hugogw / 2d-heat-equation

Simulation of the 2D diffusion equation of heat

2D-Heat-Equation

Simulation of the 2D diffusion equation of heat

In the first step, the aim is to numerically determine a solution of the one-dimensional Fourier heat equation using Fourier series with some approximations, and then we want to numerically solve this equation in 2D.

$\textbf{Solving the 1D heat equation}$

We have a 1D rod of length $L$, and we want to know the evolution of temperature $T$ as a function of time $t>0$ and position $x \in [0,L]$. The evolution of temperature with respect to position and time follows the 1D Fourier heat equation, which is given by:

$$\displaystyle (E) : ~~~~ \frac{\partial T}{\partial t} - D \frac{\partial^2 T}{\partial x^2} = 0$$

We make the assumption regarding the boundary conditions: $T(0,t) = T(L,t) = 0$, as well as regarding the initial condition : $T(x,0) = f(x) \forall x \in [0,L]$.

Fourier hypothesized that heat could be interpreted as small waves of heat. Building on this, we assume that the temperature $T$ can be interpreted as a Fourier series (to simplify matters, we can assume that $T(x,t)$ is a function $C^{\infty}$) :

$$\displaystyle T(x,t) = a_0 + \sum_{n=0}^{\infty} a_n(t) \cos \Big( \frac{\pi n}{L} x \Big) + b_n(t) \sin \Big( \frac{\pi n}{L} x \Big)$$

But we recall that $T(0,t) = T(L,t) = 0$, and only the odd part of the series satisfies this condition because $\displaystyle \sin \Big( \frac{\pi n}{L} 0 \Big) = \sin \Big( \frac{\pi n}{L} L \Big) = 0$ so we have :

$$\displaystyle T(x,t) = \sum_{n=0}^{\infty} b_n(t) \sin \Big( \frac{\pi n}{L} x \Big)$$

that we will inject into the equation $(E)$ :

$\displaystyle \frac{\partial T}{\partial t} - D \frac{\partial^2 T}{\partial x^2} = 0$

$\displaystyle \frac{\partial }{\partial t} \sum_{n=0}^{\infty} b_n(t) \sin \Big( \frac{\pi n}{L} x \Big) - D \frac{\partial^2 }{\partial x^2} \sum_{n=0}^{\infty} b_n(t) \sin \Big( \frac{\pi n}{L} x \Big) = 0$

$\displaystyle \sum_{n=0}^{\infty} b_n'(t) \sin \Big( \frac{\pi n}{L} x \Big) + D \frac{\pi^2 n^2}{L^2} b_n(t) \sin \Big( \frac{\pi n}{L} x \Big) = 0$

$\displaystyle \sum_{n=0}^{\infty} \Big( b_n'(t) + D\frac{\pi^2 n^2}{L^2} b_n(t) \Big) \sin \Big( \frac{\pi n}{L} x \Big) = 0$

For a fixed $n$, we have the differential equation $\displaystyle b_n'(t) + D\frac{\pi^2 n^2}{L^2} b_n(t) = 0$, admitting an infinite number of solutions of the form : $b_n(t) = b_n(0) e^{-\frac{D \pi^2 n^2}{L^2} t}$.

So, we have as a solution :

$$\displaystyle T(x,t) = \sum_{n=0}^{\infty} b_n(0) e^{-\frac{D \pi^2 n^2}{L^2} t} \sin \Big( \frac{\pi n}{L} x \Big)$$

We can then determine $b_n(0)$ using the initial condition : $\displaystyle T(x,0) = f(x) = \sum_{n=0}^{\infty} b_n(0) \sin \Big( \frac{\pi n}{L} x \Big)$

Since we are dealing with a Fourier series, we can calculate the coefficients $b_n(0)$ using the formulas for the Fourier coefficients :

$$\displaystyle b_n(0) = \frac{2}{L} \int_0^L f(x) \sin \Big( \frac{\pi n}{L} x \Big) dx$$

$\textbf{Solving numerically the 2D heat equation}$

First thing first, we define our parameters for the equation :

D = 150  #Thermal diffusion coefficient
Lx = 100  #length
Ly = 100  #length
time = 60
Nx = 120  #space nodes
Ny = 120  #space nodes

dx, dy = Lx / Nx, Ly / Ny  #space steps
dt = min(dx**2 / (4 * D), dy**2 / (4 * D))  #time step
Nt = int(time / dt)  #time nodes

T_max, T_min = 100, 0

center_x, center_y = Nx // 2, Ny // 2
R = 10  # Radius of the circle

I initialize my temperature by creating a matrix for T that will by update for each time step.

