3.3 Notes on similarity solutions
The transient one dimensional heat equation may be represented: \begin{equation} \frac{\partial T}{\partial \tau}=\alpha \frac{\partial^2 T}{\partial X^2} \tag{3.80} \end{equation}
where \tau and X denote time and spatial coordinates, respectively. The temperature T is a function of time and space T=T(X,\tau) , and \alpha is the thermal diffusivity. (See appendix B in Numeriske Beregninger for a derivation)
Figure 42: Beam in the right half-space in one-dimension.
In Figure 42 a one-dimensional beam in the right half-space ( 0\leq X < \infty ) is illustrated. The beam has initially a temperature T_s , but at time \tau=0 , the temperature at the left end X=0 is abruptly set to T_0 , and kept constant thereafter.
We wish to compute the temperature distribution in the beam as a function of time \tau . The partial differential equation describing this problem is given by equation (3.80), and to model the time evolution of the temperature we provide the following initial condition: \begin{equation} T(X,\tau)=T_s,\ \tau < 0 \tag{3.81} \end{equation}
along with the boundary conditions which do not change in time: \begin{equation} T(0,\tau)=T_0,\ T(\infty,\tau)=T_s \tag{3.82} \end{equation}
Before we solve the problem numerically, we scale equation (3.80) by the introduction of the following dimensionless variables: \begin{equation} u=\frac{T-T_0}{T_s-T_0},\qquad x=\frac{X}{L},\qquad t=\frac{\tau \cdot \alpha}{L^2} \tag{3.83} \end{equation}
where L is a characteristic length. By substitution of the dimensionless variables in equation (3.83) in (3.80), we get the following: \begin{equation} \frac{\partial u}{\partial t}=\frac{\partial^2 u}{\partial x^2},\qquad 0 < x < \infty \tag{3.84} \end{equation} accompanied by the dimensionless initial condition: \begin{equation} u(x,t)=1,\quad t < 0 \tag{3.85} \end{equation} and dimensionless boundary conditions: \begin{equation} u(0,t)=0,\qquad u(\infty,t)=1 \tag{3.86} \end{equation}
The particular choice of the time scale in t (3.83) has been made to make the thermal diffusivity vanish and to present the governing partial on a canonical form (3.84) which has many analytical solutions and a wide range of applications. The dimensionless time in (3.83) is a dimensionless number, which is commonly referred to as the Fourier-number.
We will now try to transform the partial differential equation (3.84) with boundary conditions (3.85) to a simpler ordinary differential equations. We will do so by introducing some appropriate scales for the time and space coordinates: \begin{equation} \bar{x}=a\, x \qquad \text{and} \qquad \bar{t}=b\, t \tag{3.87} \end{equation}
where a and b are some positive constants. Substitution of equation (3.87) into equation (3.84) yields the following equation: \begin{equation} \frac{\partial u}{\partial \bar{t}}=\frac{a^2}{b}\frac{\partial^2u}{\partial \bar{x}^2} \tag{3.88} \end{equation}
We chose b=a^2 to bring the scaled equation (3.88) on the canonical, dimensionless form of equation (3.84) with the boundary conditions: \begin{equation} u(x,t)=u(\bar{x},\bar{t})=u(ax,a^2t),\qquad \text{with} \quad b=a^2 \tag{3.89} \end{equation}
For (3.89) to be independent of a > 0 , the solution u(x,t) has to be on the form: \begin{equation} u(x,t)=f\left( \frac{x}{\sqrt{t}}\right),\ g\left( \frac{x^2}{t}\right),\ \text{etc.} \tag{3.90} \end{equation}
and for convenience we choose the first alternative: \begin{equation} u(x,t)=f\left( \frac{x}{\sqrt{t}}\right) = f \left (\eta \right ) \tag{3.91} \end{equation}
where we have introduced a new similarity variable \eta defined as: \begin{equation} \eta=\frac{x}{2 \sqrt{t}} \tag{3.92} \end{equation}
and the factor 2 has been introduced to obtain a simpler end result only. By introducing the similar variable \eta in equation (3.91), we transform the solution (and the differential equation) from being a function of x and t , to only depend on one variable, namely \eta . A consequence of this transformation is that one profile u(\eta) , will define the solution for all x and t , i.e. the solutions will be similar and is denoted a similarity solution for that reason and (3.92) a similarity transformation.
