Problem 7.1

First let us demonstrate that our integral form satisfies the heat equation. To do this, we need to take the second partial derivative of \(\theta\) with respect to \(x\). Since \(x\) is in the limit of the integral, we need to use Leibniz rule, which states:

\[ \frac{d}{dx} \left( \int_{a(x)}^{b(x)} f(x,\,y) \, \mathrm{d}y \right) = f(x,\, b(x)) \cdot \frac{d}{dx}b(x) - f(x,\, a(x)) \cdot \frac{d}{dx} a(x) + \int_{a(x)}^{b(x)} \frac{\partial}{\partial x} f(x,\,y) \, \mathrm{d}y \]

In our case we will only consider situations in which the integrand function is not a function of the variable, \(x\), with which we are taking the derivative, i.e., \(f(x,t) \rightarrow f(t)\) and \(\frac{d}{dx}f(x,t) = 0 \), so the rule simplifies to

\[ \frac{d}{dx} \left( \int_{a(x)}^{b(x)} f(y) \, \mathrm{d}y \right) = f(b(x)) \cdot \frac{d}{dx}b(x) - f(a(x)) \cdot \frac{d}{dx} a(x) \]

To show that our solution

\[ \theta(x) = \theta_1 + (1 - x)\, H_0 + \int_x^1 \int_0^y h(z) \: dz \, dy \]

satisfies the steady-state equation

\[ \partial_x^2 \theta + h = 0 \]

let us take the first derivative of the solution with respect to \(x\):

\[ \frac{d\theta}{dx} = - H_0 + \frac{d}{dx}\left[ \int_x^1 \int_0^y h(z) \: dz \, dy \right] \]

Using the simplified Leibniz rule above we identify \(a(x) = x\), \(b(x) = 1\) and \(f(y) = \int_0^y h(z) \: dz \) such that the result is

\[ \frac{d\theta}{dx} = - H_0 - \int_0^x h(z) \: \mathrm{d}z \]

Employing this a second time, in this case where \(b = x\), \(a = 0\) and \(f(y) = h(z)\) we get

\[ \frac{d^2\theta}{dx^2} = - h(x) \]

Hence, we see this satisfies our steady-state equation.