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I want to find the solution $u(x,y)$ of the PDE $$ u_x-u_y=u^2 \quad \text{where} \quad u_x = \frac{\partial u}{\partial x}\,,~u_y = \frac{\partial u}{\partial u} \, $$ The boundary condition at points on the boundary $\Gamma_u$ given by $y=2x-1$ is $u(x,y)=4$.


My approach:

$\frac{dx}{dt}=1,\frac{dy}{dt}=-1, \frac{du}{dt}=u^2$

$x=t+a,y=-t+b,-u^{-3}=3(t+c)$

We want to construct the characteristic through each point of $\Gamma_u$. Suppose that a characteristic passes through $\Gamma_u$ at $P : (s,2s-1,4)$ at $t = 0$. Then

$x=a=s,y=b=2s-1,-4^{-3}=3c,c=-\frac{1}{192}$

$x=s,y=2s-1,u=4$

$x=t+s,y=-t+2s-1,-u^{-3}=3(t+-\frac{1}{192})$

We want to eliminate s and t from these equations.

$ y=-3t-1 $
$y=-3(\frac{-u^{-3}}{3}+\frac{1}{192})-1$

but answer at the end of my book says

$u(x,y)=\frac{3}{y-2x+\frac{7}{4}}$

Could you tell me where the mistake is?


edit 1: $\frac{dx}{dt}=1,\frac{dy}{dt}=-1, \frac{du}{dt}=u^2$

$x=t+a,y=-t+b,-u^{-1}=t+c$

$x=a=s,y=b=2s-1,-4^{-1}=c$

$x=s,y=2s-1,u=4$

$x=t+s,y=-t+2s-1,-u^{-1}=t-\frac{1}{4})$

We want to eliminate s and t from these equations.

$s=x-t$, $ y=2(x-t)-t-1 $
$y=-3(-u^{-1}+\frac{1}{4})+2x-1$

$u=\frac{3}{y+\frac{7}{4}-2x}$

thanks

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  • $\begingroup$ Note that $d/du(u^{-1}) = -1/u^2$ and $\int 1/u^2\,du = -1/u$. So your solution for $du/dt = u^2$ is incorrect. $\endgroup$ – Biswajit Banerjee Apr 1 '14 at 21:29
  • $\begingroup$ @BiswajitBanerjee thanks! I did as you said but still missing smth. can u recheck again please $\endgroup$ – lyme Apr 1 '14 at 21:55
  • $\begingroup$ See the technique. $\endgroup$ – Mhenni Benghorbal Apr 1 '14 at 21:58
  • $\begingroup$ @lyme, see below. If you need a more detailed solution, feel free to ask. $\endgroup$ – Chris K Apr 1 '14 at 21:59
  • $\begingroup$ Your problem is that there is no guarantee that $y(0) = 2x(0) - 1$ and hence you cannot conclude that $c = -1/4$. Here we want to find when $y(t) = 2x(t) - 1$ and thus find c in terms of $t$, $x_{0}$ and $y_{0}$. Then everything should fall out nicely as it must. $\endgroup$ – Chris K Apr 1 '14 at 22:07
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Here's an outline of the solution. Use the method of characteristics and get by solving ODEs $x(t) = t + x_{0}$, $y(t) = -t + y_{0}$ and $u(t) = \frac{-1}{t+c}$. Now let $-t + y_{0} = y(t) = 2x(t) - 1 = 2t + 2x_{0} - 1$. Then $t = \frac{y_{0}-2x_{0}+1}{3}$. We then have $c = -1/4 - \frac{y_{0}-2x_{0}+1}{3} = -7/12 - (y+t)/3 + 2(x-t)/3$. So, $t+c = -7/12 - y/3 + 2x/3$. Finally, we have $u(x, y) = -(-7/12 - y/3 + 2x/3)^{-1} = \frac{3}{7/4 + y - 2x}$.

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