Problem 7.11

The global weak form of the dynamic heat equation is given in Eqn. 7.80 as

\[\begin{equation*} \int_0^1 \tilde{\theta} \, \partial_t \theta \: \mathrm{d}x = - \int_0^1 \alpha\, ( \partial_x\tilde{\theta} )\,( \partial_x\theta ) \: \mathrm{d}x \: + \: \int_0^1 \tilde{\theta}\, h\, \mathrm{d}x \: + \alpha\, \tilde{\theta} \, \partial_x \theta \, |_0^1 \quad. \end{equation*}\]

Writing the temperature field, \(\theta(x,\,t)\), and test function, \(\tilde{\theta}(x,\,t)\), as in Eqns 7.26 and 7.23,

\[\begin{align*} \theta(x,\,t) &= \sum_{B=1}^n d_B\,N_B(x) \:+\: \theta_1 N_{n+1}(x)\\ \tilde{\theta}(x,\,t) &= \sum_{A=1}^n c_A\,N_A(x) \end{align*}\]

we may then express the temporal gradients of these functions:

\[\begin{align*} \partial_t \theta(x,\,t) &= \sum_{B=1}^n \dot{d}_B\,N_B(x) \quad, \\ \partial_x \theta(x,\,t) &= \sum_{B=1}^n d_B\, \partial_x N_B(x) \:+\: \theta_1 \partial_x N_{n+1} \quad, \\ \partial_x \tilde{\theta}(x,\,t) &= \sum_{A=1}^n c_A\,\partial_x N_A \quad, \end{align*}\]

where a dot above a variable indicates a derivative with respect to time. Note here that only the weights \(c_A\) and \(d_B\) are time dependent, while only the shape functions \(N_i\) are spatially-varying. Finally, \(\theta_1\) has no time dependence due to the boundary condition 7.81,

\[\begin{equation*} \theta(1,\,t) = \theta_1 \quad. \end{equation*}\]

Substituting in these summations to the heat equation produces

\[\begin{align*} \int_0^1 \left(\sum_{A=1}^n c_A\,N_A \right) \, &\left( \sum_{B=1}^n \dot{d}_B\,N_B \right) \: \mathrm{d}x = \\ - &\int_0^1 \alpha\, \left( \sum_{A=1}^n c_A\,\partial_x N_A \right) \,\left(\sum_{B=1}^n d_B\, \partial_x N_B \:+\: \theta_1 \partial_x N_{n+1} \right) \: \mathrm{d}x \: \\ + \: &\int_0^1 \sum_{A=1}^n c_A\,N_A \, h\, \mathrm{d}x \\ \: + &\alpha\, \left[ \tilde{\theta} \, \partial_x \theta \,\right]_0^1 \quad. \end{align*}\]

Let us consider the different terms separately:

Left-hand side

The left-hand side can be re-ordered by removing the weights, which aren’t spatially dependent, from the integral:

\[\begin{align*} &\int_0^1 \left(\sum_{A=1}^n c_A\,N_A \right) \, \left( \sum_{B=1}^n \dot{d}_B\,N_B(x) \right) \: \mathrm{d}x \\ &= \sum_{A=1}^n \sum_{B=1}^n c_A\,\dot{d}_B\int_0^1 N_A(x) \,N_B(x) \: \mathrm{d}x \\ &= \sum_{A=1}^n \sum_{B=1}^n c_A \, M_{AB} \, \dot{d}_B \\ \end{align*}\]

Right-hand side - Term 1:

We can similarly rearrange

\[\begin{align*} - &\int_0^1 \alpha\, \left( \sum_{A=1}^n c_A\,\partial_x N_A \right) \,\left(\sum_{B=1}^n d_B\, \partial_x N_B \:+\: \theta_1 \partial_x N_{n+1} \right) \: \mathrm{d}x \\ = - &\int_0^1 \alpha\, \left( \sum_{A=1}^n c_A\,\partial_x N_A \right) \,\left(\sum_{B=1}^n d_B\, \partial_x N_B \right) \:+\: \alpha\, \left( \sum_{A=1}^n c_A\,\partial_x N_A \right) \theta_1 \partial_x N_{n+1} \: \mathrm{d}x \\ = - & \sum_{A=1}^n \sum_{B=1}^n c_A d_B \int_0^1 \alpha\, \left( \,\partial_x N_A \right) \,\left(\, \partial_x N_B \right) \: \mathrm{d}x \:-\: \sum_{A=1}^n c_A \int_0^1 \alpha\, \left( \,\partial_x N_A \right) \theta_1 \left( \partial_x N_{n+1} \right) \: \mathrm{d}x \end{align*}\]

