# Another Charpit PDE

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• Feb 8th 2011, 10:38 PM
bugatti79
Another Charpit PDE
Write down the Lagrange-Charpit equations for the equation

$\displaystyle \displaystyle \frac{\partial u}{\partial x}\frac{\partial u}{\partial y}-y\frac{\partial u}{\partial x}-x \frac{\partial u}{\partial y}=0$ and use them to show that

$\displaystyle \displaystyle \frac{d^2 p}{dt^2}=p, \frac{d^2 q}{dt^2}=q,\frac{d^2 x}{dt^2}=x,\frac{d^2 y}{dt^2}=y$

For the first term, I attempted the folllowing
$\displaystyle \displaystyle \frac{d}{dt}\frac{dp}{dt}=\frac{d}{dt}(-F_x-pF_u)=\frac{d}{dt}(-(-q))=\frac{dq}{dt}$

Cannot see any link from this to continue on....

Thanks
• Feb 9th 2011, 05:56 AM
Jester
Since $\displaystyle p_t = q$ and $\displaystyle q_t = p$, then eliminating q, i.e. $\displaystyle p_{tt} = q_t = p.$
• Feb 9th 2011, 09:52 AM
bugatti79
Quote:

Originally Posted by Danny
Since $\displaystyle p_t = q$ and $\displaystyle q_t = p$, then eliminating q, i.e. $\displaystyle p_{tt} = q_t = p.$

your answer made me relook at my derivation...I made a an error in signs, following on from my first message just for clarity....

$\displaystyle \displaystyle \frac{dp}{dt}=-F_x-pF_u=q \implies \frac{d}{dt}\frac{dq}{dt}=q$ similarly $\displaystyle \displaystyle \frac{d}{dt}\frac{dp}{dt}=p$

$\displaystyle \displaystyle \frac{d}{dt}\frac{dx}{dt}=\frac{d}{dt} (q-y)=\frac{dq}{dt}-\frac{dy}{dt}=p-(p-x)=x$ and similarly for $\displaystyle \displaystyle \frac{d}{dt}\frac{dy}{dt}$

Thanks!
• Feb 16th 2011, 02:10 PM
bugatti79
The final part of this involves solving this non linear PDE

$\displaystyle \displaystyle \frac{\partial u}{\partial x}\frac{\partial u}{\partial y}-y\frac{\partial u}{\partial x}-x\frac{\partial u}{\partial y}=0$ with IC's $\displaystyle u(x,0)=x^2$

I have attached my solution as far as determining x and y in terms of s and t. Just hoping some one could check my work before I proceed?

Thanks
• Feb 16th 2011, 02:53 PM
Jester
You have made a mistake which has snowballed. You have

$\displaystyle p_t = q$ which you said $\displaystyle p = qt + A(s).$ However, what you really have is

$\displaystyle p_t = q$ and $\displaystyle q_t = p$ or

$\displaystyle p_{tt} - p = 0$ which has the solution $\displaystyle p = A(s)e^t + B(s)e^{-t}$ which in turn gives

$\displaystyle q = p_t = A(s)e^t - B(s)e^{-t}$

Using your IC you can determine $\displaystyle A(s)$ and $\displaystyle B(s).$ The rest of the problem goes similarly.
• Feb 17th 2011, 11:34 AM
bugatti79
Quote:

Originally Posted by Danny
You have made a mistake which has snowballed. You have

$\displaystyle p_t = q$ which you said $\displaystyle p = qt + A(s).$ However, what you really have is

$\displaystyle p_t = q$ and $\displaystyle q_t = p$ or

$\displaystyle p_{tt} - p = 0$ which has the solution $\displaystyle p = A(s)e^t + B(s)e^{-t}$ which in turn gives

$\displaystyle q = p_t = A(s)e^t - B(s)e^{-t}$

Using your IC you can determine $\displaystyle A(s)$ and $\displaystyle B(s).$ The rest of the problem goes similarly.

Ok, I should have realised the first part of question wouldbe needed in the second part. Here attached is my latest x and y in terms of s and t. Solving for du/dt looks hairy....
• Feb 18th 2011, 03:49 AM
Jester
You made the same mistake in solving for $\displaystyle x$ and you did earlier. Find $\displaystyle x$ and $\displaystyle y$ the same way as $\displaystyle p$ and $\displaystyle q$.
• Feb 18th 2011, 09:45 AM
bugatti79
Quote:

Originally Posted by Danny
You made the same mistake in solving for $\displaystyle x$ and you did earlier. Find $\displaystyle x$ and $\displaystyle y$ the same way as $\displaystyle p$ and $\displaystyle q$.

Mistakes and me go hand in hand!

