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Math Help - QL PDE's - Checking IC's

  1. #1
    Senior Member bugatti79's Avatar
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    QL PDE's - Checking IC's

    All,

    I wish to check that u(x,2x)=0 for the following problem

    (y-u)\frac{\partial u}{\partial x}+(u-x)\frac{\partial u}{\partial y}=(x-y),

    to which the general solution is defined implicitly by

    x+y+u=3 \sqrt { 2( \frac {(x-y)^2}{2}-\frac{2xu +2yu-u^2}{2})

    Using IC's , I reduce it to

    3x+u=3 \sqrt{u^2-6xu-x^2}

    reducing further, I get

    3x+u=3 \sqrt {3x+- \sqrt {2} 2x}

    I am having difficulty showing u=0 from here. Mathematica shows 2 solutions u=0 and I think the other is 15x/6.

    Thanks
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  2. #2
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    How did you obtain that general solution? I obtained

    x^2 + y^2 + u^2 = f(x+y+u).
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  3. #3
    Senior Member bugatti79's Avatar
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    Quote Originally Posted by Danny View Post
    How did you obtain that general solution? I obtained

    x^2 + y^2 + u^2 = f(x+y+u).
    I get x+y+u like you do on the RHS but the LHS is based on the ratio

    \frac{d(x-y)}{(x+y-2u)}=\frac{du}{(x-y)}

    I can write the derivation later if you require.

    Thanks
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  4. #4
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    Quote Originally Posted by bugatti79 View Post
    All,

    I wish to check that u(x,2x)=0 for the following problem

    (y-u)\frac{\partial u}{\partial x}+(u-x)\frac{\partial u}{\partial y}=(x-y),

    to which the general solution is defined implicitly by

    x+y+u=3 \sqrt { 2( \frac {(x-y)^2}{2}-\frac{2xu +2yu-u^2}{2})

    Using IC's , I reduce it to

    3x+u=3 \sqrt{u^2-6xu-x^2}

    reducing further, I get

    3x+u=3 \sqrt {3x+- \sqrt {2} 2x}

    I am having difficulty showing u=0 from here. Mathematica shows 2 solutions u=0 and I think the other is 15x/6.

    Thanks
    I ask because your "general" solution has no arbitrary function in it.
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  5. #5
    Senior Member bugatti79's Avatar
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    Quote Originally Posted by Danny View Post
    I ask because your "general" solution has no arbitrary function in it.
    Sorry, the general solution should be

    x+y+u=f ( \frac {(x-y)^2}{2}-\frac{2xu +2yu-u^2}{2})

    typo on my part,

    Thanks
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  6. #6
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    Quote Originally Posted by bugatti79 View Post
    Sorry, the general solution should be

    x+y+u=f ( \frac {(x-y)^2}{2}-\frac{2xu +2yu-u^2}{2})

    typo on my part,

    Thanks
    How did you end with this part

    \dfrac {(x-y)^2}{2}-\dfrac{2xu +2yu-u^2}{2}

    in your solution?
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  7. #7
    Senior Member bugatti79's Avatar
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    Quote Originally Posted by Danny View Post
    How did you end with this part

    \dfrac {(x-y)^2}{2}-\dfrac{2xu +2yu-u^2}{2}

    in your solution?
    Danny,

    Find attached the furthest I can go with it. Difficulty seeing how to getu=0 on last line.
    Hope I havent made any mistakes!

    Thanks
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  8. #8
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    I don't get your last line - why did you do that?
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  9. #9
    Senior Member bugatti79's Avatar
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    Quote Originally Posted by bugatti79 View Post
    Danny,

    Find attached the furthest I can go with it. Difficulty seeing how to getu=0 on last line.
    Hope I havent made any mistakes!

    Thanks
    I was looking for the roots of u under the square root sign in order to simplify it further. Did I need to?
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  10. #10
    Senior Member bugatti79's Avatar
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    Quote Originally Posted by bugatti79 View Post
    I was looking for the roots of u under the square root sign in order to simplify it further. Did I need to?
    Ok, I think I have it. My above comment and the last line of attachment is incorrect. Here goes...from the second last line...

    3x+u-3\sqrt{u^2-6xu+x^2}=0
    u^2-6xu+x^2=\frac{u^2}{9}+\frac{2xu}{3}+x^2}
    \frac{8u^2}{9}-\frac{20xu}{3}=0

    Unifying u^2 and solving the quadratic yields 2 solutions
    u=0 and u=15x/2 which is what mathematica outputs.
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