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Math Help - 2nd Order Liner PDE’s Part 1

  1. #1
    Senior Member bugatti79's Avatar
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    2nd Order Liner PDE’s Part 1

    a) Determine whether the given eqn is hyperbolic, parabolic or elliptic
    b) Determine the subset of R^2
    c) Determine the characteristics curves

     <br />
u_{xx}+2xu_{yx}+u_{yy}+Cos(xy)u_{y}=u(x,y)<br />

     <br />
B^2-AC = x^2-1(1) = x^2-1<br />

    a)

    Hyperbolic if x^2-1 > 0
    ie x^2>1
     <br />
x>1, x<-1<br />

    Elliptic if x^2-1< 0
    ie x^2< 1
     <br />
-1<x<1<br />

    Parabolic if x^2-1 = 0
    ie x^2=1
     <br />
x=-1, x=+1<br />

    b) How do you write the subset query correctly?

    c)

    <br />
\frac{dy}{dx} = \frac{B \pm \sqrt{B^2-AC}}{A} = x \pm \sqrt{x^2-1}<br />

    My initial attempt to solve for y on this by letting u = x^2 -1 and use substitution doesnt seem to work?

    Thanks
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  2. #2
    MHF Contributor

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    Quote Originally Posted by bugatti79 View Post
    a) Determine whether the given eqn is hyperbolic, parabolic or elliptic
    b) Determine the subset of R^2
    c) Determine the characteristics curves

     <br />
u_{xx}+2xu_{yx}+u_{yy}+Cos(xy)u_{y}=u(x,y)<br />

     <br />
B^2-AC = x^2-1(1) = x^2-1<br />

    a)

    Hyperbolic if x^2-1 > 0
    ie x^2>1
     <br />
x>1, x<-1<br />

    Elliptic if x^2-1< 0
    ie x^2< 1
     <br />
-1<x<1<br />

    Parabolic if x^2-1 = 0
    ie x^2=1
     <br />
x=-1, x=+1<br />

    b) How do you write the subset query correctly?
    I think you already have. "Determine the subset of R^2" by itself makes no sense- but I suspect the question was asking you to determine the subsets on which the equation is "hyperbolic", "elliptic", and "parabolic".

    c)

    <br />
\frac{dy}{dx} = \frac{B \pm \sqrt{B^2-AC}}{A} = x \pm \sqrt{x^2-1}<br />

    My initial attempt to solve for y on this by letting u = x^2 -1 and use substitution doesnt seem to work?

    Thanks
    y= \int x\pm\sqrt{x^2- 1}

    To integrate \int \sqrt{x^2- 1}dx use either the trig substitution x= sec(u) or the hyperbolic substitution x= sinh(u). sec^2(x)- 1= tan^2(u) and sinh^2(u)- 1= cosh^2(u) so either will help you get rid of the square root.
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  3. #3
    Senior Member bugatti79's Avatar
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    Quote Originally Posted by HallsofIvy View Post
    I think you already have. "Determine the subset of R^2" by itself makes no sense- but I suspect the question was asking you to determine the subsets on which the equation is "hyperbolic", "elliptic", and "parabolic".


    y= \int x\pm\sqrt{x^2- 1}

    To integrate \int \sqrt{x^2- 1}dx use either the trig substitution x= sec(u) or the hyperbolic substitution x= sinh(u). sec^2(x)- 1= tan^2(u) and sinh^2(u)- 1= cosh^2(u) so either will help you get rid of the square root.
    Thanks HallsofIvy,

    I will tackle that this evening. How did you infer the correct substitution...is it based on experience?

    I mean how does one know if a substitution is correct, trial and error?

    Thanks
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  4. #4
    Senior Member bugatti79's Avatar
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    Quote Originally Posted by HallsofIvy View Post
    I think you already have. "Determine the subset of R^2" by itself makes no sense- but I suspect the question was asking you to determine the subsets on which the equation is "hyperbolic", "elliptic", and "parabolic".


    y= \int x\pm\sqrt{x^2- 1}

    To integrate \int \sqrt{x^2- 1}dx use either the trig substitution x= sec(u) or the hyperbolic substitution x= sinh(u). sec^2(x)- 1= tan^2(u) and sinh^2(u)- 1= cosh^2(u) so either will help you get rid of the square root.
    Ok,

    For the hyperbolic case x^2 -1 >0 I calculate the integral \displaystyle y = \int (x \pm \sqrt{x^2-1})dx to yield 2 solutions

    y_1 = \displaystyle \frac{1}{2}[x^2 +x \sqrt{x^2-1} -ln(x + \sqrt{x^2-1}) +C]
    and

    y_2 = \displaystyle \frac{1}{2}[x^2 -x \sqrt{x^2-1} +ln(x + \sqrt{x^2-1}) +C]


    For the parabolic case x^2-1=0 the integral yields

    y = \frac{x^2}{2} +C

    For the elliptic case x^2-1<0, I am not sure how to evaluate the integral since x^2 is less than 0 and hence there would be a negative under the sqrt...I know the solutions will involve complex numbers but I cant put my finger on it...

    Thanks
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  5. #5
    Senior Member bugatti79's Avatar
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    Quote Originally Posted by bugatti79 View Post
    For the elliptic case x^2-1<0, I am not sure how to evaluate the integral since x^2 is less than 0 and hence there would be a negative under the sqrt...I know the solutions will involve complex numbers but I cant put my finger on it...

    Thanks
    For the elliptic case my attempt is to modify the \sqrt{x^2 -1} term to handle x^<1, i change it to

    \sqrt{-(-(x^2))-1} = i \sqrt{- x^2-1}

    This doesnt look right because if x=0 then we get another imaginary term
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