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My attempt

$$u_t = -v Ʊ'$$

$$u_x = Ʊ'$$

$$u_{xx} = Ʊ''$$

$$-v Ʊ' + Ʊ Ʊ' = \nu Ʊ''$$

$$-v + Ʊ = \nu \frac{Ʊ''}{Ʊ'}$$

$$-vƱ + \frac{Ʊ^2}{2} + C = \nu \ln Ʊ'$$

$$exp(-\frac{vƱ}{\nu} + \frac{Ʊ^2}{2\nu} + \frac{C}{\nu}) = Ʊ' = \frac{dƱ}{d\xi}$$

$$d\xi = exp(\frac{vƱ}{\nu} - \frac{Ʊ^2}{2\nu} - \frac{C}{\nu}) dƱ$$

But integrating the RHS integral, I get a very ugly looking function involving the Erf function

So I don't know if the way I calculated it is the right one.

Our Prof told us that while calculating the solution, we should use the assumptions that:

$$\lim_{\xi \rightarrow -\infty} Ʊ (\xi) = u_2$$

$$\lim_{\xi \rightarrow \infty} Ʊ (\xi) = u_1$$

where $u_2 \ge u_1 \ge 0$.

But I don't see how to use it here.

anon
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1 Answers1

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$$-v Ʊ' + Ʊ Ʊ' = \nu Ʊ''$$ Integration gives $( v,\nu$ constants): $$-v Ʊ + \dfrac 12Ʊ^2 = \nu Ʊ'+C$$ How comes you get: $$-v + Ʊ = \nu \frac{Ʊ''}{Ʊ'}$$ $$-vƱ + \frac{Ʊ^2}{2} + C = \nu \ln Ʊ'$$ $Ʊ$ is a function of $\xi$. Not a variable. $$\int (-v + Ʊ) d\xi =C-v\xi+ \int Ʊ d\xi$$ $$\int (-v + Ʊ) d\xi \ne -vƱ + \frac{Ʊ^2}{2} + C $$

user577215664
  • 40,943
  • So, I should get: $$C - v \xi + \int Ʊ d\xi = \nu \ln Ʊ' $$ What should I do next to solve for $Ʊ$? – anon Nov 05 '22 at 14:38
  • You cant integrate further start from the last line where I integrated the DE instead @anon $-v Ʊ + \dfrac 12Ʊ^2 = \nu Ʊ'+C$ – user577215664 Nov 05 '22 at 14:42
  • The DE is separable @anon maybe not easy to integrate but separable – user577215664 Nov 05 '22 at 16:34
  • $$-v Ʊ + \frac{1}{2} Ʊ^2 = \nu Ʊ' + C = \nu \frac{dƱ}{d\xi} + C$$

    $$-v Ʊ + \frac{1}{2} Ʊ^2 - C = \nu \frac{dƱ}{d\xi}$$

    $$d\xi = \nu \frac{dƱ}{-v Ʊ + \frac{1}{2} Ʊ^2 - C}$$

    $$\xi + D = \nu \dfrac{2\arctan\left(\frac{Ʊ-v}{\sqrt{-v^2-2C}}\right)}{\sqrt{-v^2-2C}}$$

    So now what's left is to solve for $Ʊ$

     – anon Nov 05 '22 at 16:38
  • Put everything on the left side then take tan function on both sides @anon – user577215664 Nov 05 '22 at 16:41
  • I get: $$Ʊ = \sqrt{-v^2 - 2C} \tan (\frac{\xi +D}{2 \nu} \sqrt{-v^2 - 2C})+v)$$ – anon Nov 05 '22 at 16:42
  • But I don't think that's what my Prof wanted. Because he told us to use

    $$\lim_{\xi \rightarrow -\infty} Ʊ (\xi) = u_2$$

    $$\lim_{\xi \rightarrow \infty} Ʊ (\xi) = u_1$$

    where $u_2 \ge u_1 \ge 0$. But $\tan x$ for $x \rightarrow \pm \infty$ is indeterminate

     – anon Nov 05 '22 at 16:43
  • Or maybe this shows that it must be true that $\sqrt{-v^2 - 2C} \in \mathbb{C} \backslash \mathbb{R}$? Because $$\lim_{x \rightarrow \infty} \tan (i x) = i$$ But I still don't think it would help us solve this problem – anon Nov 05 '22 at 16:47