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Math Help - Partial Differential Equation

  1. #1
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    Partial Differential Equation

    Consider the equation
    4y^2u_{xx} + 2(1-y^2)u_{xy} - u_{yy}- \frac{2y}{1 + y^2}(2u_x - u_y) = 0

    Derive the transformation  \xi = x + 2y, \eta = x -\frac{2}{3} y^3 ,show that the equation reduces to  \omega_{\xi \eta} = 0 where  \omega(\xi, \eta) = u(x(\xi, \eta), y(\xi,eta)) , and find the general solution.

    I am a part-time maths student, so I havent done any calculus courses for about a year and a half (although I did a numerical analysis course last semester that looked at numerical methods for ODE's and PDE's) and I'm really rusty
    This is my last semester, so I would really appreciate it if someone could help me understand this.
    Thanks
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  2. #2
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    Quote Originally Posted by RoyalFlush View Post
    Consider the equation
    4y^2u_{xx} + 2(1-y^2)u_{xy} - u_{yy}- \frac{2y}{1 + y^2}(2u_x - u_y) = 0

    Derive the transformation  \xi = x + 2y, \eta = x -\frac{2}{3} y^3 ,show that the equation reduces to  \omega_{\xi \eta} = 0 where  \omega(\xi, \eta) = u(x(\xi, \eta), y(\xi,eta)) , and find the general solution.

    I am a part-time maths student, so I havent done any calculus courses for about a year and a half (although I did a numerical analysis course last semester that looked at numerical methods for ODE's and PDE's) and I'm really rusty
    This is my last semester, so I would really appreciate it if someone could help me understand this.
    Thanks
    First off, let's determine the type.

    If A = 4y^2,\;\; B = 2(1-y^2),\;\; C = -1 then

    B^2 - 4AC = 4(1-y^2)^2+ 16y^2 = 4(1+y^2)^2 > 0 so it's hyperbolic. This means under a change of variable we should be able to transform you PDE to a standard form, i.e.

    u_{rs} + lots = 0 (I'm using r and s instead of \xi and \eta. lots - lower order terms). We now introduce first and secord order transforms. If r = r(x,y), \;\;s = s(x,y) then

    u_x = u_r r_x + u_s s_x,\;(1a)
    u_y = u_r r_y + u_s s_y,\;(1b)

    u_{xx} = u_{rr} r_x^2 + 2 u_{rs} r_x s_x + u_{ss} s_x^2  +u_r r_{xx} + u_s s_{xx},\;(1c)

    u_{xy} = u_{rr} r_x r_y + u_{rs}\left( r_x s_y + r_y s_x\right) + u_{ss} s_x s_y + u_r r_{xy} + u_s s_{xy},\;(1d)

    u_{yy} = u_{rr} r_y^2 + 2 u_{rs} r_y s_y + u_{ss} s_y^2  +u_r r_{yy} + u_s s_{yy},\;(1e).

    Substitute these into your PDE and group terms

    \left( 4y^2 r_x^2 +2(1-y^2)r_x r_y - r_y^2\right) u_{rr}
    + \left( 4y^2 r_x s_x +2(1-y^2)(r_x s_y + r_y s_x) - r_y s_y\right) u_{rs} + \left( 4y^2 s_x^2 +2(1-y^2)s_x s_y - s_y^2\right) u_{ss} + lots = 0.

    In order to hit the target we need the coefficient of u_{rr} and u_{ss} to vanish. Thus,

    4y^2 r_x^2 +2(1-y^2)r_x r_y - r_y^2 = 0,\;\;\;<br />
4y^2 s_x^2 +2(1-y^2)s_x s_y - s_y^2 = 0,

    or

    \left(2r_x - r_y\right)\left(2y^2 r_x + r_y\right) = 0,\;\;\;\left(2s_x - s_y\right)\left(2y^2 s_x + s_y\right) = 0

    They are the same so choose one for r and the other for s, i.e.

    2r_x - r_y = 0,\;\;\;2y^2 s_x + s_y = 0.

    These are first oder PDEs, so by the method of characteristics, we fiind the solutions

    r = R(x + 2y),\;\;\; s = S(x - \frac{2}{3}y^3)

    which we choose the function R and S simple. So

    r = x + 2y,\;\;\; s = x - \frac{2}{3}y^3 (as you said).

    Now put these into your transformations (1) and put all of these into you PDE and simply. See how that goes.
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  3. #3
    Senior Member bugatti79's Avatar
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    Quote Originally Posted by Danny View Post
    First off, let's determine the type.

    If A = 4y^2,\;\; B = 2(1-y^2),\;\; C = -1 then

    B^2 - 4AC = 4(1-y^2)^2+ 16y^2 = 4(1+y^2)^2 > 0 so it's hyperbolic. This means under a change of variable we should be able to transform you PDE to a standard form, i.e.
    I am also self studying these types of DE. According to the book I follow, the transformations are obtained by letting
    \xi(x,y)=k be an integral of \frac{dy}{dx}=\frac{B+\sqrt{B^2-AC}}{A} (1)

    \eta(x,y)=K be an integral of \frac{dy}{dx}=\frac{B-\sqrt{B^2-AC}}{A} (2)

    I attempted to integrate (1) above
    I get \frac{dy}{dx}=\frac{(1-y^2)+\sqrt{(1-y^2)^2+4y^2}}{4y^2}
    (Note that coefficients of the general form of the PDE for uxx, uxy and uyy are A, 2B and C repsectively.)



    Is this integral solvable, suitable way to derive the transformations? Your method seems easier? It works for easier problems I solved, this is the hardest I have come across.

