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Thread: Another Charpit PDE

  1. #16
    Senior Member bugatti79's Avatar
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    Quote Originally Posted by Danny View Post
    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 $\displaystyle y$.
    I work out c=s/2 and d=s/2 therefore

    $\displaystyle \displaystyle x=\frac{s}{2}(e^t+e^{-t})$ and $\displaystyle \displaystyle y=\frac{s}{2}(e^t-e^{-t})$
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  2. #17
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    Now go after $\displaystyle u$.
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  3. #18
    Senior Member bugatti79's Avatar
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    Quote Originally Posted by Danny View Post
    Now go after $\displaystyle u$.
    I have attached my latest...
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  4. #19
    Senior Member bugatti79's Avatar
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    Quote Originally Posted by bugatti79 View Post
    I have attached my latest...
    continuing on from line 2 on previous post I simplify du/dt to

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

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

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

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

    $\displaystyle \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 \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

    $\displaystyle \displaystyle u = \frac{s^4 +(x-y)^4}{2(x-y)^2}$
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  5. #20
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    You still have an $\displaystyle s$ in your answer. Notice that

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

    Now adding and subtracting $\displaystyle x$ and $\displaystyle y$ gives $\displaystyle x+ y = s e^t,\;\;\; x - y = s e^{-t}$. Now substitute.
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  6. #21
    Senior Member bugatti79's Avatar
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    Quote Originally Posted by Danny View Post
    You still have an $\displaystyle s$ in your answer. Notice that

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

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

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

    Thanks alot Danny!
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  7. #22
    Senior Member bugatti79's Avatar
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    Exclamation

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

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

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

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

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

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

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

    $\displaystyle \displaystyle e^{-t}=\frac{s}{x+y} \neq \frac{x-y}{s}$???
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  8. #23
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    They should equal each other! You have two equations for two unknowns. From your last statement

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