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Homework Help: PDE Question

  1. Aug 22, 2011 #1
    1. The problem statement, all variables and given/known data

    sin(y)[itex]\frac{ \partial u}{ \partial x} + \frac{ \partial u}{ \partial y} = (xcos(y)-sin^2(y))u[/itex]

    where ln(u(x,[itex]\frac{\pi}{2})) = x^2 + x - \frac{\pi}{2}[/itex] for [itex]-1 \leq x \leq 3[/itex]

    determine the characteristic curves in the xy plane and draw 3 of them
    determine the general solution of this pde
    determine the particular solution and show that u(0,0) = e
    sketch the region of influence of the initial conditions

    2. Relevant equations



    3. The attempt at a solution

    dy/dx = 1/siny this gives K = x + cosy
    if i draw this out is it a cos wave along the y axis and to draw 3 of them i let k = 0,1,2?
    what happens as i add k?
     
  2. jcsd
  3. Aug 22, 2011 #2
    to get the general solution [itex] \frac{ du}{ dx} [/itex] = [itex] \frac{xcos(y) - sin^2(y))u}{ sin y}[/itex]

    for which i get ln(u) = [itex] x^2 \frac{cosy}{2siny} -xsiny +F(x-cosy)[/itex]

    so then i transfered in the given value[itex] y= \pi/2 [/itex] just to check if it works out but it doesnt. what am i doing wrong here. Am i even going about this the right way...
     
    Last edited: Aug 22, 2011
  4. Aug 22, 2011 #3

    hunt_mat

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    Write the PDE as:
    [tex]
    \dot{x}=\sin y,\quad\dot{y}=1,\quad\dot{u}=(x\cos y-\sin^{2}y)u
    [/tex]
    Integrate the y equation to obtain [itex]y=s+\pi /2[/itex], substitute this into the equation for x and integrate to obtain [itex]x=\sin s+r[/itex], where we parametrised the initial data as x=r, and [itex]u(r)=r^{2}+r-\pi /2[/itex], insert this into the final equation and integrate as a function of s, the characteristic variable. This will be your general solution.
     
  5. Aug 22, 2011 #4
    Hi Mat,
    Thanks a million.
    I dont understand where the pi/2 came from?the constant of integration..
     
  6. Aug 22, 2011 #5

    hunt_mat

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    If s is the characteristic variable, then [itex]y(0)=\pi /2[/itex].

    Do you have my notes on first order PDEs?
     
  7. Aug 22, 2011 #6
    Thanks, that makes sense.

    are your notes online?
     
  8. Aug 22, 2011 #7

    hunt_mat

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    As the initial conditions (at s=0) are paramatrised by [itex]x=r[/itex], [itex]y=\pi /2[/itex] and [itex]\log u=r^{2}+r-\pi /2[/itex].

    My notes are on this site somewhere, but if you can't find them, drop me a PM a with you e-mail and I will e-mail them to you.

    I have noticed that you do ask quite a few questions about 1st order PDEs.
     
  9. Aug 23, 2011 #8
    we use t as s here so i'm just going to leave it as t

    [itex]\frac{dx}{dt} = sin(y)[/itex]

    [itex]\frac{dy}{dt} = 1 [/itex]
    y=t+k
    @t=0 y=[itex]\frac{\pi}{2}[/itex]
    =>y=t+[itex]\frac{\pi}{2}[/itex]

    then
    [itex]\frac{dx}{dt} = sin(t+ \frac{\pi}{2})[/itex]
    Simplifies to [itex]\frac{dx}{dt} = cos(t)[/itex]
    so x = sint + k

    you parameterised x=r and subed into eq 3.

    do i sub in the values i've got for x and y into eq3? no mater what i sub in it doesn't work out.
     
  10. Aug 23, 2011 #9

    hunt_mat

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    You have dx/dt=sin(t-pi/2) and using the initial values you also have x+sin(y)=r. So what you do is substitute x and y for their respective functions and integrate.
     
  11. Aug 23, 2011 #10
    should that be dx/dt=sin(t+pi/2)
     
  12. Aug 23, 2011 #11

    hunt_mat

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    Sorry, you're right.
     
  13. Aug 23, 2011 #12
    [itex]\frac{du}{dt} = xcos(t+\frac{\pi}{2}) -sin^2(t+\frac{\pi}{2})[/itex]

    [itex]\frac{du}{dt} = -xsin(t) -sin^2(t+\frac{\pi}{2})[/itex]

    [itex]\frac{du}{dt} = -(sin(t) + k)sin(t) -sin^2(t+\frac{\pi}{2})[/itex]

    [itex]\frac{du}{dt} = -(sin^2(t) + ksin(t) -sin^2(t+\frac{\pi}{2})[/itex]

    and then integrate?
     
  14. Aug 23, 2011 #13
    sorry there should be a u at the end of all them eqs
     
  15. Aug 23, 2011 #14

    hunt_mat

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    Expand the trig function and then integrate.
     
  16. Aug 23, 2011 #15
    i do it all out and i get ln(u) = t-ksin(t)
     
  17. Aug 23, 2011 #16

    hunt_mat

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    We have [itex]y=t+\pi /2[/itex] and [itex]x=\sin t+k[/itex], inserting these in the equation for u to obtain:
    [tex]
    \frac{\dot{u}}{u}=(\sin t+k)(-\sin t)-\sin^{2}t
    [/tex]
    Now to integrate this.
     
  18. Aug 23, 2011 #17
    is that not the same as [itex]\frac{\dot{u}}{u}=(-2sin^2 t + ksin t) [/itex]

    for which i get [itex] -(t-\frac{1}{2}Sin 2t) + kcos t [/itex] and then i throw in the values for k and t but...
     
  19. Aug 23, 2011 #18

    hunt_mat

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    That is, you should get the solution:
    [tex]
    \log u=\frac{1}{2}\sin 2t-t-k\cos t+k^{2}+k-\frac{\pi}{2}
    [/tex]
    Upon using the initial condition for u. Now insert [itex]t=y-\pi /2[/itex] and [itex]k=x+\cos y[/itex] and you have your solution.
     
  20. Aug 23, 2011 #19
    is the intregral of [tex]-2sin^2 t [/tex] not [itex]\frac{1}{2}(t-\frac{1}{2}sin(2t) [/itex] and ksin(t) =kcos(t)...i dont follow where you got the k^2 from...
     
  21. Aug 23, 2011 #20

    hunt_mat

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    Oh yes, you're correct my mistake. The k^2 term comes from the initial condition for u which is:
    [tex]
    u\left( k,\frac{\pi}{2}\right) =k^{2}+k-\frac{\pi}{2}
    [/tex]
     
  22. Aug 23, 2011 #21
    right so we've integrated but i still dont get why you just added k^2+k-pi/2 on to the end of the integral...
     
  23. Aug 23, 2011 #22
    oh feck, sorry its obvious. Thanks a million
     
  24. Aug 23, 2011 #23

    hunt_mat

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    From here it should be easy to compute u(0,0) to get the answer required. My notes have notes of examples like this.
     
  25. Aug 23, 2011 #24
    u(0,0) is grand, no bothers there.
    The region of influence though, is there anything in your notes about it?
     
  26. Aug 23, 2011 #25

    hunt_mat

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    Sadly no, you will have to do a bit of googling for that.
     
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