-1)^2}, \;\; & x \in [0,2],
\end{array}   \right.
</math>

We solve it by the Fourier method

'''Step 1:''' Solve the associated eigenvalue problem

The eigenvalue problem is

<math>
\left\{ \begin{array}{ll}
\varphi''+ \lambda \varphi =0, \hspace{0.4cm} x \in (0,2), \\ \varphi(0)=\varphi(2)=0.
\end{array}   \right.
</math>

The eigenvalues and eigenfunctions are

<math>
\mu_k= \left( \frac{k\pi}{2} \right)^2 , \;\;\; \varphi_k(x)=\sin \frac{k \pi x}{2}, \;\; k=1,2, \cdot\cdot \cdot .
</math>

'''Step 2:''' Approximate the initial data by its projection on a finite dimensional subspace 

Choose Q and take

<math>
e^{-k(x-1)^2} \approx \sum_{k=1}^{Q} {c_k} \sin \frac{k \pi x}{2},
</math>

where

<math>
{c_k=\frac{\int_0^1 e^{-k(x-1)^2} \sin \frac{k \pi x}{2}dx}{\int_0^1  \sin^2 \frac{k \pi x}{2} dx}.}
</math>

'''Step 3:''' State the approximate problem for w

Approximate problem:

<math>
\left\{   \begin{array}{lll}
w_t-w_{xx}=0, \;\; & x \in (0,2), \;\; t\in (0,3], \\ w(0, t)= 0, \hspace{0.3cm} w(2,t)=0, \;\; &t \in (0, 3], \\ w(x,0)= \sum_{k=1}^{Q} {c_k} \sin \frac{k \pi x}{2}, \;\; & x \in [0,2],
\end{array}   \right.
</math>

If we assume that the solution is of the form

<math>
w_Q(x,t)=\sum_{k=1}^{Q} {T_k(t)}\sin \frac{k \pi x}{2},
</math>

then substituting in the heat equation we obtain an equation for each Fourier coefficient.

<math>
\left\{ \begin{array}{l} 
T'_k(t) + \frac{k^2 \pi^2}{4} T_k(t)=0, \\
T_k(0)={c_k}, \end{array} \right.
 \;\; k=1,2, \cdot \cdot \cdot, Q.
</math>

The solution of this equation is:

<math>
T_k(t)=c_k e^{-\frac{k^2 \pi^2}{4}  t}, \hspace{0.4cm} t \in [0,3],
</math>

and therefore

<math>
w_Q(x,t)=\sum_{k=1}^{Q} c_k e^{-\frac{k^2 \pi^2}{4}  t}\sin \frac{k \pi x}{2},
</math>

==  MATLAB code  ==
{{matlab|codigo=
% Fourier series approximatin
clear all
% Interval
a=0; b=2;
% Define the initial data that we are going to approximate
u0=@(x) (1-2*abs(x-1));

% Number of Fourier coefficients M
M=5;

% space discretization
N=100; dx=(b-a)/N;
x=a:dx:b;

% Compute the Fourier coefficients  
aproxima=0;  % aproximate function
for k=1:M
   p=sin(k*pi*x/2); % eigenfunction
   c(k)=trapz(x,u0(x).*p)/trapz(x,p.^2);
   aproxima=aproxima+c(k)*p;
end

% Draw the function and the approximation
figure(1)
plot(x,[aproxima;u0(x)])

% Draw solutions of heat equation
dt=1/100; % time mesh step
t=0:dt:2; % time mesh

[xx,tt]=meshgrid(x,t);
u=c(1)*exp(-pi^2/4*tt).*sin(pi*xx/2);
for k=2:M
   u=u+c(k)*exp(-k^2*pi^2/4*tt).*sin(k*pi*xx/2);
end
figure(2)
surf(xx,tt,u)
}}

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