Author Topic: HA1 problem 1  (Read 791 times)

Victor Ivrii

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HA1 problem 1
« on: January 20, 2015, 06:45:17 AM »
Solutions to be posted as a "Reply" only after January 22, 21:00

a. Find general solution
\begin{equation}
4 u_x -3u_y=0;
\label{eq-HA1.1}
\end{equation}
b.  Solve IVP problem $u|_{x=0}=\sin (y)$ for equation (\ref{eq-HA1.1}) in $\mathbb{R}^2$;

c.  Consider equation (\ref{eq-HA1.1}) in $\{x>0, y>0\}$ with the initial condition $u|_{x=0}=y$ ($y>0$); where this solution defined? Is it defined everywhere in $\{x>0, y>0\}$ or do we need to impose condition at $y=0$?
In the latter case impose condition $u|_{y=0}=x$ ($x>0$) and solve this IVBP;

d. Consider equation (\ref{eq-HA1.1}) in $\{x<0, y>0\}$ with the  initial condition $u|_{x=0}=y$ ($y>0$); where this solution defined? Is it defined everywhere in $\{x<0, y>0\}$ or do we need to impose condition at $y=0$? In the latter case impose condition $u|_{y=0}=x$ ($x<0$) and solve this IVBP.

Biao Zhang

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Re: HA1 problem 1
« Reply #1 on: January 22, 2015, 10:17:29 PM »
HA1-1

Jessica Chen

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Re: HA1 problem 1
« Reply #2 on: January 22, 2015, 10:18:06 PM »
a. This is a first order linear PDE with constant coefficient.
\begin{equation}
\frac{d x}{4} = \frac{d y}{3} = \frac{d u}{0};
\end{equation}
Then we get $u(x, y) = f(3x+4y)$ for some arbitrary $f$.

b.  Use IVP $u|_{x=0}=\sin (y)$, we get
\begin{equation}
f(4y) = \sin(y)\\f(w) = \sin(\frac{y}{4})
\end{equation}
Hence solution is
\begin{equation}
u(x,y) = f(3x+4y) = \sin(\frac{3}{4}x+y)
\end{equation}


c.  With initial condition $u|_{x=0}=y$, $y>0$. Hence the solution is
\begin{equation}
u(x, y) = \frac{3}{4}x+y.
\end{equation}
Since $\forall x>0, y>0\implies \frac{3}{4}x+y >0$.
Thus this solution is defined on the whole domain $\{x>0,y>0\}$.

d.  With initial condition $u|_{x=0}=y$, $y>0$. Hence the solution is
\begin{equation}
u(x, y) = \frac{3}{4}x+y.
\end{equation}
Since $f$ is define when $y>0$, the solution only is defined where $\frac{3}{4}x+y >0$, then when $y> -\frac{3}{4}x$, the solution is defined. However when $y< -\frac{3}{4}x$ we need to impose condition at $y = 0$, we get
\begin{equation}
f(3x) = x \\
 f(w) = \frac{w}{3}\\
\end{equation}
Then we get:
\begin{equation}
u(x,y) = x + \frac {4}{3}y, (y< -\frac{3}{4}x)
\end{equation}
Final solution would be:
\begin{equation}
u(x,y) = \left\{
  \begin{array}{l l}
    \frac{3}{4}x+y & \quad y>-\frac{3}{4}x\\
    x+\frac{3}{4}y & \quad y<-\frac{3}{4}x
  \end{array} \right.
\end{equation}
« Last Edit: January 23, 2015, 11:33:14 PM by Jessica Chen »

Biao Zhang

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Re: HA1 problem 1
« Reply #3 on: January 22, 2015, 10:25:39 PM »
i think equation has some typo. should be y instead of 7

Jessica Chen

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Re: HA1 problem 1
« Reply #4 on: January 22, 2015, 10:37:03 PM »
i think equation has some typo. should be y instead of 7

Yes you are right! Thanks

Victor Ivrii

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Re: HA1 problem 1
« Reply #5 on: January 23, 2015, 10:49:43 AM »
Very good, guys! With Jessica I slightly edited your post, mainly putting some $ signs (every inline math expression should have it), removed some underscores _ and put a backslash \ to have \sin instead of sin (makes it upright and provides a proper horizontal spacing).

Biao, definitely there is no abs. value sign in d.