### Author Topic: Web bonus problem : Week 2  (Read 1120 times)

#### Victor Ivrii

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##### Web bonus problem : Week 2
« on: September 21, 2015, 08:18:26 AM »
Find where solution $u(x,t)$  is properly defined for the following problem:
\begin{align}
&u_t - u u_x=0\qquad -\infty< x< \infty,\ t>0,\label{eq-1}\\
&u(x,0)= \frac{1}{2}x^2.\label{eq-2}
\end{align}

Hint
Using method of characteristics write down solution in the implicit form $F(x,t,u)=0$.

Then the border of the domain where solution is properly defined is an envelope of this family of the curves (actually, straight lines) defined by
\begin{align}
&F(x,t,u)=0,\label{eq-3}\\
&F_u (x,t,u)=0.\label{eq-4}
\end{align}
Recall that $t>0$.

#### Yeming Wen

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##### Re: Web bonus problem : Week 2
« Reply #1 on: September 22, 2015, 11:37:40 AM »
By the method of characteristics, we have

\frac{d x}{d t} = - u,\frac{d  u}{d t}=0; \label{eq1}
So,
\begin{equation*}
u=C_1,x=- u t+C_2.
\end{equation*}
By (\ref{eq1}), we know $C_1$ should be a function of $C_2$. Then we can write

u=C_1=C_1(C_2)=C_1(x+ u t)

Notice the initial condition, we have

u(x,0)=C_1(x)=\frac{1}{2}x^2
which implies that

u=C_1(x+ u t)=\frac{1}{2}(x+ u t)^2
Rearrange the equation, we have

t^2 u^2+2(tx-1) u+x^2=0.
\label{K}

So,

u=\frac{-(tx-1) \pm \sqrt{(tx-1)^2-t^2x^2}}{t^2}

« Last Edit: September 23, 2015, 05:32:09 AM by Victor Ivrii »

#### Victor Ivrii

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##### Re: Web bonus problem : Week 2
« Reply #2 on: September 22, 2015, 01:08:59 PM »
You did correctly (except renaming $u$ to $\mu$; so I replaced (batch) $\mu$ by $u$) until the last equation.
The primary purpose is not to find $u$ but to determine where it is a solution.

Follow Hint provided: your (\ref{K}) is my (\ref{eq-3}). Write system  (\ref{eq-3})-- (\ref{eq-4}) and solve it which gives the boundary of the required domain.

More clear what is the core of the problem will be after Lecture Wed, Sep 23 (see also Subsection 2.1.6 Quasilinear equations)

---

So according to (\ref{K}) $F(x,t,u)=t^2 u^2+2(tx-1) u+x^2$ and system  (\ref{eq-3})-- (\ref{eq-4}) is
\begin{align}
&t^2 u^2+2(tx-1) u+x^2,\label{L}\\
&2t^2u +2 (tx-1) =0.\label{M}
\end{align}
(\ref{M}) implies $u= -(tx-1)t^{-2}$ and plugging to (\ref{L}) we get $x=1/(2t)$. This is a border of the domain where $u$ is defined. See attached picture with cyan characteristics. Their continuations after they touched   $x=1/(2t)$ should be ignored as at the point of tangency equation breaks (as it does not belong to domain).

« Last Edit: September 25, 2015, 09:58:39 AM by Victor Ivrii »