# Find $m,n$ so that $u(x,t)=t^mf(xt^n)$ is a solution to $u_t+uu_x=\nu u_{xx}$

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.

• 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 $m, n = -1/2$. – mattos Apr 25 '20 at 8:20
• Ah, didn't think of taking a power of $t$ outside, thanks! – user30523 Apr 25 '20 at 8:57

## 1 Answer

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,

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

• The term $uu_x$ should have a power of $t^{m+2n}$ I believe, it gave me the ODE $2\nu f''+2f'f+zf'+f=0$. – user30523 Apr 26 '20 at 18:26
• @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? – Sewer Keeper Apr 27 '20 at 0:29