### Author Topic: Web Bonus Problem 4  (Read 3674 times)

#### Victor Ivrii

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##### Web Bonus Problem 4
« on: January 17, 2015, 11:56:54 AM »
Find self-similar solution

u_t=u u_x\qquad -\infty <x <\infty,\ t>0
\label{eq-1}

satisfying initial conditions

u|_{t=0}=\left\{\begin{aligned}
-1& &x\le 0,\\
1& &x> 0
\end{aligned}\right.
\label{eq-2}

#### Jessica Chen

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##### Re: Web Bonus Problem 4
« Reply #1 on: March 06, 2015, 12:13:48 AM »
Characteristic equation:
\begin{align}
\frac{dt}{1} &=\ \frac{dx}{-u} = \frac{du}{0}\\
\frac{du}{dt}&=\ 0 \implies u=f(C)\\
x &=\ C - tu\\
C &=\ x+tu
\end{align}

Impose boundary condition: $t=0$ and $x = C$,
\begin{align}
u &=\ f(C) =f(x+tu)\\
u|_{t=0}&=f(x) =\left\{\begin{aligned}
-1& &x\le 0,\\
1& &x> 0
\end{aligned}\right.
\end{align}
then,

\end{align}

« Last Edit: March 07, 2015, 08:47:21 PM by Jessica Chen »

#### Victor Ivrii

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##### Re: Web Bonus Problem 4
« Reply #2 on: March 07, 2015, 05:06:05 AM »
No, this does not fly. To construct a solution you need to fill by characteristics the whole half-plane $t>0$. So far you covered only $x<-t$ where $u=-1$ and $x>t$ where $u=1$ leaving sector $-t<x<t$ empty.

You need to apply the method of self-similar solutions to find continuous $u(x,t)$  there.

#### Jessica Chen

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##### Re: Web Bonus Problem 4
« Reply #3 on: March 07, 2015, 07:57:40 PM »
No, this does not fly. To construct a solution you need to fill by characteristics the whole half-plane $t>0$. So far you covered only $x<-t$ where $u=-1$ and $x>t$ where $u=1$ leaving sector $-t<x<t$ empty.

You need to apply the method of self-similar solutions to find continuous $u(x,t)$  there.
Sorry professor I quoted my post accidentally, could you help me delete the previous post? -t<x<t sector has two values which is definitely not right but I don't know what do you mean apply the self-similar solution, can you give me more hints please?
« Last Edit: March 07, 2015, 08:55:06 PM by Jessica Chen »

#### Victor Ivrii

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##### Re: Web Bonus Problem 4
« Reply #4 on: March 08, 2015, 07:35:52 AM »
No, sector $-t<x<t$ has not 2 values (it would be if u=\left\{\begin{aligned} &1 &&x<0,\\ -&1 &&x>0\end{aligned}\right.). So far it has NO values at all.

Now hint:
a) consider $u_s (x,t)= s^l u(s x, s^k t)$ and prove that if $u$ satisfies original problem then $u_s$ satisfies it for all $s>0$ iff $k=1$, $l=0$.
b) So, $u_s(x,t)= u(sx, st)$ satisfies it and we are interested in self-similar solution $u(x,t)=u(sx,st)$ for all $s>0$. Plugging $s=t^{-1}$ we arrive to $u(x,t)=v (xt^{-1})$ (with $v(y)=u(y, 1)$.
c) Plugging $u(x,t)=v (xt^{-1})$ into original equation we have an ODE. Which?
d) Find continuous solution of this ODE such that $v(y)=-1$ as $y<-1$ and $v(y)=1$ as $y>1$ (Think why).
e) Plug into $u$

#### Jessica Chen

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##### Re: Web Bonus Problem 4
« Reply #5 on: March 11, 2015, 09:58:31 PM »

Now hint:
a) consider $u_s (x,t)= s^l u(s x, s^k t)$ and prove that if $u$ satisfies original problem then $u_s$ satisfies it for all $s>0$ iff $k=1$, $l=0$.
b) So, $u_s(x,t)= u(sx, st)$ satisfies it and we are interested in self-similar solution $u(x,t)=u(sx,st)$ for all $s>0$. Plugging $s=t^{-1}$ we arrive to $u(x,t)=v (xt^{-1})$ (with $v(y)=u(y, 1)$.
c) Plugging $u(x,t)=v (xt^{-1})$ into original equation we have an ODE. Which?
d) Find continuous solution of this ODE such that $v(y)=-1$ as $y<-1$ and $v(y)=1$ as $y>1$ (Think why).
e) Plug into $u$

a) $s^k = s$ because one derivative with respect to t is also one derivative with respect x. Thus $k = 1$;
By total energy rule, $\int_{-\infty}^{\infty} u_s dx = s^l\int _{-\infty}^{\infty}u dx$ Thus $s^l = 1\implies l=0$

b) $u_s= u(sx,st)$. Let $s= t^{-1}$ we have $u_s= u(t^{-1}x, 1) = v(t^{-1}x)$
c) plug in $v(t^{-1}x)$
\begin{align}
ut&=-t^{-2}xv'(t^{-1}x)\\
ux&=v'(t^{-1}x)t^{-1}\\
-t^{-2}xv'(t^{-1}x)&=v(t^{-1}x)(v'(t^{-1}x)t^{-1}\\
-t^{-1}xv'(t^{-1}x)&=v(t^{-1}x)(v'(t^{-1}x)\\
\end{align}
Let $t^{-1}x=\xi$ we get

-\xi v'(\xi)=v(\xi)v'(\xi)
\label{A}

$-\xi = v(\xi)$ is the ODE?

« Last Edit: March 12, 2015, 04:40:45 AM by Victor Ivrii »

#### Victor Ivrii

You got (\ref{A}) which can be written as $v'(\xi)[v(\xi)+\xi]=0$. This is correct! But then you decided to divide by $v(\xi)$ which is not exactly justified. In fact you can conclude that either $v'(\xi)=0$ or $v(\xi)=-\xi$. The former holds as $\xi<-1$ (then $v(\xi)=1$) or $\xi>1$ (then $v(x)=-1$), so the latter holds as $-1<\xi<1$. Plugging $\xi=x/t$ we get