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Math Help - partial differential equations

  1. #1
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    partial differential equations

    Hi, I just started studying PDE's so sorry for the newbie questions

    I have two questions

    1. Solve
    au_x + bu_y + cu = 0

    2. Solve
    u_x + u_y = 1


    Here is what I have so far for #1:
    Using change of variables
    x' = ax + by
    y' = bx - ay

    u_x = au_{x'} + bu_{y'}
    u_y = bu_{x'} - au_{y'}

    a(au_{x'} + bu_{y'}) + b(bu_{x'} - au_{y'}) + cu = 0
    (a^2 + b^2)u_{x'} + cu = 0

    This is where I get stuck. How do i solve for u in this case?

    For #2, I'm not sure how to start but I want to say its unsolvable because its inhomogeneous? (probably way wrong in saying that haha)
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  2. #2
    MHF Contributor
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    Quote Originally Posted by dlbsd View Post
    Hi, I just started studying PDE's so sorry for the newbie questions

    I have two questions

    1. Solve
    au_x + bu_y + cu = 0

    2. Solve
    u_x + u_y = 1


    Here is what I have so far for #1:
    Using change of variables
    x' = ax + by
    y' = bx - ay

    u_x = au_{x'} + bu_{y'}
    u_y = bu_{x'} - au_{y'}

    a(au_{x'} + bu_{y'}) + b(bu_{x'} - au_{y'}) + cu = 0
    (a^2 + b^2)u_{x'} + cu = 0

    This is where I get stuck. How do i solve for u in this case?

    For #2, I'm not sure how to start but I want to say its unsolvable because its inhomogeneous? (probably way wrong in saying that haha)
    What you're trying to do is introduce a new coordinate system to tranform you pde to one that is easier (i.e. missing a derivative) you'll find in general, this will take you to trying to solve part of the original problem.

    For your first problem, it's separable so

    \frac{du}{u} = - \frac{c}{a^2+b^2}\, dx' so \ln u = -\frac{c}{a^2+b^2} x' + f(y')

    For your second, you can use your change of variables (a=1, b=1) to get

    2 u_{x'} = 1 - again, separable.

    Have you learned the method of characteristics yet? It's really much quicker.
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  3. #3
    MHF Contributor chisigma's Avatar
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    near Piacenza (Italy)
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    Just a minimum of theory...

    ... a linear first order PDE is written as...

    \alpha(x,y,u) \cdot u_{x} + \beta(x,y,u) \cdot u_{y} = \gamma (x,y,u) (1)

    ... and it is equivalent to the system of ordinary equations...

    \frac {dx}{\alpha(x,y,u)} = \frac {dy}{\beta(x,y,u)} =\frac {du}{\gamma(x,y,u)} (2)

    If the solution of (2) is...

    c_{1}= \varphi_{1} (x,y,u)

    c_{2}= \varphi_{2} (x,y,u) (3)

    ... where c_{1} and c_{2} are two 'arbitrary constants', the solution of 1 is...

    G \{ \varphi_{1} (x,y,u), \varphi_{2} (x,y,u) \}=0 (5)

    ... where G(*,*) is arbitrary, with G(*,*) and its partial first order derivatives are continous...

    Let's start with #2 because it is more simple...

    u_{x} + u_{y} = 1 (5)

    Here \alpha=\beta=\gamma=1, so that the system (2) is...

    dx=dy

    dy=du (6)

    ... and it is easily solved...

    c_{1}= x - y

    c_{2}= y - u (7)

    Therefore solution of (5) is...

    G \{ x - y, y - u \}=0 (8)

    And now we will 'attack' #1...

    a\cdot u_{x} + b\cdot u_{y} + c\cdot u =0 (9)

    Here is \alpha=a,\beta=b, \gamma=-c\cdot u , so that (2) is...

    \frac{dx}{a}=\frac{dy}{b}

    \frac{dy}{b}= -\frac {du}{c\cdot u} (10)

    ... the solution of which is...

    c_{1}= \frac{x}{a} - \frac{y}{b}

    c_{2}= u^{\frac{1}{c}}\cdot e^{\frac{y}{b}} (11)

    Solution of (9) is then...

    G \{ \frac{x}{a}-\frac{y}{b},u^{\frac{1}{c}}\cdot e^{\frac{y}{b}} \}=0 (12)

    Kind regards

    \chi \sigma
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