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Math Help - Tricky Partial Differential problem

  1. #1
    Senior Member bugatti79's Avatar
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    Tricky Partial Differential problem

    Hi all, I wish to verify

    u_t=ku_x_x with the shorthand notations being u_t=\frac{\partial u}{\partial t} and u_x_x=\frac{\partial^2 u}{\partial x^2} for the following expression

    u(x,t)=\frac{1}{2\sqrt {\pi kt}} \int^{\infty}_{-\infty} e^{-(x-\xi)^2/4kt} f(\xi)d\xi where k is a constant and f is continous on the real line ( I actually dont understrand what that 'f' bit means)

    Can anyone give me a clue as how to tackle this one? Do I integrate the integrand first and then find the derivatives u_t=ku_x_x? If so, how does one tackle the integrand with the infinity signs etc?

    Thanks, tips will be appreciated.
    Looking forward to hearing from some one.
    Bugatti79
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  2. #2
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    Use "Leibniz' rule":
    \frac{\partial}{\partial x} \int_{\alpha(x,t)}^{\beta(x,t)} f(x, t, \xi)d\xi= \frac{\partial\beta}{\partial x} f(x, t, \beta(x,t))- \frac{\partial\alpha}{\partial x} f(x, t, \alpha(x,t))+ \int_{\alpha(x,t)}^{\beta(x,t)}\frac{\partial f}{\partial x} d\xi.
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  3. #3
    Senior Member bugatti79's Avatar
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    Hi HallsofIvy,

    This exercise is taken out of 'beginning partial differential equations' by Peter V O Neil. It states the solution is achieved using chain rule method. No mention of Leibniz method unless both are same thing?
    1) Is this equivalent to the chain rule method?

    2) Does anyone know this book? I find it difficult to grasp. I was hoping to find an easier book on beginning partial differential equations. Any suggestions will be appreciated.


    Thanks
    bugatti79
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  4. #4
    Senior Member bugatti79's Avatar
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    Tricky Partial Differential problem

    Loooking at that Leibniz expression...i dont know how one would tackle the problem because I believe the derivative of infinity is undefined ( I am referring to \frac{\partial\beta}{\partial x} ). A function must be continous to be differentiable....
    bugatti79
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  5. #5
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    Quote Originally Posted by bugatti79 View Post
    Hi all, I wish to verify

    u_t=ku_x_x with the shorthand notations being u_t=\frac{\partial u}{\partial t} and u_x_x=\frac{\partial^2 u}{\partial x^2} for the following expression

    u(x,t)=\frac{1}{2\sqrt {\pi kt}} \int^{\infty}_{-\infty} e^{-(x-\xi)^2/4kt} f(\xi)d\xi where k is a constant and f is continous on the real line ( I actually dont understrand what that 'f' bit means)

    Can anyone give me a clue as how to tackle this one? Do I integrate the integrand first and then find the derivatives u_t=ku_x_x? If so, how does one tackle the integrand with the infinity signs etc?

    Thanks, tips will be appreciated.
    Looking forward to hearing from some one.
    Bugatti79
    So you just want to verify that u_t=ku_{xx}?

    You should not take the integral, you should just differentiate. Since you're differentiating with respect to t,x then you can move the derivatives inside the integral. WRT to t, you just have simple product rule. And then similarly WRT to x. Compute it out (it will be slightly long/messy) and see that they're equal.

    EDIT: Leibniz' Rule gives you the same thing here, since the first two terms in the RHS are zero, leaving just what I said.
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  6. #6
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    Quote Originally Posted by bugatti79 View Post
    Loooking at that Leibniz expression...i dont know how one would tackle the problem because I believe the derivative of infinity is undefined ( I am referring to \frac{\partial\beta}{\partial x} ). A function must be continous to be differentiable....
    bugatti79
    Why do you think this?
    \frac{d}{dx}\lim _ {a \to \infty} a=\lim _ {a \to \infty}\frac{d}{dx}a=\lim _ {a \to \infty}0=0
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  7. #7
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    Quote Originally Posted by bugatti79 View Post
    Hi all, I wish to verify

    u_t=ku_x_x with the shorthand notations being u_t=\frac{\partial u}{\partial t} and u_x_x=\frac{\partial^2 u}{\partial x^2} for the following expression

    u(x,t)=\frac{1}{2\sqrt {\pi kt}} \int^{\infty}_{-\infty} e^{-(x-\xi)^2/4kt} f(\xi)d\xi where k is a constant and f is continous on the real line ( I actually dont understrand what that 'f' bit means)

    Can anyone give me a clue as how to tackle this one? Do I integrate the integrand first and then find the derivatives u_t=ku_x_x? If so, how does one tackle the integrand with the infinity signs etc?

    Thanks, tips will be appreciated.
    Looking forward to hearing from some one.
    Bugatti79
    For \partial _{t} I think an argument using the dominated convergence theorem will give you the interchanging of integral and partial derivative. For \partial _{xx} just note that if t is fixed your integral is just a convolution of functions, and so you can derive only the exponential inside the integral.
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  8. #8
    Senior Member bugatti79's Avatar
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    Im confused as to where to start
    If i dont need to evalute the integral then i can re-arrange
    u(x,t)=\frac{1}{2\sqrt {\pi kt}} \int^{\infty}_{-\infty} e^{-(x-\xi)^2/4kt} f(\xi)d\xi to

    u(x,t)= \int^{\infty}_{-\infty} \frac{e^{-(x-\xi)^2/4kt}}{2\sqrt {\pi kt}} f(\xi)d\xi which looks likes solving using quotient rule? Is that correct? How is this a product?

