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Math Help - Differential Heat Equations Question

  1. #1
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    Differential Heat Equations Question

    Can anyone help with the following maths question:



    Show that the function:

    u(t,x)=f(y), where y=((2x)/sqrt(t))

    is a solution of the heat equation:

    partialderivative(u) / partialderivative(t) = partialderivative(u)^2 / partialderivative(x)^2


    (where the partialderivative(u or x)^2 is the second derivative)

    provided that f(y) satisfies 8f''(y) + yf'(y)=0



    Thanks in advance!
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  2. #2
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    Re: Differential Heat Equations Question

    Quote Originally Posted by AlanC877 View Post
    Can anyone help with the following maths question:



    Show that the function:

    u(t,x)=f(y), where y=((2x)/sqrt(t))

    is a solution of the heat equation:

    partialderivative(u) / partialderivative(t) = partialderivative(u)^2 / partialderivative(x)^2


    (where the partialderivative(u or x)^2 is the second derivative)

    provided that f(y) satisfies 8f''(y) + yf'(y)=0



    Thanks in advance!
    You just need to use the chain rule

     \frac{\partial u}{\partial t} =f'(y) \cdot \frac{\partial }{\partial t}\left(\frac{2x}{\sqrt{t}} \right) =-\frac{x}{t^{-\frac{3}{2}}}f'(y)=-\frac{y}{2t}f'(y)

    Now see if you can do the same thing with the x dervaitves.
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  3. #3
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    Re: Differential Heat Equations Question

    Quote Originally Posted by TheEmptySet View Post
    You just need to use the chain rule

     \frac{\partial u}{\partial t} =f'(y) \cdot \frac{\partial }{\partial t}\left(\frac{2x}{\sqrt{t}} \right) =-\frac{x}{t^{-\frac{3}{2}}}f'(y)=-\frac{y}{2t}f'(y)

    Now see if you can do the same thing with the x dervaitves.
    Thanks for the help.

    All my ds in the following are partials.

    I got that du/dt = -f'(y)xt^(-3/2)

    I then found that du/dx = 2f'(y)t^(-1/2)

    From this I said that d^2u/dx^2 = 2f''(y)t^(-1/2)

    I then re-arranged 8f''(y) + yf'(y) = 0 to f''(y) = (-yf'(y))/8

    Substituting into d^2u/dx^2 gives (-yf'(y)t^(-1/2))/4

    Letting y = 2xt^-(1/2) means d^2u/dx^2 = (-f'(y)xt^(-1))/2



    Guessing I'm doing something stupid somewhere?!
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  4. #4
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    Re: Differential Heat Equations Question

    Quote Originally Posted by AlanC877 View Post
    Thanks for the help.

    All my ds in the following are partials.

    I got that du/dt = -f'(y)xt^(-3/2)

    I then found that du/dx = 2f'(y)t^(-1/2)

    From this I said that d^2u/dx^2 = 2f''(y)t^(-1/2)

    I then re-arranged 8f''(y) + yf'(y) = 0 to f''(y) = (-yf'(y))/8

    Substituting into d^2u/dx^2 gives (-yf'(y)t^(-1/2))/4

    Letting y = 2xt^-(1/2) means d^2u/dx^2 = (-f'(y)xt^(-1))/2



    Guessing I'm doing something stupid somewhere?!
    I agree that


    \frac {\partial u}{\partial x}=f'(y)\cdot \left( \frac{2}{\sqrt{t}}\right)=\frac{2}{\sqrt{t}}f'(y)

    but you have forgotten to use the chain rule when you take the 2nd dervaitive. You should get

    \frac{\partial^2 u}{\partial x^2}=\frac{2}{\sqrt{t}}f''(y)\cdot \left( \frac{2}{\sqrt{t}}\right)=\frac{4}{t}f''(y)
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  5. #5
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    Re: Differential Heat Equations Question

    Quote Originally Posted by TheEmptySet View Post
    I agree that


    \frac {\partial u}{\partial x}=f'(y)\cdot \left( \frac{2}{\sqrt{t}}\right)=\frac{2}{\sqrt{t}}f'(y)

    but you have forgotten to use the chain rule when you take the 2nd dervaitive. You should get

    \frac{\partial^2 u}{\partial x^2}=\frac{2}{\sqrt{t}}f''(y)\cdot \left( \frac{2}{\sqrt{t}}\right)=\frac{4}{t}f''(y)
    Ah yes, stupid me :P Thanks a million!
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