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Solving a PDE with elementary methods

  1. Sep 10, 2013 #1
    1. The problem statement, all variables and given/known data

    OK, a PDE: $$a\frac{\partial u}{\partial t} + b \frac{\partial u}{\partial x} = u$$

    we want the general solution.

    2. The attempt at a solution

    So, I'll set up a couple of equations thus:
    r = m11x + m21t
    s = m12x + m22t

    (We have a nice matrix of m here if we want).

    partially differentiating both:[itex] \frac{\partial r}{\partial x}= m_{11},\frac{\partial r}{\partial t}= m_{12}, \frac{\partial s}{\partial x}= m_{21}, \frac{\partial s}{\partial t}= m_{22} [/itex]

    and plugging them back into the original PDE we get: $$ \frac{\partial u}{\partial t}= \frac{\partial u}{\partial r}\frac{\partial r}{\partial t}+\frac{\partial u}{\partial s}\frac{\partial s}{\partial t}= m_{12} \frac{\partial u}{\partial r}+ m_{22}\frac{\partial u}{\partial s} $$ and $$ \frac{\partial u}{\partial x}= \frac{\partial u}{\partial s}\frac{\partial s}{\partial x}+\frac{\partial u}{\partial r}\frac{\partial r}{\partial x}= m_{21} \frac{\partial u}{\partial s}+ m_{11}\frac{\partial u}{\partial r}=u$$

    from there I substitute in to the original PDE and got: $$a\left(m_{12} \frac{\partial u}{\partial r}+ m_{22} \frac{\partial u}{\partial s} \right) + b\left(m_{11} \frac{\partial u}{\partial r}+ m_{21} \frac{\partial u}{\partial s} \right) \Rightarrow (am_{12}+bm_{11}) \frac{\partial u}{\partial r}+ (am_{22}+bm_{21}) \frac{\partial u}{\partial s} =u$$

    So I want to get rid of one of the partials. The simplest way to do that is to set one of the coefficients equal to zero. So:
    am12+bm11=0. That makes m12=-b and m11=a. So my r from above is r=ax-bt.

    That leaves me with
    [itex]\left(am_{22}+bm_{21}\right)\frac{\partial u}{\partial s}=u[/itex]

    and I can set [itex]\left(am_{22}+bm_{21}\right)= 1[/itex] leaving [itex]m_{22} = \frac{1}{a}[/itex] and [itex]m_{21}=0[/itex]. From the equation for [itex]s[/itex]above that leaves us with [itex]s=\frac{1}{a}t[/itex] and we go back to the partials.

    Since we set the coefficient of [itex]\frac{\partial u}{\partial s}[/itex] equal to 1 we are left with [itex]\frac{\partial u}{\partial s}=u[/itex] and that is a simple ODE that reduces to [itex]u=Ce^{-s}[/itex] and
    Since $$u=f(r)e^{-s} = f(ax-bt)e^{-s}= f(ax-bt)e^{-\frac{t}{a}}$$ as our general solution.

    So how did I do? We're still in the beginning of the class, and this was one of the problems that is supposed to explore "easy" methods; I know that the method of characteristics would be better.
     
  2. jcsd
  3. Sep 10, 2013 #2

    pasmith

    User Avatar
    Homework Helper

    The solution of [itex]\frac{\partial u}{\partial s} = u[/itex] is [itex]u = C(r)e^s[/itex]. Aside from that your work is correct.
     
  4. Sep 10, 2013 #3
    Thanks a lot, I wanted to be sure I wasn't doing something dumb. One other question: I noticed that if I put the m terms into a matrix I get: [itex]
    \begin{pmatrix}
    a & -b \\
    0 & \frac{1}{a}\\
    \end{pmatrix}
    [/itex]

    and the determinant is 1. I get the sense that when looking for solutions with this method that could be important. Is there a good way to use this fact to come up with (relatively) easy coefficients (the mmn values)?
     
  5. Sep 10, 2013 #4

    pasmith

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    Homework Helper

    The determinant must be non-zero, or the substitution will not be invertible.
     
  6. Sep 10, 2013 #5
    so I could put any numbers there in the matrix as long as they didn't come out to zero, right? (I realize it would be more complicated sometimes but I am just checking).
     
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