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Partial Diff. Eqn 1st order watch

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    I have to solve 3 * du/dx - 4*du/dy + u = 1 + x/3
    with initial condition u(0,y)= y ^2 using the method of characteristics

    The above derivatives are partial derivatives


    My working until now:

    We have dx/ds = 3 =>x=3s+x(0) => s=(x-x(0))/3

    dy/ds = -4 => y= -4s + y(0) => y= -4(x-x(0))/3 + y(0) {using that s=(x-x(0))/3}

    du/ds = 1+ x/3 - u

    But since s=(x-x(0))/3 => du/ds = 1+s+x(0)/3 - u

    The above were our characteristic equations.

    From our initial conditions now, x(0)=0 so x=3s

    Also y= -4(x-x(0))/3 + y(0) from above becomes y+ 4/3 * x = y(0)

    Furthermore du/ds = 1+s - u {since x(0)=0}

    Solving using the integrating factor yields that u(s) = s + c*e^(-s), where c is an arbitrary constant!

    Moreover u(s=0) = u(0) = y(0) ^2 {using the initial condition u(0,y) = y^2}

    Finally, since x = 3s from above, we deduce u(x/3) = x/3 + (y + 4/3 * x)^2 * e^(-x/3)


    Then I don't know how to proceed to find u(x,y)


    May someone review my work and help me please?

    Thanks!!!
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    (Original post by Darkprince)

    Finally, since x = 3s from above, we deduce u(x/3) = x/3 + (y + 4/3 * x)^2 * e^(-x/3)


    Then I don't know how to proceed to find u(x,y)


    May someone review my work and help me please?

    Thanks!!!
    This is the answer! You've got u only in terms of x and y.
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    But I have to deduce u(x,y) not u(x/3)! How do I do that?
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    It's your notation which is confusing you.

    Let

     P(x,y,u)\frac{\partial u}{\partial x} + Q(x,y,u)\frac{\partial u}{\partial y} = R(x,y,u)

    We parameterise along characteristics with s and our intial data with t so our initial data is given by
     \gamma (t) = (x(t),y(t),u(t)) . We set s=0 wlog on the characteristics so  x(0,t)=x(t),  y(0,t)=y(t),  u(0,t)=u(t)

    We find  x=x(s,t), y=y(s,t), z=z(s,t) by solving

     \frac{\partial x}{\partial s} = P

     \frac{\partial y}{\partial s} = Q

     \frac{\partial u}{\partial s} = R


    For this problem  \gamma (t) = (0,t,t^2)

    we solve for x,y, denoting our constants of intergration A(t) and B(t)

     x(s,t) = 3s + A(t)

y(s,t) = -4s + B(t)

    Using in our initial data (ie setting s = 0) we find A(t) = 0, B(t)=t

     x = 3s

y = -4s + t  

so t = y + \frac{4x}{3}

    so now solve for u

     u(s,t) = s + C(t)e^{-s}

    using our initial data  u(0,t) = C(t) = t^2 so  u(s,t) = s + t^2e^{-s}

    but  s = \frac{x}{3} and  t = y + \frac{4x}{3} then sub in to u(s,t) to get  u(x,y) = \frac{x}{3} +  (y + \frac{4x}{3})^2e^{-\frac{x}{3}} which is the answer you have got.
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    To be honest your notation confuses me even more, may you review my answer using my notation please? At the stage
    u(x/3) = x/3 + (y + 4/3 * x)^2 * e^(-x/3)
    since s=x/3

    what should I say?

    Thanks again for your time, I appreciate it



    With a bit of reconsideration if I use my notation and just say that u=s+c*e^(-s)
    then say s=x/3
    then say u(s=0) = u(0) = c = y(0)^2 and then that y(0)=y+4/3 * x, so c= (y+4/3 * x)^2

    And then I just say u(x,y)=..... {without saying before u(x/3)=..... etc }

    If I use my notation and the way I just used, am I correct?
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    When finding the characteristic equation for u you are solving

     \frac{\partial u(s,t)}{\partial s} = 1+ \frac{x}{3} - u

    where t is the parameter along your initial data curve, so  u=u(s,t), not just a function of s!

    You've written u as a function of x and y which is what you want. You do not need to do anything else!
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    So if I just use my notation and the exact way I used in my first post am I correct? Nothing more?
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    (Original post by Darkprince)
    So if I just use my notation and the exact way I used in my first post am I correct? Nothing more?
    You are correct.
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    Thanks for your time and help!

    Kind regards
 
 
 
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