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    Hi, I'm new to this forum, just to be able to ask you this question.

    I have a PDE that I want to solve and has been looking for a solution through the Method of Characteristics, but fail to complete one of the steps.

    Can you please help me:

    Here is my equation:
    d(u(x,t))/dt = (ax^2-(a+b)x+b)d(u(x,t))/dx

    don't know how to write it in math characters but maybe you could,
    here is a picture of it though:
    Name:  Skärmavbild 2013-04-11 kl. 02.59.15.png
Views: 386
Size:  7.8 KB

    Thanks! // Marcus from Sweden
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    Oh I found out you could write in LaTeX, so here it is again:

    [latex]
    \frac{\partial}{\partial t}u(x,t) = (ax^2-(a+b)x+b) \frac{\partial}{\partial x}u(x,t)
    [\latex]

    And when I started with MOC, and created my new variables s,r

    I use the chainrule and get:
    [latex] \frac{du}{dr} = u_t \frac{dt}{dr} + u_x \frac{dx}{dr}
    [\latex]
    which is my initial PDE, if

    [latex] \frac{dt}{dr} = 1 [\latex] and [latex] \frac{dx}{dr} = ax^2-(a+b)x+b
    [\latex]
    but that last equation is quite a challenge, or is it?
    ..because x in this case depend on r and s, right?

    So I don't know how to continue from there! Please help me if you can!
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    Why didn't the Tex code output what it should have??
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    The slashes for the Latex tags are the opposite way around (plus there are some other bugs). Here's a fixed version.

    (Original post by vidomir)
    Oh I found out you could write in LaTeX, so here it is again:

     \frac{\partial}{\partial t} u(x,t) = (ax^2-(a+b)x+b) \frac{\partial}{\partial x} u(x,t)

    And when I started with MOC, and created my new variables s,r

    I use the chainrule and get:

     \frac{du}{dr} = u_t \frac{dt}{dr} + u_x \frac{dx}{dr}

    which is my initial PDE, if

     \frac{dt}{dr} = 1

    and

     \frac{dx}{dr} = ax^2-(a+b)x+b

    but that last equation is quite a challenge, or is it?
    ..because x in this case depend on r and s, right?

    So I don't know how to continue from there! Please help me if you can!
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    (Original post by vidomir)
    Hi, I'm new to this forum, just to be able to ask you this question.

    I have a PDE that I want to solve and has been looking for a solution through the Method of Characteristics, but fail to complete one of the steps.

    Can you please help me:

    Here is my equation:
    d(u(x,t))/dt = (ax^2-(a+b)x+b)d(u(x,t))/dx

    don't know how to write it in math characters but maybe you could,
    here is a picture of it though:
    Name:  Skärmavbild 2013-04-11 kl. 02.59.15.png
Views: 386
Size:  7.8 KB

    Thanks! // Marcus from Sweden
    From equation
    \displaystyle \frac{\partial u}{\partial t}\cdot \frac{\partial x}{\partial u}=ax^2-(a+b)x+b
    that is

    \displaystyle \frac{dx}{dt}=ax^2-(a+b)x+b
    \displaystyle \frac{1}{ax^2-(a+b)x+b}dx=dt
    Integrating
    \displaystyle \frac{1}{a} \int \frac{1}{\frac{b}{a}+x^2-\frac{a+b}{a}x} dx=\int dt
    \displaystyle \int \frac{1}{\frac{b}{a}+\left (x-\frac{a+b}{2a}\right )^2-\frac{(a+b)^2}{4a^2}} dx=a\int dt
    \displaystyle \int \frac{1}{\frac{(a-b)^2}{4a^2}-\left (x-\frac{a+b}{2a}\right )^2} dx=-a\int dt
    \displaystyle \int \frac{1}{1-\left (\frac{x-\frac{a+b}{2a}}{\frac{a-b}{2a}}\right )^2} dx=-\frac{(a-b)^2}{4a} \int dt
    \frac{a-b}{2a} artanh \left (\frac{2a}{a-b}x-\frac{a+b}{a-b}\right ) =-\frac{(a-b)^2}{4a}t +C
    and simplify
    use
    artanh y=\frac{1}{2} ln\left |\frac{1+y}{1-y}\right |
    for |y|<1
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    Thanks ztibor, I wasn't able to see your answer until today and will have no time to go through it right now but I will definitely take a look at it soon. Seems to be the way to go with that.
 
 
 
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