To compute the elements of the diffusivity matrix, let us use the formula of Eqn. 7.31:
\[\begin{equation*}
K_{AB} = a(N_A,\,N_B) = \int_0^1 \left( \partial_x N_A \right) \left( \partial_x N_B \right) \, \mathrm{d}x
\end{equation*}\]
Computing the kernel requires taking the integral over the domain, which we can break up into an integral over each of the elements. Let us determine the partial derivative of each of the shape functions in each element:
\[\begin{align*}
\partial_x N_1(x) &=
\begin{cases}
-2 , & 0 \leq x\leq \frac{1}{2} \\
0 , & \frac{1}{2} \leq x \leq 1
\end{cases} \\
\partial_x N_2(x) &=
\begin{cases}
2 , & 0 \leq x\leq \frac{1}{2} \\
-2 , & \frac{1}{2} \leq x \leq 1
\end{cases} \\
\partial_x N_3(x) &=
\begin{cases}
0 , & 0 \leq x\leq \frac{1}{2} \\
2 , & \frac{1}{2} \leq x \leq 1
\end{cases} \\
\end{align*}\]
Computing Global Heat Supply Vetor (\(F_A\))
\(F_A\) is computed using Eqn. 7.30:
\[\begin{equation*}
F_A = \int_0^1 N_A \, h \, \mathrm{d}x + N_A(0) \, H_0 - a(N_A, \, N_{n+1}) \theta_1
\end{equation*}\]
For \(F_1\) therefore this is
\[\begin{equation*}
F_1 = \int_0^1 N_1 \, h \, \mathrm{d}x + N_1(0) \, H_0 - a(N_1, \, N_3) \theta_1
\end{equation*}\]
where \(a(N_1, \, N_3)\) is the diffusivity matrix element \(K_{13}\). Note that this could be computed using the Eqn. 7.31 however, since \(\partial_x N_1\) is 0 from 0.5 to 1, while \(\partial_x N_3\) is 0 from 0 to 0.5 it is clear that this will be zero. So \(F_1\) reduces to
\[\begin{equation*}
F_1 = \int_0^1 N_1 \, h \, \mathrm{d}x + N_1(0) \, H_0
\end{equation*} \]
where \(N_1(0)\) is defined to be 1, such that this further reduces to
\[\begin{equation*}
F_1 = \int_0^1 N_1 \, h \, \mathrm{d}x + \, H_0 \quad.
\end{equation*}\]
Writing the shape function in the integral explicity for the region it is non-zero (0 to 0.5), and using \(y\) as the dummy variable of integration, we can write this as
\[\begin{equation*}
F_1 = \int_0^{\frac{1}{2}} (1-2x) \, h(y) \, \mathrm{d}y + \, H_0 \quad.
\end{equation*}\]
Similarly for computing \(F_2\) let us write the equation from 7.30:
\[\begin{equation*}
F_2 = \int_0^1 N_2 \, h \, \mathrm{d}x + N_2(0) \, H_0 - a(N_2, \, N_{3}) \theta_1
\end{equation*}\]
First, let us note that \(N_2(0) = 0\) to remove that term:
\[\begin{equation*}
F_2 = \int_0^1 N_2 \, h \, \mathrm{d}x - a(N_2, \, N_{3}) \theta_1
\end{equation*}\]
Next we must evaluate \(a(N_2, \, N_{3})\):
\[\begin{equation*}
a(N_2, \, N_{3}) = \int_0^1 \left( \partial_x N_2 \right) \left( \partial_x N_3 \right) \, \mathrm{d}x = \int_\frac{1}{2}^1 \left( \partial_x N_2 \right) \left( \partial_x N_3 \right) \, \mathrm{d}x
\end{equation*}\]
since \(\partial_x N_3\) is 0 between 0 and 0.5. Substituting in the correct values, this yields
\[\begin{equation*}
a(N_2, \, N_{3}) = \int_\frac{1}{2}^1 \left(-2 \right) \left( 2 \right) \, \mathrm{d}x = -2
\end{equation*}\]
such that the force is
\[\begin{equation*}
F_2 = \int_0^1 N_2 \, h \, \mathrm{d}x + 2 \theta_1
\end{equation*}\]
Finally, substituting in the \(N_2\) shape function for the two elements this yields:
\[\begin{equation*}
F_2 = 2 \int_0^{\frac{1}{2}} x \, h \, \mathrm{d}x + 2\int_{\frac{1}{2}}^1 (1-x) \, h \, \mathrm{d}x + 2 \theta_1
\end{equation*}\]
