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

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

Folks,

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

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

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

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

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

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

Thanks

2. ## Re: Charpits: Non Linear 1st order PDE: Particular IC

So now you have one equation

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

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

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

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

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

3. ## Re: Charpits: Non Linear 1st order PDE: Particular IC

Originally Posted by Danny
So now you have one equation

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

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

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

BTW - what do you mean @ $\displaystyle t= 0$. There's no $\displaystyle t$ 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). ...?

4. ## 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.

5. ## Re: Charpits: Non Linear 1st order PDE: Particular IC

I'm good. I usually use $\displaystyle r$ and $\displaystyle s$ and leave $\displaystyle t$ for a time variable :-)

6. ## Re: Charpits: Non Linear 1st order PDE: Particular IC

Originally Posted by Danny
I'm good. I usually use $\displaystyle r$ and $\displaystyle s$ and leave $\displaystyle t$ 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 :-)

7. ## Re: Charpits: Non Linear 1st order PDE: Particular IC

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

8. ## Re: Charpits: Non Linear 1st order PDE: Particular IC

Originally Posted by Danny
Actually you can! If $\displaystyle u_x^2+u_y^2 = 1$ for all $\displaystyle x$ and $\displaystyle y$ then certainly along some curve $\displaystyle y = f(x)$ (as long as the derivatives exist).
OK, thanks. I will give it a go and keep you posted. :-)

9. ## Re: Charpits: Non Linear 1st order PDE: Particular IC

Originally Posted by Danny
Actually you can! If $\displaystyle u_x^2+u_y^2 = 1$ for all $\displaystyle x$ and $\displaystyle y$ then certainly along some curve $\displaystyle y = f(x)$ (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

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

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

Are these right?

10. ## Re: Charpits: Non Linear 1st order PDE: Particular IC

Looks good be I might be careful on the $\displaystyle \pm$.