Question
The temperature profile $\theta(x, t)$ in a semi-infinite rod obeys the heat
diffusion equation
$$
\frac{\partial }{\partial t} \theta(x, t) =
\frac{\partial^2}{\partial x^2} \theta(x, t)
$$
With initial temperature distribution $\theta(x) = T_0$
At $t=0$, the temperature at the left end of the rod is changed instantaneously
to $\theta(0, t) = 0$ and kept at this temperature for all $t > 0$. Use the
Laplace transform method with respect to $t$ to find the solution to the
differential equation.
Working
The laplace of the left hand side :
\begin{equation}
\begin{aligned}
L\left(
\frac{\partial }{\partial t} \theta(x, t)
\right)
&= s \Theta(x, s) – \theta(x, 0) \\
&= s \Theta(x, s) – T_0
\end{aligned}
\end{equation}
And of the right hand side :
\begin{equation}
\begin{aligned}
L\left(
\frac{\partial^2}{\partial x^2} \theta(x, t)
\right)
= \frac{\partial^2}{\partial x^2} \Theta(x, s)
\end{aligned}
\end{equation}
Which gives
\begin{equation}
\begin{aligned}
s \Theta(x, s) – T_0
&=
\frac{\partial^2}{\partial x^2} \Theta(x, s)
\\
\frac{\partial^2}{\partial x^2} \Theta(x, s) – s \Theta(x, s) &= T_0 \\
\frac{\partial^2 \Theta}{\partial x^2} – s \Theta &= T_0
\end{aligned}
\end{equation}
I'm a bit unsure how to proceed from here though, in terms of what I can do to solve this.
What I'm considering is
\begin{equation}
\begin{aligned}
\Theta'' – s\Theta = T_0
\end{aligned}
\end{equation}
Then this has the particular integral of
\begin{equation}
\begin{aligned}
\Theta'' – s\Theta = 0
\end{aligned}
\end{equation}
With auxiliary equation
\begin{equation}
\begin{aligned}
m^2 – s & = 0 \\
m &= \pm \sqrt{s}
\end{aligned}
\end{equation}
And from here this is solved by considering cases for $s$ , those being
$s<0,s=0,s>0$.
For $s<0$, $m$ is imaginary and the solution for $\Theta$ is
\begin{equation}
\begin{aligned}
\Theta &= c_1 \cos(\sqrt{s}x) + c_2 \sin(\sqrt{s}x)
\end{aligned}
\end{equation}
But this must be wrong as I've not considered any separation of variables.
I'm unsure how to proceed, and whether what I've done is valid.
Posts I've viewed
Solving Heat Equation with Laplace Transform, I didn't really follow some of the notation here, such as:
I am setting $\mathcal{L}_t(u(x,t)) = U(x,s)|_s$
$\mathcal{L}(u'')=\mathcal{L}(\dot u) \rightarrow U''(x,s)=\frac s 4U(x,s)-\frac 14u(x,0)$
Problem with Heat Equation and Laplace
Transform,
this is more relating to Fourier transforms it seems. I couldn't follow it and
apply it to my confusion here at least.
Best Answer
You're on the right track. What you should have now is $\Theta_{xx} - s\Theta = -T_0$ (note: you have a sign error). For each fixed $s$ this is a constant-coefficient second-order linear ODE in $x$. The general solution is given by the sum of the general solution to the homogeneous equation and any particular solution to the inhomogeneous equation. There shouldn't be a need to consider $s < 0$, as the Laplace variable is usually assumed $>0$ by definition. So for $s>0$ the solution to the homogeneous equation is given as a sum of exponentials $c_1e^{\sqrt{s}x} + c_2e^{-\sqrt{s}x}$ and by inspection $\Theta_p(x) = \frac{T_0}{s}$ is a solution to the inhomogeneous equation. You can use the initial-value theorem for the Laplace transform ($f(0^+) = \lim_{s\to\infty} sF(s)$) to show that $c_1 = 0$. The boundary condition $\theta(0,t) = 0$ implies $\Theta(0,s) = 0$ for all $s>0$, which then implies $c_2 = -\frac{T_0}{s}$. Altogether from here we obtain the Laplace transform $$ \Theta(x,s) = -T_0\frac{e^{-\sqrt{s}x}}{s} + \frac{T_0}{s}.$$ We then invert this Laplace transform. The second term just gives a unit step function, and while the inverse Laplace transform of the first term can't be expressed in terms of elementary functions, we can express it using the rule $$F(s)/s = \mathcal{L}\left[\int_0^t f(\tau)~d\tau\right](s)$$ which gives $$ \theta(x,t) = -T_0\int_0^t \mathcal{L}^{-1}[e^{-\sqrt{s}x}](\tau)~d\tau + T_0u(t), $$ where $u(t)$ is the unit step function. The inverse Laplace transform in the integral is a bit messy: Wolfram gives $$ \mathcal{L}^{-1}[e^{-\sqrt{s}x}](\tau) = \frac{xe^{-\frac{x^2}{4\tau}}}{2\sqrt{\pi}\tau^{3/2}}. $$ Making the change of variables $\eta = x/2\sqrt{\tau}$, or equivalently $\tau = x^2/4\eta^2$, $d\tau = -\frac{x^2}{2\eta^3}d\eta$ gives $$ \int_0^t \mathcal{L}^{-1}[e^{-\sqrt{s}x}](\tau)~d\tau = \frac{2}{\sqrt{\pi}}\int_{x/2\sqrt{t}}^\infty e^{-\eta^2} ~d\eta = \operatorname{erfc}(x/2\sqrt{t}), $$ where $\operatorname{erfc}$ denotes the complementary error function. Therefore the solution comes out to $$ \theta(x,t) = -T_0\operatorname{erfc}(x/2\sqrt{t}) + T_0u(t). $$ To check that this is consistent with the initial and boundary conditions: at $t=0$, $\operatorname{erfc}(x/\sqrt{t}) = 0$ for all $x>0$, so $\theta(x,0) = T_0$ for all $x>0$, while at $x=0$ one can explicitly calculate the value of $\operatorname{erfc}(0)$ (it is half of the famous Gaussian integral) to find that $\theta(0,t) = -T_0 + T_0 = 0$.