Solve $u_t + u u_x = \nu u_{xx}$ using $u(t,x) = Ʊ (\xi)$ if $\xi = x - vt$ and $v > 0$

Question

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.

Answer 1

$$-v Ʊ' + Ʊ Ʊ' = \nu Ʊ''$$ 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. $$\int (-v + Ʊ) d\xi =C-v\xi+ \int Ʊ d\xi$$ $$\int (-v + Ʊ) d\xi \ne -vƱ + \frac{Ʊ^2}{2} + C $$