# Use Charpits to solve this PDE

• Feb 7th 2011, 02:40 PM
bugatti79
Use Charpits to solve this PDE
Use Charpits Equations to Solve the following given that u=1 on the line x+2y=2.
Is the solution unique?

$\displaystyle 4u \frac{\partial {u}}{\partial x}=(\frac{\partial {u}}{\partial y})^2$

$\displaystyle 4up-q^2=0, F(x,y,u,p,q)=0$

$\displaystyle \frac{\partial F}{\partial x}=0, \frac{\partial {F}}{\partial y}=0,\frac{\partial {F}}{\partial u}=4p$

$\displaystyle \frac{dx}{dt}=F_p, \frac{dy}{dt}=F_q, \frac{dp}{dt}=-F_x-pF_u, \frac{dq}{dt}=-F_y-qF_u, \frac{du}{dt}=pF_P+qF_q$

Initial Condition are

$\displaystyle t=0,u=1, x=s,y=1-\frac{s}{2}, \frac{d}{ds} (u(s,1-\frac{s}{2}))=\frac{d}{ds}(1)$

Therefore

$\displaystyle \frac{\partial u}{\partial x}+ \frac{\partial u}{\partial y}( \frac{-1}{2})=0$

Therefore
$\displaystyle p - \frac{q}{2}=0, q=2p, @t=0, u=1 \implies 4p-4p^2=0$

2 solutions p=0, p=1

$\displaystyle \frac{dp}{dt}=-4p^2 => p=\frac{1}{4t-A}$

No solution for p=0, therefore continuing with p=1 at t=0 to find A

$\displaystyle \frac{dp}{dt}=-4p^2 => p=\frac{1}{4t-A}$

$\displaystyle \implies A=\frac{-1}{4}$.

Therefore $\displaystyle p=\frac{1}{4t+1}$. Now q=2p therefore $\displaystyle q=\frac{2}{4t+1}$.

Since 4up=q^2 implies u is $\displaystyle \frac{4t+1}{16t^2+1}$

How am I doing so far? (Thinking)
• Feb 8th 2011, 04:30 AM
Jester
A couple of comments. First, you say there's no solution for $\displaystyle p = 0$. Why can't $\displaystyle p = q = 0$ always? (although it does lead to a trivial solution). Second, when you integrate the Chapit equation you should obtain arbitrary functions of $\displaystyle s$ and when $\displaystyle t = 0$ then

$\displaystyle x = s, \;y = 1-s/2, \;u = 1, \;p = 1, \;q = 2.$
• Feb 8th 2011, 04:42 AM
bugatti79
Quote:

Originally Posted by Danny
A couple of comments. First, you say there's no solution for $\displaystyle p = 0$. Why can't $\displaystyle p = q = 0$ always? (although it does lead to a trivial solution). Second, when you integrate the Chapit equation you should obtain arbitrary functions of $\displaystyle s$ and when $\displaystyle t = 0$ then

$\displaystyle x = s, \;y = 1-s/2, \;u = 1, \;p = 1, \;q = 2.$

What is a trivial solution?

For this particular problem, the charpit equations do not have s in them...so obviously I have gone wrong somewhere...pardon my lack of understanding

Thanks
• Feb 8th 2011, 04:57 AM
Jester
I see another problem. You say that $\displaystyle q =2p$. This is true but only when $\displaystyle t = 0$ - not in general!
• Feb 8th 2011, 05:15 AM
Jester
Quote:

Originally Posted by bugatti79
What is a trivial solution?

For this particular problem, the charpit equations do not have s in them...so obviously I have gone wrong somewhere...pardon my lack of understanding

Thanks

If $\displaystyle p = 0$ and $\displaystyle q = 0$ always then $\displaystyle u = u_0$ (constant) and the IC gives that $\displaystyle u = 1$ (always).

When integrating say $\displaystyle p_t = -4p^2$ this gives $\displaystyle \dfrac{1}{p} = 4t + A(s)$. From your IC $\displaystyle p = 1$ when $\displaystyle t = 0$ giving $\displaystyle A(s) = 1$ so

$\displaystyle \dfrac{1}{p} = 4t + 1\;\;\;\Rightarrow\;\;\;p = \dfrac{1}{4t+1}$.

Then continue in this fashion.
• Feb 8th 2011, 05:33 AM
bugatti79
Quote:

Originally Posted by Danny
If $\displaystyle p = 0$ and $\displaystyle q = 0$ always then $\displaystyle u = u_0$ (constant) and the IC gives that $\displaystyle u = 1$ (always).

When integrating say $\displaystyle p_t = -4p^2$ this gives $\displaystyle \dfrac{1}{p} = 4t + A(s)$. From your IC $\displaystyle p = 1$ when $\displaystyle t = 0$ giving $\displaystyle A(s) = 1$ so

$\displaystyle \dfrac{1}{p} = 4t + 1\;\;\;\Rightarrow\;\;\;p = \dfrac{1}{4t+1}$.

Then continue in this fashion.

Ok, thanks Danny..
I proceed to find q using $\displaystyle q_t = -q*4*[\frac{1}{4t+1}]$? I will tackle it later this evening..
• Feb 8th 2011, 06:41 AM
Jester
Here's the solution that I got (to compare)

Spoiler:
$\displaystyle u = e^{x+2y-2}$
• Feb 8th 2011, 02:55 PM
bugatti79
Quote:

Originally Posted by Danny
Here's the solution that I got (to compare)

Spoiler:
$\displaystyle u = e^{x+2y-2}$

$\displaystyle \frac{dq}{q}=\frac{-4}{4t+1} \implies q={e^{{ln(4t+1)^{-1}} +B'(s)} =\frac{B(s)}{(4t+1)}$

At t=0 q=2p where $\displaystyle p=\frac{1}{(4t+1)}$ therefore B(s)=2 $\displaystyle \implies q=\frac{2}{(4t+1)}$

$\displaystyle \frac{du}{dt}=\frac{4u}{(4t+1)}-\frac{8}{(4t+1)^2}$.

Using an integrating factor I get $\displaystyle u=\frac{1}{4t+1}+(4t+1)C$.

At t=0 u=1 $\displaystyle \implies u=\frac{1}{4t+1}$

$\displaystyle \frac{dx}{dt}=4u$ and using x=s at t=0 I get $\displaystyle x=ln(4t+1)+s$

Similarly $\displaystyle y=-ln(4t+1)+1-\frac{s}{2}$

Eliminating s and t I get $\displaystyle u(x,y)=\frac{1}{e^{(-x-2y+2)}-1+1}$

Finally $\displaystyle u(x,y)=e^{x+2y-2}$???(Rofl)
• Feb 8th 2011, 10:13 PM
bugatti79
The question asks Is the solution unique?

I am guessing that it is not unique because it has 2 valid solutions ie 1) p=0 and 2) p=1?

However, Im confused because when solving for p using $\displaystyle \Displaystyle \frac{dp}{dt}$ given the condition that p=0 at t=0 I get

A(s)=0 which is not possible since A is constant, right?
• Feb 9th 2011, 05:45 AM
Jester
But you have $\displaystyle p_t = -4p^2$ which has two solutions, $\displaystyle p = 0$ and $\displaystyle p = \dfrac{1}{4t + A(s)}$.
• Feb 9th 2011, 07:56 AM
bugatti79
Quote:

Originally Posted by Danny
But you have $\displaystyle p_t = -4p^2$ which has two solutions, $\displaystyle p = 0$ and $\displaystyle p = \dfrac{1}{4t + A(s)}$.

It is not unique because it has 2 solutions?

Thanks