solution
Let \(\bar{x}\) be the non-dimensional space coordinate and \(\bar{t}\) the non-dimensional time coordinate. Therefore we need \begin{align*} \bar{x} & =\frac{x}{l_{0}}\\ \bar{t} & =\frac{t}{t_{0}}\\ \bar{u} & =\frac{u}{u_{0}} \end{align*}
Where \(l_{0}\) is the physical characteristic length scale (even if this infinitely long domain, \(l_{0}\) is given) whose dimensions is \(\left [ L\right ] \) and \(t_{0}\) of dimensions \(\left [ T\right ] \) is the characteristic time scale and \(\bar{u}\left ( \bar{x},\bar{t}\right ) \) is the new dependent variable, and \(u_{0}\) characteristic value of \(u\) to scale against (typically this is the initial conditions) but this will cancel out. We now rewrite the PDE \(\frac{\partial u}{\partial t}=\upsilon \frac{\partial ^{2}u}{\partial x^{2}}\) in terms of the new dimensionless variables.\begin{align} \frac{\partial u}{\partial t} & =\frac{\partial u}{\partial \bar{u}}\frac{\partial \bar{u}}{\partial \bar{t}}\frac{\partial \bar{t}}{\partial t}\nonumber \\ & =u_{0}\frac{\partial \bar{u}}{\partial \bar{t}}\frac{1}{t_{0}} \tag{1} \end{align}
And\begin{align*} \frac{\partial u}{\partial x} & =\frac{\partial u}{\partial \bar{u}}\frac{\partial \bar{u}}{\partial \bar{x}}\frac{\partial \bar{x}}{\partial x}\\ & =u_{0}\frac{\partial \bar{u}}{\partial \bar{x}}\frac{1}{l_{0}} \end{align*}
And\begin{equation} \frac{\partial ^{2}u}{\partial x^{2}}=u_{0}\frac{\partial ^{2}\bar{u}}{\partial \bar{x}^{2}}\frac{1}{l_{0}^{2}} \tag{2} \end{equation} Substituting (1) and (2) into \(\frac{\partial u}{\partial t}=\upsilon \frac{\partial ^{2}u}{\partial x^{2}}\) gives\begin{align*} u_{0}\frac{\partial \bar{u}}{\partial \bar{t}}\frac{1}{t_{0}} & =\upsilon u_{0}\frac{\partial ^{2}\bar{u}}{\partial \bar{x}^{2}}\frac{1}{l_{0}^{2}}\\ \frac{\partial \bar{u}}{\partial \bar{t}} & =\left ( \upsilon \frac{t_{0}}{l_{0}^{2}}\right ) \frac{\partial ^{2}\bar{u}}{\partial \bar{x}^{2}} \end{align*}
The above is now non-dimensional. Since \(\upsilon \) has units \(\left [ \frac{L^{2}}{T}\right ] \) and \(\frac{t_{0}}{l_{0}^{2}}\) also has units \(\left [ \frac{T}{L^{2}}\right ] \), therefore the product \(\upsilon \frac{t_{0}}{l_{0}^{2}}\) is non-dimensional quantity.
If we choose \(t_{0}\) to have same magnitude (not units) as \(l_{0}^{2}\), i.e. \(t_{0}=l_{0}^{2}\), then \(\frac{t_{0}}{l_{0}^{2}}=1\) (with units \(\left [ \frac{T}{L^{2}}\right ] \)) and now we obtain the same PDE as the original, but it is non-dimensional. Where now \(\bar{u}\equiv \bar{u}\left ( \bar{t},\bar{x}\right ) \).
