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Solving z'(t) = az(t)

  1. Nov 4, 2014 #1
    So [itex]z'(t) = x'(t) + i y'(t)[/itex], I just want to make sure that what I'm going to do is OK. I'm trying to solve [itex]z' = az [/itex] where [itex]a = \alpha + i \beta [/itex]

    [itex] z'(t) = az(t) \Rightarrow x'(t) + iy'(t) = a(x(t) + iy(t)) \Rightarrow x'(t) = a x(t), y'(t) = a y(t) [/itex]

    If you could give me a justification that'll be nice. I feel like what I'm doing is instinctual rather than "I'm following the rules" if that makes any sense.

    I mean I've already broken this into the system

    [itex] \begin{pmatrix} x' \\ y' \end{pmatrix} = \begin{pmatrix} \alpha & -\beta\\ \beta & \alpha \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} [/itex]

    and I know what the general solution looks like here. I'm just trying to compare the two methods. I'm taking an introductory course in Dynamical Systems. Also any suggested books that discusses Solving Differential equations with complex variables will be great.
     
  2. jcsd
  3. Nov 4, 2014 #2

    Mark44

    Staff: Mentor

    Is there some reason you can't solve this directly?

    If z' = az, then all solutions are given by z = Keat.
     
  4. Nov 4, 2014 #3
    Rules for differential equations usually extend to the complex plane. You don't need to split z into its real and complex parts. If you really want to, you could split the solution into a real and a complex part using Euler's formula.
     
  5. Nov 4, 2014 #4
    By directly do you mean separation of variables? [itex] \dfrac{dz}{z} = a dt [/itex]. First we need to define complex integration which would require line integration. I think that's what you mean right? Thing is, antiderivatives are a little more trickier in [itex]\mathbb{C}[/itex] than they are with straight up real valued functions. If someone could elaborate on this more that'll be great. I guess what I'm trying to say is, how does one justify the separation of variables, since log(z) is multivalued?

    I mean you do get [itex]z(t) = z_0 e^{a t}[/itex] when you solve [itex]x' = ax, y'= ay[/itex]. I'm just trying to justify this all.

    Bigfoot I don't understand what you mean?

    I mean suppose we don't know that [itex]z(t) = z_0 e^{at}[/itex], what do you mean by euler's formula? Do you mean let [itex]z=re^{i \theta}[/itex]?
     
  6. Nov 4, 2014 #5

    Mark44

    Staff: Mentor

    If z(t) = z0eat, it's easy to verify that z' = az, which should be a reasonable justification that z(t) is a solution of the diff. equation. As far as the ln function having multiple values when the argument is complex, you can choose the principal value.
    ##e^{at} = e^{(\alpha + i\beta) t} = e^{\alpha t} \cdot e^{i \beta t} = e^{\alpha t}(cos(\beta t) + i sin(\beta t)##. I believe that's what @bigfooted was referring to.
     
  7. Nov 4, 2014 #6
    Thanks Mark, shortly after I saw that I did the work and Now I'm convinced x' = ax and y' = ay is now justified.

    Now suppose we chose the principal value log, will that then justify the separation of variables? I feel like what I'm about to do is a little naive, but let's give it a shot

    [itex] \displaystyle{\int \dfrac{dz}{z}} = Log (z) = \ln r + i \Theta [/itex]

    So here's where I'm at

    [itex] \displaystyle{\int \dfrac{dz}{z}} = \int a dt \Rightarrow Log(z) = at + C [/itex] , let [itex]z = re^{i\theta}, C = x_0 + iy_0[/itex]

    [itex] \ln r + i\Theta = at + C = (\alpha + i \beta)t + x_0 + iy_0[/itex]

    So

    [itex] \ln r = \alpha t + x_0 \Rightarrow r = e^{\alpha t + x_0}, \Theta = \beta t + y_0[/itex]

    [itex] z = e^{\alpha t + x_0}e^{i(\beta t + y_0)} = e^{x_0 + iy_0} e^{(\alpha +i \beta)t} = z_0 e^{a t} [/itex]

    I suppose I get what I want if I reduce the argument to the principal value. In other words, I may justify separation of variables almost.
     
  8. Nov 5, 2014 #7
    Not my area of expertise, but here it goes:
    the complex logarithm is defined as [itex]Log(z)=\ln r + i\theta = \ln |z| + i \arg z[/itex].
    Now let's show that some results 'look' the same in complex and real variables:
    [itex] e^{Log z} = e^{\ln |z| + i \arg z} = e^{\ln |z|} e^{i \arg z} = |z|e^{i \arg z}=z[/itex]

    OK. We'll need that later.

    So we know that [itex]e^{Log z} = z[/itex]. Now let's take the derivative on both sides:
    [itex](e^{Log z})' = (z)'[/itex]
    [itex]e^{Log z}\cdot(Log z)' = 1[/itex]
    So [itex](Log z)' = \frac{1}{e^{Log z}} = \frac{1}{z}[/itex]

    so: [itex]\int {\frac{1}{z}dz} = Log z[/itex]

    The derivative of a complex exponential can be derived in a similar way:
    [itex](e^{zt})' = ze^{zt}[/itex]

    You can now find the solution of the ODE [itex]z'=\alpha z[/itex]
     
  9. Nov 5, 2014 #8
    Yeah that looks better!
     
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