### 3.5 HW5

3.5.1 Part a
3.5.2 Part (b)
3.5.3 Part (c)
3.5.4 Part (d)
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#### 3.5.1 Part a

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$$\frac{\partial ^{2}u}{\partial x^{2}}=u_{0}\frac{\partial ^{2}\bar{u}}{\partial \bar{x}^{2}}\frac{1}{l_{0}^{2}} \tag{2}$$ 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 $$\bar{u}\equiv \bar{u}\left ( \bar{t},\bar{x}\right )$$.

#### 3.5.2 Part (b)

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$$u\equiv f\left ( x,t,\upsilon ,u_{0}\right ) \tag{1}$$ 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$$\left [ u\right ] =\left [ x^{a}t^{b}\upsilon ^{c}u_{0}^{d}\right ] \tag{2}$$

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 $$a=1 \tag{2}$$ And obtain$$\bar{u}\equiv f\left ( \frac{x}{\sqrt{\upsilon t}}\right ) \tag{3}$$ In the above $$\frac{x}{\sqrt{\upsilon t}}$$ is now non-dimensional quantity, which we call, the similarity variable $$\xi =\frac{x}{\sqrt{\upsilon t}} \tag{4}$$ 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.

#### 3.5.3 Part (c)

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$f\left ( \xi \right ) =c_{2}+c_{1}\int _{0}^{\xi }e^{\frac{-s^{2}}{4}}ds$

#### 3.5.4 Part (d)

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 $$u\left ( x,t\right ) =\frac{C}{2}+\frac{C}{2}\operatorname{erf}\left ( \frac{x}{\sqrt{4\upsilon t}}\right ) \tag{1}$$ 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 $$u\left ( x,t\right )$$$$u\left ( x,t\right ) =c_{3}\operatorname{erf}\left ( \frac{x}{\sqrt{4\upsilon t}}\right ) +c_{2} \tag{2}$$ Comparing (1) and (2), we see they are the same. Constants of integration are arbitrary and can be found from initial conditions.