The Transformation in equation (3.92) is often referred to as the Boltzmann-transformation.
The original PDE in equation (3.84) has been transformed to an ODE, which will become clearer from the following. We have introduced the following variables: \begin{equation} t=\frac{\tau\cdot\alpha}{L^2},\qquad \eta=\frac{x}{2 \sqrt{t}}=\frac{X}{2 \sqrt{\tau\alpha}} \tag{3.93} \end{equation}
Let us now solve equation (3.84) analytically and introduce: \begin{equation} u=f(\eta) \tag{3.94} \end{equation}
with the boundary conditions: \begin{equation} f(0)=0,\ f(\infty)=1 \tag{3.95} \end{equation}
Based on the mathematical representation of the solution in equation (3.94) we may express the partial derivatives occurring in equation (3.84) as: \begin{align*} \frac{\partial u}{\partial t} & =\frac{\partial u}{\partial\eta}\left( \frac{\partial n}{\partial t}\right)=f'(\eta)\cdot\left( -\frac{x}{4t\sqrt{t}}\right)=-f'(\eta)\frac{\eta}{2t} \\ \frac{\partial u}{\partial x}&=\frac{\partial u}{\partial\eta}\left( \frac{\partial\eta}{\partial x}\right)=f'(\eta)\frac{1}{2\sqrt{t}},\qquad \frac{\partial^2u}{\partial x^2}=\frac{\partial}{\partial x}\left( f'(\eta)\frac{1}{2\sqrt{t}}\right)=f''(\eta)\frac{1}{4t} \end{align*}
By substituting the expressions for \dfrac{\partial u}{\partial t} and \dfrac{\partial^2u}{\partial x^2} in equation (3.84) we transform the original PDE in equation (3.84) to an ODE in equation (3.96) as stated above: \begin{align*} f''(\eta)\frac{1}{4t}+f'(\eta)\frac{\eta}{2t}=0 \end{align*}
or equivalently: \begin{equation} f''(\eta)+2\eta f'(\eta)=0 \tag{3.96} \end{equation}
The ODE in equation (3.96) may be solved directly by integration. First, we rewrite equation (3.96) as: \begin{equation*} \frac{f''(\eta)}{f'(\eta)}=-2\eta\ \end{equation*} which may be integrated to yield: \begin{equation} \ln f'(\eta)=-\eta^2+\ln C_1 \tag{3.97} \end{equation} which may be simplified by and exponential transformation on both sides: \begin{equation*} f'(\eta)=C_1e^{-\eta^2} \end{equation*} and integrated once again to yield: \begin{equation} f(\eta)=C_1 \int_0^\eta e^{-t^2}\,dt \tag{3.98} \end{equation}
where we have used the boundary condition f(0)=0 from (3.95).
The integral in (3.98) is related with the error function: \begin{equation*} \int_0^x e^{-t^2}dt=\frac{\sqrt{\pi}}{2} \text{erf(x)} \end{equation*} where the error function erf(x) is defined as: \begin{equation} \frac{2}{\sqrt{\pi}}\int_0^x e^{-t^2}dt \tag{3.99} \end{equation}
Substitution of equation (3.99) in equation (3.98) yields: \begin{equation} f(\eta)=C_1\frac{\sqrt{\pi}}{2}\text{erf}(\eta ) \tag{3.100} \end{equation}
As the error function has the property \mathrm{erf(\eta)} \to 1 for \eta\to\infty we get: C_1=\dfrac{2}{\sqrt{\pi}} , and subsequent substitution of equation (3.100) yields:
In Figure 43 the error function \text{erf}(\eta) , which is a solution of the one-dimensional heat equation (3.84), is plotted along with the complementary error function \text{erfc}(\eta) defined as: \begin{equation} \text{erfc}(\eta)=1-\text{erf}(\eta)=\frac{2}{\sqrt{\pi}}\int_{\eta}^\infty e^{-t^2} \;dt \tag{3.103} \end{equation}
Figure 43: The error function is a similarity solution of the one-dimensional heat equation expressed by the similarity variable \eta .