However,

$\displaystyle \displaystyle \frac{d}{dt}\frac{dx}{dt}=\frac{d}{dt} (q-y)=\frac{dq}{dt}-\frac{dy}{dt}=p-(p-x)=x$ (1)

$\displaystyle \displaystyle \frac{d}{dt}\frac{dy}{dt}=\frac{d}{dt} (p-x)=\frac{dp}{dt}-\frac{dx}{dt}=q-(q-y)=y$ (2)

Solving these 2nd order linear ODE's I will end up with

$\displaystyle \displaystyle x= C(s)e^t +D(s)e^{-t}$ and $\displaystyle \displaystyle y= E(s)e^t+ F(s)e^{-t}$ IC's would give me s=C+D for x equation and 0=E+F for y equation??

Unless I try (from 1 and 2 above)

$\displaystyle \displaystyle \frac{d}{dt} (q-y)=x$

$\displaystyle \displaystyle \frac{d}{dt} (p-x)=y$ because we know p and q....?(Wondering)
• Feb 18th 2011, 09:54 AM
Jester
What you have is good! Use your expressions for $\displaystyle x$ and $\displaystyle y$ (you know $\displaystyle p$ and $\displaystyle q$ already) and subs into

$\displaystyle x_t = q - y,\;\;\;y_t = p - x$ and equate terms involving $\displaystyle e^t$ and $\displaystyle e^{-t}$. This will more conditions on your functions $\displaystyle C, D, E$ and $\displaystyle F.$
• Feb 18th 2011, 10:33 AM
bugatti79
Quote:

Originally Posted by Danny
What you have is good! Use your expressions for $\displaystyle x$ and $\displaystyle y$ (you know $\displaystyle p$ and $\displaystyle q$ already) and subs into

$\displaystyle x_t = q - y,\;\;\;y_t = p - x$ and equate terms involving $\displaystyle e^t$ and $\displaystyle e^{-t}$. This will more conditions on your functions $\displaystyle C, D, E$ and $\displaystyle F.$

$\displaystyle \displaystyle x_t=s(e^t-e^{-t})-E(s) e^t- F(s)e^{-t}$

$\displaystyle \displaystyle y_t=s(e^t+e^{-t})-C(s) e^t- D(s)e^{-t}$

Not sure how to equate terms involving powers ofe considering we have a derivative of x and y on the LHS..?
• Feb 18th 2011, 10:35 AM
Jester
Now subs your expressions for x and y into these.
• Feb 18th 2011, 10:46 AM
bugatti79
Quote:

Originally Posted by Danny
Now subs your expressions for x and y into these.

These expressions?

$\displaystyle \displaystyle \frac{d}{dt} (q-y)=x$

$\displaystyle \displaystyle \frac{d}{dt} (p-x)=y$

Im lost at this stage (Doh)
• Feb 18th 2011, 12:32 PM
Jester
This is what I'd do. You have $\displaystyle x_{tt} = x$ which you solved giving

$\displaystyle x = C(s)e^{t} + D(s)e^{-t}$.

From $\displaystyle x_t = q - y$ then $\displaystyle y = q - x_t$ and since you know $\displaystyle x$ and $\displaystyle q$ then

$\displaystyle y = se^t - se^{-t} - C(s)e^t +D(s)e^{-t}$.

If you use your IC for $\displaystyle x$ and $\displaystyle y$ this will give $\displaystyle C$ and $\displaystyle D$.
• Feb 18th 2011, 01:09 PM
bugatti79
Quote:

Originally Posted by Danny
This is what I'd do. You have $\displaystyle x_{tt} = x$ which you solved giving

$\displaystyle x = C(s)e^{t} + D(s)e^{-t}$.

From $\displaystyle x_t = q - y$ then $\displaystyle y = q - x_t$ and since you know $\displaystyle x$ and $\displaystyle q$ then

$\displaystyle y = se^t - se^{-t} - C(s)e^t +D(s)e^{-t}$.

If you use your IC for $\displaystyle x$ and $\displaystyle y$ this will give $\displaystyle C$ and $\displaystyle D$.

Ok, so since we have x we need to differentiate it with respect to t and then sub into eqn for y.

At t=0 the IC's give D(s)=C(s) implies

$\displaystyle \displaystyle y = se^t - se^{-t} + C(s)(e^t+ e^{-t})$

For the x eqn I get th IC's to give s=E(s)-F(s)
• Feb 19th 2011, 04:25 AM
Jester
From post #13, when $\displaystyle t = 0$ then $\displaystyle x = s$ and $\displaystyle y = 0$ so

$\displaystyle y = -C + D = 0, \; \text{and}\; x = C + D = s$.

Solve for $\displaystyle C$ and $\displaystyle D$ and sub into $\displaystyle x$ and $\displaystyle y$.
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