Find the error function of a value.

erf(0.76)

ans = 0.7175

Find the error function of the elements of a vector.

V = [-0.5 0 1 0.72];
erf(V)

ans = 1×4

   -0.5205         0    0.8427    0.6914

Find the error function of the elements of a matrix.

M = [0.29 -0.11; 3.1 -2.9];
erf(M)

ans = 2×2

    0.3183   -0.1236
    1.0000   -1.0000

The cumulative distribution function (CDF) of the normal, or Gaussian, distribution with standard deviation $$\sigma$$ and mean $$\mu$$ is

Note that for increased computational accuracy, you can rewrite the formula in terms of erfc . For details, see Tips.

Plot the CDF of the normal distribution with $$\mu =0$$ and $$\sigma =1$$.

x = -3:0.1:3;
y = (1/2)*(1+erf(x/sqrt(2)));
plot(x,y)
grid on
title('CDF of normal distribution with \mu = 0 and \sigma = 1')
xlabel('x')
ylabel('CDF')

Where $$u(x,t)$$ represents the temperature at position $$x$$ and time $$t$$, the heat equation is

where $$c$$ is a constant.

For a material with heat coefficient $$k$$, and for the initial condition $$u(x,0)=a$$ for $$x>b$$ and $$u(x,0)=0$$ elsewhere, the solution to the heat equation is

For k = 2, a = 5, and b = 1, plot the solution of the heat equation at times t = 0.1, 5, and 100.

x = -4:0.01:6;
t = [0.1 5 100];
a = 5;
k = 2;
b = 1;
figure(1)
hold on
for i = 1:3
    u(i,:) = (a/2)*(erf((x-b)/sqrt(4*k*t(i))));
    plot(x,u(i,:))
end
grid on
xlabel('x')
ylabel('Temperature')
legend('t = 0.1','t = 5','t = 100','Location','best')
title('Temperatures across material at t = 0.1, t = 5, and t = 100')