I Will use the Buckingham \(\pi \) theorem for finding expression for the solution in the form \(u\left ( x,t\right ) =f\left ( \xi \right ) \) where \(\xi \) is the similarity variable. Starting with \(\frac{\partial u}{\partial t}=\upsilon \frac{\partial ^{2}u}{\partial x^{2}}\), in this PDE, the diffusion substance is heat with units of Joule \(J\). Hence the concentration of heat, which is what \(u\) represents, will have units of \(\left [ u\right ] =\frac{J}{L^{3}}\). (heat per unit volume). From physics, we expect the solution \(u\left ( x,t\right ) \) to depend on \(x,t,\upsilon \) and initial conditions \(u_{0}\) as these are the only relevant quantities involved that can affect the diffusion. Therefore, by Buckingham theorem we say\begin{equation} u\equiv f\left ( x,t,\upsilon ,u_{0}\right ) \tag{1} \end{equation} We have one dependent quantity \(u\) and \(4\) independent quantities\(.\)The units of each of the above quantities is\begin{align*} \left [ u\right ] & =\frac{J}{L^{3}}\\ \left [ x\right ] & =L\\ \left [ t\right ] & =T\\ \left [ \upsilon \right ] & =\frac{L^{2}}{T}\\ \left [ u_{0}\right ] & =\frac{J}{L^{3}} \end{align*}
Hence using Buckingham theorem, we write\begin{equation} \left [ u\right ] =\left [ x^{a}t^{b}\upsilon ^{c}u_{0}^{d}\right ] \tag{2} \end{equation}
We now determine \(a,b,c,d\), by dimensional analysis. The above is\begin{align*} \frac{J}{L^{3}} & =L^{a}T^{b}\left ( \frac{L^{2}}{T}\right ) ^{c}\left ( \frac{J}{L^{3}}\right ) ^{d}\\ \left ( J\right ) \left ( L^{-3}\right ) & =\left ( L^{a+2c-3d}\right ) \left ( T^{b-c}\right ) \left ( J^{d}\right ) \end{align*}
Comparing powers of same units on both sides, we see that\begin{align*} d & =1\\ b-c & =0\\ a+2c-3d & =-3 \end{align*}
From second equation above, \(b=c\), hence third equation becomes\begin{align*} a+2c-3d & =-3\\ a+2c & =0 \end{align*}
Since \(d=1\). Hence\begin{align*} c & =-\frac{a}{2}\\ b & =-\frac{a}{2} \end{align*}
Therefore, now that we found all the powers, (we have one free power \(a\) which we can set to any value), then from equation (1)\begin{align*} \left [ u\right ] & =\left [ x^{a}t^{b}\upsilon ^{c}u_{0}^{d}\right ] \\ \frac{u}{u_{0}} & =\bar{u}=x^{a}t^{b}\upsilon ^{c} \end{align*}
Therefore \(\bar{u}\) is function of all the variables in the RHS. Let this function be \(f\) (This is the same as \(H\) in problem statement). Hence the above becomes\begin{align*} \bar{u} & =f\left ( x^{a}t^{-\frac{a}{2}}\upsilon ^{-\frac{a}{2}}\right ) \\ & =f\left ( \frac{x^{a}}{\upsilon ^{\frac{a}{2}}t^{\frac{a}{2}}}\right ) \end{align*}
Since \(a\) is free variable, we can choose \begin{equation} a=1 \tag{2} \end{equation} And obtain\begin{equation} \bar{u}\equiv f\left ( \frac{x}{\sqrt{\upsilon t}}\right ) \tag{3} \end{equation} In the above \(\frac{x}{\sqrt{\upsilon t}}\) is now non-dimensional quantity, which we call, the similarity variable \begin{equation} \xi =\frac{x}{\sqrt{\upsilon t}} \tag{4} \end{equation} Notice that another choice of \(a\) in (2), for example \(a=2\) would lead to \(\xi =\frac{x^{2}}{\upsilon t}\) instead of \(\xi =\frac{x}{\sqrt{\upsilon t}}\) but we will use the latter for the rest of the problem.
Using \(u\equiv f\left ( \xi \right ) \) where \(\xi =\frac{x}{\sqrt{\upsilon t}}\) then \begin{align*} \frac{\partial u}{\partial t} & =\frac{df}{d\xi }\frac{\partial \xi }{\partial t}\\ & =\frac{df}{d\xi }\frac{\partial }{\partial t}\left ( \frac{x}{\sqrt{\upsilon t}}\right ) \\ & =-\frac{1}{2}\frac{df}{d\xi }\left ( \frac{x}{\sqrt{\upsilon }t^{\frac{3}{2}}}\right ) \end{align*}
