28 Problem set 3
In Chapter 8 we discussed how the zero-dimensional energy equation occurs in the form of \[ (\rho c_p V) \frac{\mathrm{d}T}{\mathrm{d}t} = E_{\text{in}}(t, T) - E_{\text{out}}(t, T). \] This is oversimplistic since in actuality, your temperature function, \(T\), should also depend on space. For instance, if you are interested in modelling \(T\) with a depth coordinate, then it would be \(T(z, t)\). In this case, we know that temperature would be governed by a second-order differential equation (in space) such as the kind that appears in (Equation 1.1).
In this problem set, we will make up a toy example of a differential equation that might be encountered in heat flow. Consider now (in non-dimensional coordinates): \[ \frac{\mathrm{\partial}T}{\mathrm{\partial}t} = \epsilon \frac{\mathrm{\partial}^2 T}{\mathrm{\partial}z^2} + 2 \frac{\mathrm{\partial}T}{\mathrm{\partial}z} + T = 0, \] which is determined for some function \(T(z, t)\). Here, the small parameter \(\epsilon > 0\), \(\epsilon \ll 1\), is linked to the heat diffusion.
We will study the steady-state version of the problem. So consider only \(T = T(z)\). We will make up some boundary conditions as well. \[ \begin{aligned} \epsilon T'' + 2T' + T &= 0, \\ T(0) &= 0, \\ T(1) &= 1. \end{aligned} \tag{28.1}\] The challenge is to study the above problem for small values of \(\epsilon\).
Q1. Conversion to first-order system
By using the procedure reviewed in Section 6.2, convert (Equation 28.1) to a first-order system of equations.
Q2. Numerical solutions
By adapting the code studied in lectures (script lecture12-SolvingBVPs), write a numerical code to solve (Equation 28.1) using Python’s built-in functions. Use your code to investigate the solution profiles for different values of \(\epsilon\).
Q3. Investigation of the boundary layer
Using your code, use the following command in order to investigate the maxima \((x_m, T_m)\) as \(\epsilon\) varies:
ind = np.argmax(Y[0])
print(ep, z[ind], Y[0, ind])You may want to consider, as an example, the values \(\epsilon = \{0.05, 0.1, 0.15, 0.2\}\) and fill the following table.
| \(\epsilon\) | \(x_m\) | \(T_m\) |
|---|---|---|
| 0.05 | ||
| 0.1 | ||
| 0.15 | ||
| 0.2 |
Create a plot of \((\epsilon, x_m)\) and discuss the observed trend and its implications.
Q4. Outer asymptotic solutions
Begin by setting \[ T = T_{\text{outer}} = T_0(z) + \epsilon T_1(z) + \epsilon^2 T_2(z) + \ldots \] Substitute the above expansion into the system and solve for the first two orders.
You may verify that the solution is given by
\[\begin{align} T_0 &= e^{1/2} e^{-z/2}, \\ T_1 &= -\frac{1}{8} e^{1/2} e^{-z/2} (z - 1). \end{align}\]
Q5. Inner asymptotic solutions
There will be a boundary layer near \(z = 0\). Set \(z = \epsilon^{\alpha} s\) and \(T(z) = U(s)\). Follow the same procedure, as in Chapter 8 in order to determine the correct choice of \(\alpha\) for the inner region. This choice should balance the two terms \(\epsilon T''\) and \(2T'\).
Q6. Matching and comparison
Expanding the inner solution as \(U = U_0(s) + \epsilon U_1(s) + \ldots\), write down the equation and boundary conditions that \(U_0\) must satisfy. You will notice that \(U_0\) is governed by a second-order differential equation and therefore needs two boundary conditions. One boundary condition comes from \(z = 0\), i.e. \[ U_0(0) = 0. \]
The other boundary condition is a matching condition: \[ \lim_{s \to \infty} U_0(s) = \lim_{z \to 0} T_0(z), \] which imposes that the inner solution, as it leaves the boundary layer, matches the outer solution, it tends into the inner region.
Solve for \(U_0\).