0
$\begingroup$

So I have the following PDE: for z(x,y) $\frac{\partial z}{\partial x}$ + $\frac{\partial z}{\partial x}$ = $-z^2$;

Boundary/Initial conditions: $z(x,0)=1$ & $z(0,y)=0$

Ideally,using the method of characteristics I could see,

$\frac{dx}{dt}=1$ & $\frac{dy}{dt}=1$ so $x=t+\phi_1(s)$ & $y=t+\phi_2(s)$.

Next, $\frac{dz}{dt}=-z^2$, solving gives $\frac{1}{z}=t+\phi_3(s)$ or $z=\frac{1}{t+\phi_3(s)}$.

The problem I have now is satisfying both boundary conditions. I can see $\phi_1(s)=s$,$\phi_2(s)=0$ gives $\phi_3=1$ so we get $z=\frac{1}{y+1}$, as $x=t+s$ and $y=t$. However this satisfies $\frac{\partial z}{\partial x}$ + $\frac{\partial z}{\partial x}$ = $-z^2$, as well as $z(x,0)=1$, but does not satisfy & $z(0,y) = 0$ as $z(0,y)=\frac{1}{y+1}$.

I cannot figure out how to satisfy the second boundary condition at all, yet minding solving both at the same time. Any help is greatly appreciated.

1 Answers 1

0

Note that the conditions $z(x,0)=1$ and $z(0,y)=0$ are not consistent at $(0,0)$ since $z(0,0)=0=1$ is non-sens.

So, $z(x,y)$ cannot be a continuous function. As a consequence, the answer is :

Either there is no solution on the form of continuous function, or one have to consider a piecewise function : $$z(x,y)=\begin{cases} \frac{1}{y+1} & x\neq 0 \\ 0 & x=0\end{cases}$$

On $x=0$ , both $\frac{\partial z}{\partial x}=0$ and $\frac{\partial z}{\partial y}=0$ , thus $\frac{\partial z}{\partial x}+\frac{\partial z}{\partial y}= -z^2=0$ is fulfilled.

  • 0
    Very insightful, I should have realized right away there was a discontuinity at z(0,0). Can you think of a way to display this piecewise with respect to y? i.e., $z(x,y) = \begin{cases} f(x) \text{ if } y \neq 0 \\ 0 \text{ if } y=0 \end{cases}$ – 2017-01-11
  • 0
    Another manner to express the result is $z(x,y)=\frac{1-\delta(x)}{y+1}$ where $\delta(x)$ is the Dirac delta function. The graphical representation is problematic. One can use approximates, for example such as shown in http://mathworld.wolfram.com/DeltaFunction.html – 2017-01-11
  • 0
    Huh, I didn't even think of that. How would one be able to derive that form where the dirac delta is in the numerator? I cannot see it intuitively. I do see how z(0,y)=0, and z(x,0)=1 (provided x $\neq 0$). – 2017-01-11
  • 0
    It's more intuitively physical than mathematical. If the problem doesn't come from a physical question, better use the form of piecewise function. Anyways, the question of no derivative at the discontinuity remains the same. – 2017-01-11