With Eqn. 7.86 this can then be written as

\[\begin{align*} = - & \sum_{A=1}^n \sum_{B=1}^n c_A K_{AB}\, d_B \:-\: \sum_{A=1}^n c_A \,\theta_1 \int_0^1 \alpha\, \left( \,\partial_x N_A \right) \left( \partial_x N_{n+1} \right) \: \mathrm{d}x \end{align*}\]

Right-hand side - Term 2:

\[\begin{align*} &\int_0^1 \sum_{A=1}^n c_A\,N_A \, h\, \mathrm{d}x \\ = & \sum_{A=1}^n c_A \int_0^1 \,N_A \, h\, \mathrm{d}x \\ \end{align*}\]

Right-hand side - Term 3:

Upon imposing the constraint of 7.21 that

\[\begin{equation*} \tilde{\theta}(1) = 0\quad, \end{equation*}\]

this term reduces to

\[\begin{align*} \alpha\, \left[ \tilde{\theta} \, \partial_x \theta \,\right]_0^1 &= \alpha(1)\, \tilde{\theta}(1) \, \partial_x \theta(1) - \alpha(0)\, \tilde{\theta}(0) \, \partial_x \theta(0) \\ &= - \alpha(0) \, \tilde{\theta}(0) \, \partial_x \theta(0) \end{align*}\]

and using the boundary condition in Eqn 7.82 that

\[\begin{equation*} \partial_x\theta(0,\,t) = - H_0 \quad, \end{equation*}\]

this becomes

\[\begin{align*} &= \alpha(0) \, H_0 \, \tilde{\theta}(0) \\ &= \sum_{A=1}^n c_A \, N_A(0)\, \alpha(0) \, H_0 \end{align*}\]

Re-assembling the equation

We can now write our global heat equation as

\[\begin{align*} \sum_{A=1}^n \sum_{B=1}^n c_A \, M_{AB} \, \dot{d}_B = - & \sum_{A=1}^n \sum_{B=1}^n c_A K_{AB}\, d_B \:-\: \sum_{A=1}^n c_A \,\theta_1 \int_0^1 \alpha\, \left( \,\partial_x N_A \right) \left( \partial_x N_{n+1} \right) \: \mathrm{d}x \\ + & \sum_{A=1}^n c_A \int_0^1 \,N_A \, h\, \mathrm{d}x + \sum_{A=1}^n c_A \, N_A(0)\, \alpha(0) \, H_0 \quad. \end{align*}\]

By moving the term containing the diffusivity matrix to the LHS, we may write this as

\[\begin{align*} \sum_{A=1}^n c_A \left\{ \sum_{B=1}^n \, M_{AB} \, \dot{d}_B + \sum_{B=1}^n K_{AB}\, d_B \right\} = \sum_{A=1}^n c_A \Big\{&- \,\theta_1 \int_0^1 \alpha\, \left( \,\partial_x N_A \right) \left( \partial_x N_{n+1} \right) \: \mathrm{d}x \\ &+ \int_0^1 \,N_A \, h\, \mathrm{d}x + \, N_A(0)\, \alpha(0) \, H_0 \Big\} \quad. \end{align*}\]

which can be simplified to

\[\begin{align*} \sum_{A=1}^n c_A \left\{ \sum_{B=1}^n \, M_{AB} \, \dot{d}_B + \sum_{B=1}^n K_{AB}\, d_B \right\} = \sum_{A=1}^n c_A F_A \quad. \end{align*}\]

in which

\[\begin{equation*} F_A = - \,\theta_1 \int_0^1 \alpha\, \left( \,\partial_x N_A \right) \left( \partial_x N_{n+1} \right) \: \mathrm{d}x + \int_0^1 \,N_A \, h\, \mathrm{d}x + \, N_A(0)\, \alpha(0) \, H_0 \end{equation*}\]

Note here the minor differences between the definition above and Eqn 7.87

Probing the test function

For the weak form to hold, the equations must hold for any test function \(\tilde{\theta}\). Hence, we can probe the system of equations for different weak functions. This means that for any \(C_A\)

\[\begin{align*} \sum_{A=1}^n c_A \left\{ \sum_{B=1}^n \, M_{AB} \, \dot{d}_B + \sum_{B=1}^n K_{AB}\, d_B \right\} = \sum_{A=1}^n c_A F_A \quad. \end{align*}\]

must hold. To probe this, let us consider each case \(i = 1, n\) for which

\[\begin{equation*} C_A = \begin{cases} 1 & A=i \\ 0 & A\ne i \end{cases} \quad. \end{equation*}\]

For this to hold, it requires that

\[\begin{align*} \sum_{B=1}^n \, M_{AB} \, \dot{d}_B + \sum_{B=1}^n K_{AB}\, d_B = F_A \quad \end{align*}\]

for every \(A\).