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Math Help - pde satisfy with conditions

  1. #1
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    pde satisfy with conditions

    Hello everybody.

    I am new and I have one new question

    i am trying these question, i must to show:
    y=log((x-a)^2+(z-b)^2))

    satisfies

    \frac{\partial^2y}{\partial{x^2}}+\frac{\partial^2  {y}}{\partial{z^2}}=0

    except at (a,b)

    I differentiated twice for z and for x (chain rule and quotient) and put in equation to get:
    \frac{2x+2y+4(x-a)^2+4(z-b)^2}{((x-a)^2 +(z-b)^2)^2}

    I think it is ok and it does not work for a, b (as it goes to infinity) but is it enough to do this and say equation is satisfied? sorry if is obvious question! or have i done calculation wrong?
    Thank you
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  2. #2
    Super Member flyingsquirrel's Avatar
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    Hello
    Quote Originally Posted by jescavez View Post
    [...]
    or have i done calculation wrong?
    [...]
    Yes, I think you made a mistake when you differentiated (x-a)^2 with respect to x and (z-b)^2 with respect to z.

    Remember the chain rule : (u\circ v)'=v'\times u'\circ v. Here, v(x)=x-a \implies  v'(x)=1 and u(x)=x^2 \implies  u'(x)=2x :

    \frac{\partial }{\partial x}\left[(x-a)^2 \right]=v'(x)\times u'\circ v(x)= 1\cdot 2(x-a)=2(x-a)\neq 2x

    The same goes for \frac{\partial }{\partial z}\left[(z-b)^2 \right].

    Does it help ?
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  3. #3
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    Hello flying squirrel

    thank you for the reply, yes I get same for \frac{\partial }{\partial x} term but i need differentiate twice. then i multiply by \frac{\partial{y}}{\partial{u}}=\frac{1}{((x-a)^2+(z-b)^2)} where u=(x-a)^2+(z-b)^2
    and i have \frac{2(x-a)}{(x-a)^2+(z-b)^2)} for dy/dx. i quotient rule this and have (i am correcting now as i notice a mistake before) \frac{\partial^2{y}}{\partial{x^2}}=\frac{2((x-a)^2+(z-b)^2)+4(x-a)^2}{((x-a)^2+(z-b)^2)^2}
    do you think this is right now (i do same for z). so 2x term (and 4(x-a)^2) is from second derivative. so finish formula is:
    \frac{4((x-a)^2+(z-b)^2)+4(x-a)^2+4(z-b)^2}{((x-a)^2+(z-b)^2)^2}
    i think is right now but now for x=a and z=b it not fail.
    what do you think?
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  4. #4
    Super Member flyingsquirrel's Avatar
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    Quote Originally Posted by jescavez View Post
    Hello flying squirrel

    thank you for the reply, yes I get same for \frac{\partial }{\partial x} term but i need differentiate twice. then i multiply by \frac{\partial{y}}{\partial{u}}=\frac{1}{((x-a)^2+(z-b)^2)} where u=(x-a)^2+(z-b)^2
    and i have \frac{2(x-a)}{(x-a)^2+(z-b)^2)} for dy/dx. i quotient rule this and have (i am correcting now as i notice a mistake before) \frac{\partial^2{y}}{\partial{x^2}}=\frac{2((x-a)^2+(z-b)^2)+4(x-a)^2}{((x-a)^2+(z-b)^2)^2}
    do you think this is right now (i do same for z). so 2x term (and 4(x-a)^2) is from second derivative. so finish formula is:
    \frac{4((x-a)^2+(z-b)^2)+4(x-a)^2+4(z-b)^2}{((x-a)^2+(z-b)^2)^2}
    i think is right now but now for x=a and z=b it not fail.
    what do you think?
    I think that you forgot that the quotient rule tells us that \left(\frac{f}{g}\right)'=\frac{f'g{\color{red}-}g'f}{g^2}.

    With this minus sign \frac{4((x-a)^2+(z-b)^2)+4(x-a)^2+4(z-b)^2}{((x-a)^2+(z-b)^2)^2} becomes \frac{4((x-a)^2+(z-b)^2){\color{red}-}4(x-a)^2{\color{red}-}4(z-b)^2}{((x-a)^2+(z-b)^2)^2}=0 as expected.
    Last edited by flyingsquirrel; June 22nd 2008 at 01:43 PM. Reason: ...
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  5. #5
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by flyingsquirrel View Post
    I thinks that you forgot that the quotient rule tells us that ({\color{red}\bold{uv}})'=\frac{u'v{\color{red}-}v'u}{v^2}.

    With this minus sign \frac{4((x-a)^2+(z-b)^2)+4(x-a)^2+4(z-b)^2}{((x-a)^2+(z-b)^2)^2} becomes \frac{4((x-a)^2+(z-b)^2){\color{red}-}4(x-a)^2{\color{red}-}4(z-b)^2}{((x-a)^2+(z-b)^2)^2}=0 as expected.
    ..
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  6. #6
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    Thank you flying squirrel and maths stud, I see now.

    How do I do the thank you for a post?
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  7. #7
    Super Member flyingsquirrel's Avatar
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    Quote Originally Posted by jescavez View Post
    Thank you flying squirrel and maths stud, I see now.

    How do I do the thank you for a post?
    Click on the button of the post you want to... thank. (it's at the bottom right of the post)
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  8. #8
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    Ok. am i being blind i just see quote button no thanks button.
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  9. #9
    Super Member flyingsquirrel's Avatar
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    Quote Originally Posted by jescavez View Post
    Ok. am i being blind i just see quote button no thanks button.
    The "thanks" button does not appear on your posts since you can't thank yourself (): try to find it on someone else's posts.
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  10. #10
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    I was not logged in! thank you now :-)
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