  T = np.zeros((Nx, Ny)) + 20  # Initial temperature of the plate : 15°C

Next, my goal is to fixed the initial conditions and boundary conditions. All the map is at 15K (because it's impossible to reach 0K). Thus, I want to create a circle at 100K at the center of the map and few elements at 100K too at the edge of the map :

T[(np.arange(Nx)[:, np.newaxis] - center_x)**2 + (np.arange(Ny) - center_y)**2 == R**2] = 100

# Boundary conditions
T[0, :] = np.linspace(T_min, T_max, Ny)
T[-1, :] = np.linspace(T_min, T_max, Ny)
T[:, 0] = np.linspace(T_min, T_max, Nx)
T[:, -1] = np.linspace(T_min, T_max, Nx)

We now come to the heart of the work. We aim to simulate the heat equation using the finite difference method. Therefore, we will discretize this equation as follows: :

$$\displaystyle \frac{\partial T}{\partial t} - D \bigg(\frac{\partial^2 T}{\partial x^2} + \frac{\partial^2 T}{\partial y^2} \bigg) = 0$$

Let's discretize $T \longrightarrow T_{i,j}^n$ where $n$ is the $n$-th time step, $i$ is the $i$-th node for $x$, and $j$ is the $j$-th node for $y$ and for the derivates :

$\displaystyle \frac{\partial T}{\partial t} \longrightarrow \frac{T_{i,j}^{n+1} - T_{i,j}^n}{dt}$

$\displaystyle \frac{\partial^2 T}{\partial x^2} \longrightarrow \frac{T_{i+1,j}^n - 2T_{i,j}^n + T_{i-1,j}^n}{dx^2}$

$\displaystyle \frac{\partial^2 T}{\partial y^2} \longrightarrow \frac{T_{i,j+1}^n - 2T_{i,j}^n + T_{i,j-1}^n}{dy^2}$

So, we find :

$$\displaystyle T_{i,j}^{n+1} = T_{i,j}^{n} + dt D \bigg(\frac{T_{i+1,j}^n - 2T_{i,j}^n + T_{i-1,j}^n}{dx^2} + \frac{T_{i,j+1}^n - 2T_{i,j}^n + T_{i,j-1}^n}{dy^2} \bigg)$$

This means that for each time step $dt$, we compute the new value $T_{i,j}^{n+1}$ using the equation above, and then proceed with another iteration. The term $T_{i,j}^{n+1}$ becomes the current term $T_{i,j}^{n}$, and we compute the new term $T_{i,j}^{n+1}$ based on the previous one, repeating the process for each time step. At each iteration, we plot $T_{i,j}^{n+1}$ to visualize the overall evolution of temperature.

counter = 0
while counter < time:
    for i in range(1, Nx - 1):
        for j in range(1, Ny - 1):
            d2x = (T[i - 1, j] - 2 * T[i, j] + T[i + 1, j]) / dx**2
            d2y = (T[i, j - 1] - 2 * T[i, j] + T[i, j + 1]) / dy**2
            T[i, j] = dt * D * (d2x + d2y) + T[i, j]

    counter += dt
    T[(np.arange(Nx)[:, np.newaxis] - center_x)**2 + (np.arange(Ny) - center_y)**2 == R**2] = 100
    pcm.set_array(T)
    ax.set_title("Time: {:.3f} s, Avg Temperature: {:.2f}".format(counter, np.average(T)))
    plt.pause(0.01)

plt.show()

It's a relatively short but efficient code. We can have fun changing the initial conditions, the colormap style, or the boundary conditions, the temperatures of certain points to create patterns and figures that are more or less beautiful or symmetrical. Your only limit is your imagination.