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When we have a linear homogeneous ODE, suppose $$M\; \ddot{x} + K \;x =0$$

Where $x$ is a real function of some parameter. We can analyze, for a complex function of the same parameter $$M\; \ddot{z} + K \;z =0$$

As the coefficients are real, if $z_0$ is a solution then so is $\bar{z}_0$. From linearity we construct the real solutions ($x$'s) by taking the linear combinations, $\large \frac{z_0+\bar{z}_0}{2}$, $\large\frac{z_0-\bar{z}_0}{2i}$.

Can someone explain to me why the second expression in the last paragraph ($\large\frac{z_0-\bar{z}_0}{2i}$) is valid, as up until now, the ODE was linear over the Reals, and if $M$ or $K$ were chosen to be complex, then $\bar{z}_0$ would not be a solution.

My question is, why is it valid to create a linear combination with complex coefficients?

Addition:

From the comments I get that I havent been able to explain my doubt, so I will try more.

  1. The solutions $x_i$ of the ODE form a linear vector space over the field of reals, i.e if $x_1$ and $x_2$ are solutions then so is $ax_1+bx_2$ for some $a,b\in\mathbb{R}$

  2. The solutions of $M\; \ddot{z} + K \;z =0$ form a linear vector space over the field of complex numbers, i.e for $z_1$ and $z_2$ to be solutions of the ODE, $\alpha z_1 + \beta z_2$ where $\alpha,\beta\in \mathbb{C}$ is also a solution.

  3. It is apparent that we can choose $\alpha,\beta = -i/2$ for $z_i$ and $\bar{z}_i$ to create some $x_i$

Everything is fine in this sequence. Except that I am having trouble understanding how step 3 relates to step 1.


Thanks to anon for answering in the comment.

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  • $\begingroup$ What do you mean by "the linearity of $x$ was over the Reals"? $x$ is not supposed to be linear in any way : if $M$ and $K$ are real, then $M = \overline M$ and $K = \overline K$, this is the only thing that allows us to conclude that $z_0$ is a solution implies that $\overline {z_0}$ is too. $\endgroup$ – Patrick Da Silva Aug 9 '11 at 1:37
  • $\begingroup$ @Patrick This is a linear ODE in $x$ with real coefficients. That is given. However, if the coefficients are complex, then for each complex solution, its conjugate is not a solution. I have edited the question to make it clearer. $\endgroup$ – kuch nahi Aug 9 '11 at 1:39
  • $\begingroup$ Yeah I know that, but you said "As the coefficients are real", so I assumed it. Is that somehow not assumed? Because if it is, there is absolutely no problem. If the coefficients might be complex, then this technique obviously does not work. $\endgroup$ – Patrick Da Silva Aug 9 '11 at 1:42
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    $\begingroup$ When constructing real solutions, you only use a linear combination over the reals. The $1/2i$ is not considered part of the coefficient: it's part of the solution being multiplied by a real scalar. If you have no qualms with $z$ and $\overline{z}$ being in the numerator of a real solution when they are both complex, why get in a tizzy over the denominator? You might as well say an expression like $a(b/a)$ isn't an integer because $b/a$ isn't whole. $\endgroup$ – anon Aug 9 '11 at 3:12
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    $\begingroup$ @anon, may I suggest you expand your comment to an answer so OP can accept it? $\endgroup$ – Gerry Myerson Aug 9 '11 at 3:21
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The two solutions $\frac{z+\overline{z}}{2}$ and $\frac{z-\overline{z}}{2i}$ are both real and linearly independent, so all real solutions must be linear combinations of them with coefficients over the field of reals. The $1/2i$ part is a cancelling factor that is intrinsic to the second given solution in order to make it real, it is not considered part of the real coefficients. One might as well call the two given solutions $u$ and $v$ and then write all real solutions as combinations $au+bv,$ and $a,b\in\mathbb{R}$ so one doesn't have to explicitly see an $i$ in the expression (or $z$ or $\overline{z}$, both of which are complex, for that matter).

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