    Thanks
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  4. #4
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    At first glance your problem seems difficult but notice that

    (1-y^2)^2 + 4y^2 = (1+y^2)^2

    That helps tremendously!
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  5. #5
    Senior Member bugatti79's Avatar
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    Quote Originally Posted by Danny View Post
    At first glance your problem seems difficult but notice that

    (1-y^2)^2 + 4y^2 = (1+y^2)^2

    That helps tremendously!
    Well spotted, thanks. So I can get the same transformations. I have a query regarding the last part of your method. How did you arrive at the following (particularly the captial R and S) using the method of characterictics?


    I have used the method of characteristics for different problems such as

    2u_x-u_y=0

    where u_x=u_rr_x+ u_ss_x but I dont know how you would use this method for

    2r_x - r_y = 0,\;\;\;2y^2 s_x + s_y = 0

    Thanks
    BTW, the substitution of the derived terms back into the PDE is extremely labourious!!
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  6. #6
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    Quote Originally Posted by bugatti79 View Post
    Well spotted, thanks. So I can get the same transformations. I have a query regarding the last part of your method. How did you arrive at the following (particularly the captial R and S) using the method of characterictics?


    I have used the method of characteristics for different problems such as

    2u_x-u_y=0

    where u_x=u_rr_x+ u_ss_x but I dont know how you would use this method for

    2r_x - r_y = 0,\;\;\;2y^2 s_x + s_y = 0

    Thanks
    BTW, the substitution of the derived terms back into the PDE is extremely labourious!!
    To solve a PDE of the form

    a(x,y)u_x + b(x,y)u_y = 0

    the method of characteristics is to solve

    \dfrac{dx}{a(x,y)} = \dfrac{dy}{b(x,y)},\;\;\;du = 0

    Solve the first (an ODE) and second gives

    I(x,y) = c_1,\;\;\;  u = c_2

    and solution of your PDE is

    c_2 = F(c_1) or u = F(I(x,y)).
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  7. #7
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    Quote Originally Posted by Danny View Post
    First off, let's determine the type.

    If A = 4y^2,\;\; B = 2(1-y^2),\;\; C = -1 then

    B^2 - 4AC = 4(1-y^2)^2+ 16y^2 = 4(1+y^2)^2 > 0 so it's hyperbolic. This means under a change of variable we should be able to transform you PDE to a standard form, i.e.

    u_{rs} + lots = 0 (I'm using r and s instead of \xi and \eta. lots - lower order terms). We now introduce first and secord order transforms. If r = r(x,y), \;\;s = s(x,y) then

    u_x = u_r r_x + u_s s_x,\;(1a)
    u_y = u_r r_y + u_s s_y,\;(1b)

    u_{xx} = u_{rr} r_x^2 + 2 u_{rs} r_x s_x + u_{ss} s_x^2  +u_r r_{xx} + u_s s_{xx},\;(1c)

    u_{xy} = u_{rr} r_x r_y + u_{rs}\left( r_x s_y + r_y s_x\right) + u_{ss} s_x s_y + u_r r_{xy} + u_s s_{xy},\;(1d)

    u_{yy} = u_{rr} r_y^2 + 2 u_{rs} r_y s_y + u_{ss} s_y^2  +u_r r_{yy} + u_s s_{yy},\;(1e).

    Substitute these into your PDE and group terms

    \left( 4y^2 r_x^2 +2(1-y^2)r_x r_y - r_y^2\right) u_{rr}
    + \left( 4y^2 r_x s_x +2(1-y^2)(r_x s_y + r_y s_x) - r_y s_y\right) u_{rs} + \left( 4y^2 s_x^2 +2(1-y^2)s_x s_y - s_y^2\right) u_{ss} + lots = 0.

    In order to hit the target we need the coefficient of u_{rr} and u_{ss} to vanish. Thus,

    4y^2 r_x^2 +2(1-y^2)r_x r_y - r_y^2 = 0,\;\;\;<br />
4y^2 s_x^2 +2(1-y^2)s_x s_y - s_y^2 = 0,

    or

    \left(2r_x - r_y\right)\left(2y^2 r_x + r_y\right) = 0,\;\;\;\left(2s_x - s_y\right)\left(2y^2 s_x + s_y\right) = 0

    They are the same so choose one for r and the other for s, i.e.

    2r_x - r_y = 0,\;\;\;2y^2 s_x + s_y = 0.

    These are first oder PDEs, so by the method of characteristics, we fiind the solutions

    r = R(x + 2y),\;\;\; s = S(x - \frac{2}{3}y^3)

    which we choose the function R and S simple. So

    r = x + 2y,\;\;\; s = x - \frac{2}{3}y^3 (as you said).

    Now put these into your transformations (1) and put all of these into you PDE and simply. See how that goes.
    I am really having trouble with this one, and I have a mid-term that I am worried about. I'm still having trouble getting out the general solution and showing that the equation reduces to \omega_{\xi \eta} = 0 where \omega(\xi, \eta) = u(x(\xi, \eta), y(\xi,eta))

    My teacher said we will have a question very similar to this on the mid-term, and it will be the most difficult on the test. Where do I go from here?
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  8. #8
    Senior Member bugatti79's Avatar
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    Quote Originally Posted by RoyalFlush View Post
    I am really having trouble with this one, and I have a mid-term that I am worried about. I'm still having trouble getting out the general solution and showing that the equation reduces to \omega_{\xi \eta} = 0 where \omega(\xi, \eta) = u(x(\xi, \eta), y(\xi,eta))

    My teacher said we will have a question very similar to this on the mid-term, and it will be the most difficult on the test. Where do I go from here?
    Where exactly are you having trouble?
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