    Thanks
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  9. #9
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    The quotient and product rule are the same, which ever you prefer.
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  10. #10
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    Quote Originally Posted by bugatti79 View Post
    Hi all, I wish to verify

    u_t=ku_x_x with the shorthand notations being u_t=\frac{\partial u}{\partial t} and u_x_x=\frac{\partial^2 u}{\partial x^2} for the following expression

    u(x,t)=\frac{1}{2\sqrt {\pi kt}} \int^{\infty}_{-\infty} e^{-(x-\xi)^2/4kt} f(\xi)d\xi where k is a constant and f is continous on the real line ( I actually dont understrand what that 'f' bit means)

    Can anyone give me a clue as how to tackle this one? Do I integrate the integrand first and then find the derivatives u_t=ku_x_x? If so, how does one tackle the integrand with the infinity signs etc?

    Thanks, tips will be appreciated.
    Looking forward to hearing from some one.
    Bugatti79
    BTW - f(x) is the intial condition so u(x,0) = f(x).
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  11. #11
    Senior Member bugatti79's Avatar
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    Tricky Partial Differential problem

    I dont know where I am gone wrong. u(x,t)= \int^{\infty}_{-\infty} \frac{e^{-(x-\xi)^2/4kt}}{2\sqrt {\pi kt}} f(\xi)d\xi

    Using the chain rule I calculate for u_x, u_x_x
    u_x=\frac{\frac{-2(x-\xi)e^{-(x-\xi)^2/4kt}}{4kt}}{2\sqrt {\pi k t}} and

    u_x_x=\frac{\frac{4(x-\xi)e^{-(x-\xi)^2/4kt}}{4kt}}{2\sqrt {\pi k t}}

    and the product rule for u_t since t is on both numerator and denominator
    \frac {2\sqrt {\pi k t} (\frac{(x-\xi)^2}{4kt^2}e^{-(x-\xi)^2/4kt}) - (e^{-(x-\xi)^2/4kt} {(\pi k )}{\pi k t}^{-1/2})}{4\pi kt}

    Can anyone confirm im going correct so far?
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  12. #12
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    I think you forgot some parentheses in u_t
    I didn't check u_{xx}, but it looks like it simplified nicely for you.
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  13. #13
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    Quote Originally Posted by bugatti79 View Post
    I dont know where I am gone wrong. u(x,t)= \int^{\infty}_{-\infty} \frac{e^{-(x-\xi)^2/4kt}}{2\sqrt {\pi kt}} f(\xi)d\xi

    Using the chain rule I calculate for u_x, u_x_x
    u_x=\frac{\frac{-2(x-\xi)e^{-(x-\xi)^2/4kt}}{4kt}}{2\sqrt {\pi k t}} and

    u_x_x=\frac{\frac{4(x-\xi)e^{-(x-\xi)^2/4kt}}{4kt}}{2\sqrt {\pi k t}}

    and the product rule for u_t since t is on both numerator and denominator
    \frac {2\sqrt {\pi k t} (\frac{(x-\xi)^2}{4kt^2}e^{-(x-\xi)^2/4kt}) - (e^{-(x-\xi)^2/4kt} {(\pi k )}{\pi k t}^{-1/2})}{4\pi kt}

    Can anyone confirm im going correct so far?
    Check your u_{xx} again (product rule).
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  14. #14
    Senior Member bugatti79's Avatar
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    Tricky Partial Differential problem

    Still cant get u_t=ku_x_x

    I dont know wheter im getting my algebra in a twist or my derivatives are still wrong.
    u(x,t)= \int^{\infty}_{-\infty} \frac{e^{-(x-\xi)^2/4kt}}{2\sqrt {\pi kt}} f(\xi)d\xi

    I recalculate
    u_x=\frac{\frac{-2(x-\xi)e^{-(x-\xi)^2/4kt}}{4kt}}{2\sqrt {\pi k t}}
    and for u_x_x I use product rule in the numerator

    u_x_x=\frac{\frac{(-2(x-\xi))}{4kt}\frac{(-2(x-\xi))}{4kt}(e^\frac{-(x-\xi)^2}{4kt})+e^\frac{-(x-\xi)^2}{4kt}(\frac{-2}{4kt})}{2(\pi kt)^\frac{1}{2}}
    and qoutient rule for u_t
    \frac {2\sqrt {\pi k t} (\frac{(x-\xi)^2}{4kt^2}e^{-(x-\xi)^2/4kt}) - (e^{-(x-\xi)^2/4kt} {(\pi k )}({\pi k t})^{-1/2})}{4\pi kt}

    If these derivatives can be confirmed correct then I know the rest is basic algebra....
    Thanks
    bugatti79
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  15. #15
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    I can - I got the same.
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