# Thread: Use Charpits to solve this PDE

1. ## 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?

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

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

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

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

Initial Condition are

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

Therefore

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

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

2 solutions p=0, p=1

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

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

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

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

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

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

How am I doing so far?

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

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

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

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

4. I see another problem. You say that $q =2p$. This is true but only when $t = 0$ - not in general!

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

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

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

Then continue in this fashion.

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

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

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

Then continue in this fashion.
Ok, thanks Danny..
I proceed to find q using $q_t = -q*4*[\frac{1}{4t+1}]$? I will tackle it later this evening..

7. Here's the solution that I got (to compare)

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

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

Spoiler:
$u = e^{x+2y-2}$
$
\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 $p=\frac{1}{(4t+1)}$ therefore B(s)=2 $\implies q=\frac{2}{(4t+1)}$

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

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

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

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

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

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

Finally $u(x,y)=e^{x+2y-2}$???

9. 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 \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?

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

11. Originally Posted by Danny
But you have $p_t = -4p^2$ which has two solutions, $p = 0$ and $p = \dfrac{1}{4t + A(s)}$.
It is not unique because it has 2 solutions?

Thanks