# Math Help - Another Charpit PDE

1. Originally Posted by Danny
From post #13, when $t = 0$ then $x = s$ and $y = 0$ so

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

Solve for $C$ and $D$ and sub into $x$ and $y$.
I work out c=s/2 and d=s/2 therefore

$\displaystyle x=\frac{s}{2}(e^t+e^{-t})$ and $\displaystyle y=\frac{s}{2}(e^t-e^{-t})$

2. Now go after $u$.

3. Originally Posted by Danny
Now go after $u$.
I have attached my latest...

4. Originally Posted by bugatti79
I have attached my latest...
continuing on from line 2 on previous post I simplify du/dt to

$\displaystyle \frac{du}{dt}=s (e^t + e^{-t}) \left(s (e^t - e^{-t}) - \frac{s}{2} (e^t - e^{-t})) \right) + s (e^t - e^{-t}) \left(s (e^t + e^{-t}) - \frac{s}{2} (e^t + e^{-t})\right)$

$\displaystyle \frac{du}{dt}=s^2 e^{-2 t} \left(e^{4 t}-1\right)\right\}$

Integrating with respect to t and imposing the IC's giving the constant of integration =0 I get

$\displaystyle u=\frac{s^2}{2} (e^{-2t}+e^{2t})$ For clarity I rewrite u, x and y in terms of s and t as

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

Since $\displaystyle e^t=\frac{s}{x-y} \implies e^{-t}=\frac{x-y}{s}, e^{-2t}=\left [\frac{x-y}{s}\right]^2, e^{2t}=\left [\frac{s}{x-y}\right]^2$

Substituting relevant terms into u i get

$\displaystyle u = \frac{s^4 +(x-y)^4}{2(x-y)^2}$

5. You still have an $s$ in your answer. Notice that

$u = \dfrac{s^2 e^{-2t}}{2} + \dfrac{s^2 e^{2t}}{2} = \dfrac{1}{2} \left(s e^{-t}\right)^2 + \dfrac{1}{2} \left(s e^{t}\right)^2$

Now adding and subtracting $x$ and $y$ gives $x+ y = s e^t,\;\;\; x - y = s e^{-t}$. Now substitute.

6. Originally Posted by Danny
You still have an $s$ in your answer. Notice that

$u = \dfrac{s^2 e^{-2t}}{2} + \dfrac{s^2 e^{2t}}{2} = \dfrac{1}{2} \left(s e^{-t}\right)^2 + \dfrac{1}{2} \left(s e^{t}\right)^2$

Now adding and subtracting $x$ and $y$ gives $x+ y = s e^t,\;\;\; x - y = s e^{-t}$. Now substitute.
No wonder I couldnt get the full solution...I had errors in 2nd last line in post 19

$\displaystyle u(x,y)=x^2+y^2$ which satisfies both IC's and original PDE.

Thanks alot Danny!

7. This is spin off algebraic query and is wrecking my head.

Given $\displaystyle x=\frac{s}{2} (e^t+e^{-t})$ and $\displaystyle y=\frac{s}{2} (e^t-e^{-t})$ Adding we get

$x+y = s e^t$ (1)

and subtracting we get $x-y = s e^{-t}$ (2)

but focusing on (1), I want to find $e^{-t}$ from this...I get

$\displaystyle ln (e^{t})= ln [\frac{x+y}{s}]=ln(x+y)-ln (s) \implies$

$\displaystyle -t = ln (s) - ln(x+y)=ln [\frac{s}{x+y}] \implies$

$\displaystyle e^{-t}=\frac{s}{x+y} \neq \frac{x-y}{s}$???

8. They should equal each other! You have two equations for two unknowns. From your last statement

$e^{-t} = \dfrac{s}{x+y} = \dfrac{x-y}{s}$ gives $s^2 = x^2-y^2$ - there's nothing wrong with this.

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