4
$\begingroup$

Consider the viscous Burgers’ equation $$u_t + uu_x = \nu u_{xx},\nu > 0.$$ Identify the exponents $n,m$ such that self-similar solutions of the form $u(x,t) = t^mf(xt^n)$ can be obtained. Write down the resulting ODE for the function $f$.

Letting $z=xt^n$ and using the chain rule I get the ODE $$mt^{m-1}f(xt^n)+nxt^{n+m-1}f'(xt^n)+t^{m+n}f(xt^n)f'(xt^n)-\nu t^{m+2n}f''(xt^n)=0,$$ and using $z=xt^n$ I get $$mt^{m-1}f+nzt^{m-1}f'+t^{2m+n}ff'-\nu t^{m+2n}f''=0.$$ Since I want an ODE in just $z$, I want the powers of $t$ to vanish. So I must have $m=1$, $n=-2$ which gives a power of $t^{-3}$ in the last term which is inconsistent. Can anyone see where I went wrong?

EDIT: Forgot a term of $f$ in the next last term, thanks to @mattos for pointing it out.

$\endgroup$
2
  • 1
    $\begingroup$ You should have an $f f'$ term. I get $$t^{m-1} (m f + n z f' + t^{m + n + 1} ff') = t^{m - 1} (\nu t^{2n + 1} f'')$$ and hence it should be $m, n = -1/2$. $\endgroup$ – mattos Apr 25 '20 at 8:20
  • $\begingroup$ Ah, didn't think of taking a power of $t$ outside, thanks! $\endgroup$ – user30523 Apr 25 '20 at 8:57
0
$\begingroup$

Let $u(x,t) = t^m f(xt^n)$ and $ z = x t^n$.

Following mattos's comment we get $$ \begin{split} u_t &= mt^{m-1}f+nxt^{n+m-1}f' \\ u_x &= t^{m+n}f'\\ u_{xx}&= t^{m+2n}f'' \end{split} $$ So the equation becomes $$\left( mt^{m-1}f+nxt^{n+m-1} f' \right)+ \left(t^{m+n}f' \right) \ t^m f = \nu t^{m+2n}f''$$

And, if we divide by $t^{m-1}$ we get $$ mf + nxt^nf'+t^{m+n+1}f \cdot f ' = \nu \, t^{2n+1} f'' $$ Remembering that $z= xt^m$ we obtain $$ mf + nzf'+t^{m+n+1}f \cdot f ' = \nu \, t^{2n+1} f'' $$ Now we want to anihilate the exponent hence we should impose $m+n = -1$ and $2n+1=0$.

Hence $n = m = - \frac{1}{2}$ and we obtain $$ -\frac{1}{2}f - \frac{1}{2}zf'+f \cdot f ' = \nu \, f'' $$ and, finally,

$$ \boxed{2 \nu f''(z) -2f f' +zf' +f = 0} $$

$\endgroup$
2
  • $\begingroup$ The term $uu_x$ should have a power of $t^{m+2n}$ I believe, it gave me the ODE $2\nu f''+2f'f+zf'+f=0$. $\endgroup$ – user30523 Apr 26 '20 at 18:26
  • $\begingroup$ @user30523 Thanks, I wrote $t^n f(x t^m)$ instead of $t^m f(x t^n)$ :P. I believe there is a miuns in front of the factor $f f'$, right? $\endgroup$ – Sewer Keeper Apr 27 '20 at 0:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.