### Author Topic: Solving the Burgers equation  (Read 318 times)

#### Shaghayegh A

• Full Member
• Posts: 21
• Karma: 0
##### Solving the Burgers equation
« on: September 23, 2016, 07:29:07 PM »
In example 7 of chapter 2.1, we wish to solve $$u_{t}+uu_{x}=0.$$
The textbook says
$$\frac{dt}{1}=\frac{dx}{u}=\frac{du}{0}.$$
So far correct  The rest here are just your fantasies V.I.

We know $$\frac{\partial x}{\partial t}=u\;and\;\frac{du}{dt}=0\;so\;du=0\implies\frac{du}{0}=1$$
Why is $$\frac{dx}{u}=1?$$
« Last Edit: September 24, 2016, 08:24:29 AM by Victor Ivrii »

#### Shentao YANG

• Full Member
• Posts: 24
• Karma: 0
##### Re: Solving the Hopf equation
« Reply #1 on: September 23, 2016, 10:10:38 PM »
Here is my thinking:

Since u is a constant along the integral curve, then in solving
$$\frac{dt}{1}=\frac{dx}{u}.$$
we can treat the u as a constant, since the above equation already ensures / implies that u is on the integral line.
Therefore, we basically have
$$dx = u dt\implies\ x - ut = c .$$(probably you can think of u as some constant like 1, 2, 3...)
And noted that $$u = f(c)$$
for some function f of one variable.
Therefore, by combining the above two equation, we have: $$u = f (x-ut).$$
Which is the answer that Prof. Ivrii gave us in the class.
Waiting for Prof. Ivrii's correction...
« Last Edit: September 23, 2016, 10:20:12 PM by Shentao YANG »

#### Victor Ivrii

• Elder Member
• Posts: 1332
• Karma: 0
##### Re: Solving the Hopf equation
« Reply #2 on: September 24, 2016, 08:21:29 AM »
Shentao is completely correct. Therefore we have an implicit equation for $u$:

u=f(x-ut).\label{A}

Remember implicit function theorem? We need $F_u\ne 0$ where $F=F(x,t,u)=u-f(x,t,u)$. Since $F(x,0,u)=u)$ we have $F_u(x,0,u)=1>0$ and (\ref{A}) is fulfilled as long $F_u(x,t,u)>0$; so solution breaks when

F_u= 1-t f(x-ut)=0.
\label{B}

Let $s=x-ut$, then (\ref{A}) implies $u=f(s)$, (\ref{B}) implies $t=-1/f'(s)$, then $x=s+ut= s-f(s)/f'(s)$. So

\left\{\begin{aligned}
&x=s-f(s)/f'(s),\\
&t=f'(s)
\end{aligned}
\right.
\label{C}

Figure below shows case $f(s)=-\tanh (s)$: orange lines are $u=f(s)$, $x-ut=s$ for different values of of $s$. Blue line is an envelope of orange lines, and it shows where solution breaks. The "beak" corresponds to $s=0$ when $-f'(s)=1/cosh^2(s)$ reaches it maximum. Continuous solution exists below blue line but not above it (where orange lines overlap).

« Last Edit: September 26, 2016, 05:47:58 AM by Victor Ivrii »

#### Victor Ivrii

• Elder Member
• Posts: 1332
• Karma: 0
##### Re: Solving the Burgers equation
« Reply #3 on: September 24, 2016, 08:47:53 AM »
Remark
Later in Chapter 12 we consider discontinuous solutions of

u_t+(\frac{u^2}{2})_x=0
\label{D}

which coincides with $u_t+uu_x=0$ for continuous solutions (but it also has sense for discontinuous solutions) and the "correct" solution would be like on picture below:
blue line shows where the discontinuity is.
« Last Edit: September 24, 2016, 08:50:30 AM by Victor Ivrii »

#### Roro Sihui Yap

• Full Member
• Posts: 30
• Karma: 16
##### Re: Solving the Hopf equation
« Reply #4 on: September 25, 2016, 10:06:30 PM »
Remember implicit function theorem? We need $F_u\ne 0$ where $F=F(x,t,u)=u-f(x,t,u)$. Since $F(x,0,u)=u)$ we have $F_u(x,0,u)=1>0$ and (\ref{A}) is fulfilled as long $F_u(x,t,u)>0$; so solution breaks when

F_u= 1-t f(x-ut)=0.

Let $s=x-ut$, then (\ref{A}) implies $u=f(s)$, (\ref{B}) implies $t=1/f(s)$, then $x=s+ut= s-f(s)/f'(s)$. So

\left\{\begin{aligned}
&x=s-f(s)/f'(s),\\
&t=f'(s)
\end{aligned}
\right.

Should the following be equation (2) instead ?
Since $F = u - f(x-ut)$
$F_u = 1 - fâ€™(x-ut)\cdot(-t) = 1+t f'(x-ut)=0$

Then, substitution of $s = x-ut$ implies $t = -1/f'(s)$
And subsequently $x = s + ut = s + f(s)\cdot\frac{-1}{f'(s)} = s - \frac{f(s)}{f'(s)}$
« Last Edit: September 26, 2016, 05:46:15 AM by Victor Ivrii »