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Math Help - very basic PDE question

  1. #1
    Super Member Random Variable's Avatar
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    very basic PDE question

    I want to find the general solution to  \frac{\partial u}{\partial t} - 2 \ \frac{\partial u}{\partial x} = 2

    so I let  \alpha = ax + bt and  \beta = cx+dt

    then  \frac{\partial u}{\partial t} = \frac{\partial u}{\partial \alpha} \frac{\partial \alpha}{\partial t} + \frac{\partial u}{\partial \beta} \frac{\partial \beta}{\partial t} = a \frac{\partial u}{\partial \alpha} + c \ \frac{\partial u}{\partial \beta}

    similarly,  \frac{\partial u}{\partial x} = b \frac{\partial u}{\partial \alpha} + d \ \frac{\partial u}{\partial \beta}


    so  (b-2a) \ \frac{\partial u}{\partial \alpha} + (d-2c) \ \frac{\partial u}{\partial \beta} =2

    Would be OK to now do any of the following:

    Let a=1,b=0,c=1, and d=2 (then  \frac{\partial u}{\partial \alpha} = -1 )

    Let a=0,b=1, c=1,and d=2 (then  \frac{\partial u}{\partial \alpha} = 2 )

    Let a=1, b=0, c=1/2, d=1 (then then  \frac{\partial u}{\partial \alpha} = -1 )

    Let a=1, b=2, c= -1, d=0 (then then  \frac{\partial u}{\partial \beta} = 1 )

    etc. ?


    Are there any values that would somehow lead me to the incorrect solution?
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  2. #2
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    Quote Originally Posted by Random Variable View Post
    I want to find the general solution to  \frac{\partial u}{\partial t} - 2 \ \frac{\partial u}{\partial x} = 2

    so I let  \alpha = ax + bt and  \beta = cx+dt

    then  \frac{\partial u}{\partial t} = \frac{\partial u}{\partial \alpha} \frac{\partial \alpha}{\partial t} + \frac{\partial u}{\partial \beta} \frac{\partial \beta}{\partial t} = a \frac{\partial u}{\partial \alpha} + c \ \frac{\partial u}{\partial \beta}

    similarly,  \frac{\partial u}{\partial x} = b \frac{\partial u}{\partial \alpha} + d \ \frac{\partial u}{\partial \beta}


    so  (b-2a) \ \frac{\partial u}{\partial \alpha} + (d-2c) \ \frac{\partial u}{\partial \beta} =2

    Would be OK to now do any of the following:

    Let a=1,b=0,c=1, and d=2 (then  \frac{\partial u}{\partial \alpha} = -1 )

    Let a=0,b=1, c=1,and d=2 (then  \frac{\partial u}{\partial \alpha} = 2 )

    Let a=1, b=0, c=1/2, d=1 (then then  \frac{\partial u}{\partial \alpha} = -1 )

    Let a=1, b=2, c= -1, d=0 (then then  \frac{\partial u}{\partial \beta} = 1 )

    etc. ?


    Are there any values that would somehow lead me to the incorrect solution?
    They will all lead to the same solution. The only choice would be if the Jacobian of the transformation vanishes. The method of characteristics is probably the quickest way

    \frac{dt}{1} = \frac{dx}{-2} = \frac{du}{2}

    and pick in pair

    1) \frac{dt}{1} = \frac{dx}{-2}\;\; \Rightarrow\;\; I_1 = x + 2t

    2) \frac{dt}{1} = \frac{du}{2}\;\; \Rightarrow\;\; I_2 = u - 2t

    Solution I_2 = f(I_1) \; \text{or}\; u = 2t + f(x+2t).
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  3. #3
    Super Member Random Variable's Avatar
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    The solution could also be written as -x + f(x+2t), correct?
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  4. #4
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    Quote Originally Posted by Random Variable View Post
    The solution could also be written as -x + f(x+2t), correct?
    Absolutely! As f(\lambda) is arbitrary you could let f(\lambda) = \tilde{f}(\lambda) - \lambda where mine becomes yours.
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