Messages

16

HA1 / Re: HA1 problem 4

« on: January 22, 2015, 11:23:04 PM »

I got a very nasty equation for part b
b.
First solve a homogeneous: we get $u(x, y) = \phi (4x^2 +y^2)$ and $C = 4x^2 + y^2$
Then to find the general solution:
\begin{equation}
\frac{d x}{y} = \frac{d x}{x^2} = \frac{d u}{0};\\
\frac{x^2}{y}d x = d u;\\ \text{Since } y = \sqrt{C-4x^2},\\
\frac{x^2}{\sqrt{C-4x^2}}d x=d u\\
\frac{x^2}{\sqrt{C\sqrt{1-\frac{4x^2}{c}}}}dx = du;\\
\end{equation}

We substitute
\begin{equation}
\frac{2x}{\sqrt{c}} = \sin \theta

\end{equation}
Then we get:
\begin{equation}
du=\frac{\frac{\sqrt{c}}{4}\sin^2\theta}{\sqrt{1-\sin^2\theta}}d\sin\theta \frac{\sqrt{2}}{2}\\
du=\frac{\frac{c}{8}\sin^2 \theta }{\cos \theta }d\sin \theta\\
du=\frac{c}{8}\frac{\sin^2 \theta}{\cos \theta} \cos \theta d\theta \\
du=\frac{c}{16}(\theta - \frac{1}{2}\sin 2\theta )\\
\end{equation}
Use trig identity we have $u = \frac{c}{16} \arcsin \frac{2x}{\sqrt{c}} - (\frac{1}{2}(2\times\frac{2x}{\sqrt{c}}\sqrt{1-\frac{4x^2}{c}})$
The general solution is $u = \phi(4x^2+y^2)+ \frac{c}{16} \arcsin \frac{2x}{\sqrt{c}} - (\frac{2x}{\sqrt{c}}\sqrt{1-\frac{4x^2}{c}})$

17

HA1 / Re: HA1 problem 2

« on: January 22, 2015, 10:51:12 PM »

Not sure if I interpreted it right either.

c.
The difference between a and b is
part a) the characteristic lines look like a parabola, all trajectories have (0, 0) as the limit point;
part b) the characteristic lines look like a delta function, only (x=0, y=0) has (0,0) as the limit point.

18

HA1 / Re: HA1 problem 1

« 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

19

HA1 / Re: HA1 problem 6

« on: January 22, 2015, 10:35:35 PM »

a. This is a quasilinear equation.
\begin{equation}
\frac{d t}{1} = \frac{d x}{u} = \frac{d u}{0};
\end{equation}
Then we get $ut-x= C$ for some arbitrary constant C.
Consider the initial condition $u|_{t=0}=x$, take an initial point $(0, x_0)$ such that
\begin{equation}
u(x, 0) = x
\end{equation}
 Therefore we have $u = f(x_0) = f(x-ut)$ along characteristics, so
\begin{equation}
u(x,y) = f(x-ut) = x-ut\\
u(x, y) = \frac{x}{1+t}
\end{equation}


b.
When $t>-1$, the solution is clearly hold.
However when $t<-1$ the solution breaks because the characteristics lines cannot be interpreted if they are intersected or undefined.

20

HA1 / Re: HA1 problem 1

« 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}