# Charpits: Non Linear 1st order PDE: Particular IC

• Aug 14th 2011, 07:50 AM
bugatti79
Charpits: Non Linear 1st order PDE: Particular IC
Folks,

I am struggling to get the initial conditions @ t=0 for p and q.

Given $(u_x)^2+(u_y)^2-1=0$ for $u=0$ on $x^2+y^2=1$
My attempt:

Parameterise x such that $x=s$ and $y=\sqrt{1-s^2}$ and differentiate the given IC

$\displaystyle \frac{\partial u}{\partial x} \frac{d(s)}{ds}+\frac{\partial u}{\partial y} \frac{d(\sqrt{1-s^2})}{ds}=\frac{d(0)}{ds}$ This gives

$\displaystyle \frac{\partial u}{\partial x}+\frac{\partial u}{\partial y} \frac{-s}{\sqrt{1-s^2}}=0$

Not sure if this right or how to proceed further to find pand q?

Thanks
• Aug 15th 2011, 06:49 AM
Jester
Re: Charpits: Non Linear 1st order PDE: Particular IC
So now you have one equation

$p - \frac{s}{\sqrt{1-s^2}}q = 0$.

Your second equation is the PDE itself $p^2+q^2=1$.

Two equations for two unknowns. Now solve for $p$ and $q$.

BTW - what do you mean @ $t= 0$. There's no $t$ in this problem.

Because of the symmetry of the BC, I might also suggest switching to polar coordinates.
• Aug 15th 2011, 10:40 AM
bugatti79
Re: Charpits: Non Linear 1st order PDE: Particular IC
Quote:

Originally Posted by Danny
So now you have one equation

$p - \frac{s}{\sqrt{1-s^2}}q = 0$.

Your second equation is the PDE itself $p^2+q^2=1$.

Two equations for two unknowns. Now solve for $p$ and $q$.

BTW - what do you mean @ $t= 0$. There's no $t$ in this problem.

Because of the symmetry of the BC, I might also suggest switching to polar coordinates.

All the problems involving Charpits I have been solving are in parametric form where the IC's are taken to be at t=0. I believe the curve is parametric in x(t), y(t) and u(t). ...?
• Aug 15th 2011, 11:18 AM
bugatti79
Re: Charpits: Non Linear 1st order PDE: Particular IC
Maybe I should clarify that in this problem we have u=0 when x^2+y^2=1. These are the IC's I take at t=0. I hope that makes sense. This is the approach I take for all other charpit problems based on my notes.
• Aug 15th 2011, 12:16 PM
Jester
Re: Charpits: Non Linear 1st order PDE: Particular IC
I'm good. I usually use $r$ and $s$ and leave $t$ for a time variable :-)
• Aug 15th 2011, 12:21 PM
bugatti79
Re: Charpits: Non Linear 1st order PDE: Particular IC
Quote:

Originally Posted by Danny
I'm good. I usually use $r$ and $s$ and leave $t$ for a time variable :-)

Based on your post no 2, you mention about 2 eqns ans 2 unknowns. However, I dont believe I can use the second equation (actual PDE) because it does not represent an initial condition like the first equation does, so how does one proceed? Perhaps I am wrong :-)
• Aug 15th 2011, 01:18 PM
Jester
Re: Charpits: Non Linear 1st order PDE: Particular IC
Actually you can! If $u_x^2+u_y^2 = 1$ for all $x$ and $y$ then certainly along some curve $y = f(x)$ (as long as the derivatives exist).
• Aug 15th 2011, 01:38 PM
bugatti79
Re: Charpits: Non Linear 1st order PDE: Particular IC
Quote:

Originally Posted by Danny
Actually you can! If $u_x^2+u_y^2 = 1$ for all $x$ and $y$ then certainly along some curve $y = f(x)$ (as long as the derivatives exist).

OK, thanks. I will give it a go and keep you posted. :-)
• Aug 22nd 2011, 12:59 PM
bugatti79
Re: Charpits: Non Linear 1st order PDE: Particular IC
Quote:

Originally Posted by Danny
Actually you can! If $u_x^2+u_y^2 = 1$ for all $x$ and $y$ then certainly along some curve $y = f(x)$ (as long as the derivatives exist).

OK, recall that I parameterise x as equal to s so that t=0 corresponds to the initial conditions.

Based on this I calculate, using the 2 equations and 2 unknowns

$p= \sqrt{1-s^2}$ and $q=s @ t=0.$

I should be able continue and find u(x,y)

Are these right?(Nerd)
• Aug 22nd 2011, 01:47 PM
Jester
Re: Charpits: Non Linear 1st order PDE: Particular IC
Looks good be I might be careful on the $\pm$.