And\begin{align*} \frac{\partial u}{\partial x} & =\frac{df}{d\xi }\frac{\partial \xi }{\partial x}\\ & =\frac{df}{d\xi }\frac{\partial }{\partial x}\left ( \frac{x}{\sqrt{\upsilon t}}\right ) \\ & =\frac{df}{d\xi }\frac{1}{\sqrt{\upsilon t}} \end{align*}
And\begin{align*} \frac{\partial ^{2}u}{\partial x^{2}} & =\frac{\partial }{\partial x}\left ( \frac{df}{d\xi }\frac{1}{\sqrt{\upsilon t}}\right ) \\ & =\frac{1}{\sqrt{\upsilon t}}\frac{\partial }{\partial x}\left ( \frac{df}{d\xi }\right ) \\ & =\frac{1}{\sqrt{\upsilon t}}\left ( \frac{d^{2}f}{d\xi ^{2}}\frac{\partial \xi }{\partial x}\right ) \\ & =\frac{1}{\sqrt{\upsilon t}}\left ( \frac{d^{2}f}{d\xi ^{2}}\frac{1}{\sqrt{\upsilon t}}\right ) \\ & =\frac{1}{\upsilon t}\frac{d^{2}f}{d\xi ^{2}} \end{align*}
Hence the PDE \(\frac{\partial u}{\partial t}=\upsilon \frac{\partial ^{2}u}{\partial x^{2}}\) becomes\begin{align*} -\frac{1}{2}\frac{df}{d\xi }\left ( \frac{x}{\sqrt{\upsilon }t^{\frac{3}{2}}}\right ) & =\upsilon \frac{1}{\upsilon t}\frac{d^{2}f}{d\xi ^{2}}\\ \frac{1}{t}\frac{d^{2}f}{d\xi ^{2}}+\frac{1}{2}\frac{x}{\sqrt{\upsilon }t^{\frac{3}{2}}}\frac{df}{d\xi } & =0\\ \frac{d^{2}f}{d\xi ^{2}}+\frac{1}{2}\frac{x}{\sqrt{\upsilon t}}\frac{df}{d\xi } & =0 \end{align*}
But \(\frac{x}{\sqrt{\upsilon t}}=\xi \), hence we obtain the required ODE as\begin{align*} \frac{d^{2}f\left ( \xi \right ) }{d\xi ^{2}}+\frac{1}{2}\xi \frac{df\left ( \xi \right ) }{d\xi } & =0\\ f^{\prime \prime }+\frac{\xi }{2}f^{\prime } & =0 \end{align*}
We now solve the above ODE for \(f\left ( \xi \right ) \). Let \(f^{\prime }=z\), then the ODE becomes\[ z^{\prime }+\frac{\xi }{2}z=0 \] Integrating factor is \(\mu =e^{\int \frac{\xi }{2}d\xi }=e^{\frac{\xi ^{2}}{4}}\), hence\begin{align*} \frac{d}{d\xi }\left ( z\mu \right ) & =0\\ z\mu & =c_{1}\\ z & =c_{1}e^{\frac{-\xi ^{2}}{4}} \end{align*}
Therefore, since \(f^{\prime }=z\), then\[ f^{\prime }=c_{1}e^{\frac{-\xi ^{2}}{4}}\] Integrating gives\[ f\left ( \xi \right ) =c_{2}+c_{1}\int _{0}^{\xi }e^{\frac{-s^{2}}{4}}ds \]
For initial conditions of step function \[ u\left ( x,0\right ) =\left \{ \begin{array} [c]{cc}0 & x<0\\ C & x>0 \end{array} \right . \] The solution found in class was \begin{equation} u\left ( x,t\right ) =\frac{C}{2}+\frac{C}{2}\operatorname{erf}\left ( \frac{x}{\sqrt{4\upsilon t}}\right ) \tag{1} \end{equation} Where \(\operatorname{erf}\left ( \frac{x}{\sqrt{4\upsilon t}}\right ) =\frac{2}{\sqrt{\pi }}\int _{0}^{\frac{x}{\sqrt{4\upsilon t}}}e^{-z^{2}}dz\). The solution found in part (c) earlier is \[ f\left ( \xi \right ) =c_{1}\int _{0}^{\xi }e^{\frac{-s^{2}}{4}}ds+c_{2}\] Let \(s=\sqrt{4}z\), then \(\frac{ds}{dz}=\sqrt{4}\), when \(s=0,z=0\) and when \(s=\xi ,z=\frac{\xi }{\sqrt{4}}\), therefore the integral becomes\[ f\left ( \xi \right ) =c_{1}\sqrt{4}\int _{0}^{\frac{\xi }{\sqrt{4}}}e^{-z^{2}}dz+c_{2}\] But \(\frac{2}{\sqrt{\pi }}\int _{0}^{\frac{\xi }{\sqrt{4}}}e^{-z^{2}}dz=\operatorname{erf}\left ( \frac{\xi }{\sqrt{4}}\right ) \), hence \(\int _{0}^{\frac{\xi }{\sqrt{4}}}e^{-z^{2}}dz=\frac{\sqrt{\pi }}{2}\operatorname{erf}\left ( \frac{\xi }{\sqrt{4}}\right ) \) and the above becomes\begin{align*} f\left ( \xi \right ) & =c_{1}\sqrt{\pi }\operatorname{erf}\left ( \frac{\xi }{\sqrt{4}}\right ) +c_{2}\\ & =c_{3}\operatorname{erf}\left ( \frac{\xi }{\sqrt{4}}\right ) +c_{2} \end{align*}
Since \(\xi =\frac{x}{\sqrt{\upsilon t}}\), then above becomes, when converting back to \(u\left ( x,t\right ) \)\begin{equation} u\left ( x,t\right ) =c_{3}\operatorname{erf}\left ( \frac{x}{\sqrt{4\upsilon t}}\right ) +c_{2} \tag{2} \end{equation} Comparing (1) and (2), we see they are the same. Constants of integration are arbitrary and can be found from initial conditions.