I have the following similarity solutions problem and solution:


$u_t = ku_{xx}$ for all $x > 0$, with $u_x (0, t) = 1$, $u(x, t) \to 0$ as $x \to \infty$, and $u(x, 0) = 0$ for $x > 0$.


Introducing $u′ = \dfrac{u}{u_0}$, $t′ = \dfrac{t}{t_0}$ and $x′ = \dfrac{x}{x_0}$, we will reduce the original PDE to

$$\dfrac{\partial{u′}}{\partial{t′}} = \dfrac{\partial^2{u'}}{\partial{x'}^2}$$

when choosing $x^2_0 = k t_0$.

The boundary condition $u_x(0, t) = 1$ will make us choose $x_0 = u_0$ so that $u'_{x'} (0, t') = 1$.

Choosing $t_0 = t$ we have

$$\sqrt{kt}u' \left( \dfrac{x}{\sqrt{kt}}, 1 \right) = \sqrt{kt} f(\eta), \eta = \dfrac{x}{\sqrt{kt}}$$

The PDE will reduce to the ODE

$$f''(\eta) + \dfrac{\eta}{2}f'(\eta) − \dfrac{1}{2}f(\eta) = 0$$

subject to $f'(0) = 0$ and $f(\eta) \to 0$ as $\eta \to \infty$.

I was told that, despite how this problem was posed, my instructor would provide us with the change of variables $u′ = \dfrac{u}{u_0}$, $t′ = \dfrac{t}{t_0}$ and $x′ = \dfrac{x}{x_0}$.

However, I'm struggling to understand the reasoning behind each of the steps taken in the solution. For all of these steps, I find myself asking "Why was this step taken/How did they know to take this step?". Can someone please take the time to annotate this solution with explanations of each step, so that I can understand this type of problem and be able to do it myself in the future? In other words, please walk me through the process, explaining why each step was taken.

For instance, why will the boundary condition $u_x(0, t) = 1$ make us choose $x_0 = u_0$? Where did $u'_{x'} (0, t') = 1$ come from?

And so on for the rest of the solution ...

I would really appreciate it if someone could please take the time to do this.

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    $\begingroup$ I found these notes to be quite useful. $\endgroup$ – Mattos Sep 16 '18 at 11:51
  • $\begingroup$ @Mattos Hi Mattos. Can you please check over the solution and comment on whether it is correct? As I said, I have a suspicion that there are errors. Thanks for the notes. I will read over them. Do you, by any chance, have any textbook recommendations for similarity solutions problems? I've struggled to find such a textbook. $\endgroup$ – The Pointer Sep 16 '18 at 11:58
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    $\begingroup$ I'm not sure how you managed to get a $t_{0}$ in the denominator in your first question. Using your approach, we have $$u(x,t) = u_{0}u'(x_{0}x', t_{0}t')$$ and so \begin{align} u_{t}&=u_{0}u'_{t} \\ &=u_{0}u'_{t'}t'_{t} \\ &=\frac{u_{0}}{t_{0}}u'_{t'} \end{align} Similarly, \begin{align} u_{x}&=u_{0}u'_{x} \\ &=u_{0}u'_{x'} x'_{x} \\ &=\frac{u_{0}}{x_{0}}u'_{x'} \\ \implies u_{xx}&=\frac{u_{0}}{x^{2}_{0}}u'_{x'x'} \end{align} Putting together yields $$\frac{u_{0}}{t_{0}} u'_{t'} = k \frac{u_{0}}{x^{2}_{0}} u'_{x'x'}$$ and hence we require $x^{2}_{0} = k t_{0}^{2}$. So the notes are correct. $\endgroup$ – Mattos Sep 16 '18 at 12:25
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    $\begingroup$ Use the derivation in my previous comment that shows $u_{x}=\frac{u_{0}}{x_{0}}u'_{x'}$. Then set $x = x_{0}x' = 0$. $\endgroup$ – Mattos Sep 16 '18 at 12:29
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    $\begingroup$ Yes, I meant $x_{0}^{2} = kt$. Using the scalings, you have $$t' = \frac{t}{t_{0}} = \frac{t}{t} = 1$$ Similarly, you have $$x' = \frac{x}{x_{0}} = \frac{x}{\sqrt{kt}}$$ and so $$u'(x', t') = u'\left(\frac{x}{\sqrt{kt}}, 1\right)$$ $\endgroup$ – Mattos Sep 17 '18 at 23:50

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