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I'm trying to show that for the equation $y''+a_1y' + a_2y = Ae^{i\omega x}$, there is a solution of $\phi(x)$ of the form $$\phi(x)=\frac{A}{\vert p(i\omega)\vert}e^{i(\omega x - \alpha)}$$

where $ p(i\omega)=\vert p(i\omega)\vert e^{ix}$, and $p(i\omega) \neq 0$. The last statement is confusing and I'm unsure how to interpret it. Why would $p(i\omega)=\vert p(i\omega)\vert e^{ix}$? I think that $$p(i \omega)=\omega^2+a_1i\omega+a_2$$

which is a complex number with $Re = \omega ^2+a_2$, and $Im = a_1\omega$. It does not seem like we will ever end up to such an identity, with the standard definition of complex norm. To be precise, we have that $$\vert p(i\omega)\vert=\sqrt{(\omega^2 +a_2)^2 + (a_1\omega)^2}$$

As for validating the solution, one would compute $L(\frac{A}{\vert p(i\omega)\vert}e^{i(\omega x - \alpha)})$, take the term with the norm as a common factor. I imagine the norm is cancelled at some point. What am I missing here?

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  • $\begingroup$ The function $\phi (x)$ is meant to be the particular solution. The problem is to show that this is the case. $\endgroup$ Jun 15, 2021 at 7:06
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    $\begingroup$ Well, no. I didn't really consider doing that. But IF I had the full analytic solution you suggest, I could show that it can be cast in the form of $\phi (x)$. It does seem like a worthwhile idea. $\endgroup$ Jun 15, 2021 at 12:21
  • $\begingroup$ If $p$ is a function of $\mathrm iω$, how can $p(\mathrm iω)=|p(\mathrm iω)|\mathrm e^{\mathrm ix}$ since the RHS depends on another variable $x$? And could $α$ also depend on $ω$? Also, the ODE in the title is different from that in the body. $\endgroup$ Jul 10, 2021 at 11:25

2 Answers 2

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I'm sure $p(i\omega)=\vert p(i\omega)\vert e^{ix}$ should be $p(i\omega)=\vert p(i\omega)\vert e^{i\alpha}$

The form $e^{i\omega x}$ is preserved by derivatives so for the particular solution try:

$$y_p = K e^{i\omega x} \tag{1}$$

into

$$ y''+a_1y' + a_2y = Ae^{i\omega x} \tag{2}$$

factor out $e^{i\omega x}$

$$-K {\omega}^2 + iKa_1\omega + Ka_2 = A \tag{3}$$

$$K = \frac{A}{-{\omega}^2 + ia_1\omega + a_2} \tag{4} $$

$$ y_p = \frac{Ae^{i\omega x}}{-{\omega}^2 + ia_1\omega + a_2} \tag{5} $$

The characteristic equation of this DE is: $$p(z) = z^2 + a_1z + a_2 \tag{6}$$

So

$$p(i\omega) = -{\omega}^2 + ia_1\omega + a_2 \tag{7}$$

In $re^{i\theta}$form:

$$p(i\omega) = \sqrt{(a_2-{\omega}^2)^2 + (a_1\omega)^2 } \, e^{i \arctan{\left(\frac{a_1\omega}{a_2-{\omega}^2}\right)}} \tag{8}$$

Substitute $(8)$ into $(5)$

$$ y_p = \frac{Ae^{i\omega x}}{\sqrt{(a_2-{\omega}^2)^2 + (a_1\omega)^2 } \, e^{i \arctan{\left(\frac{a_1\omega}{a_2-{\omega}^2}\right)}}} \tag{9} $$

$$ y_p = \frac{Ae^{i(\omega x - \alpha)}}{|p(i\omega)|} \tag{10} $$

Where

$$\alpha = arg(p(i\omega)) = \arctan{\left(\frac{a_1\omega}{a_2-{\omega}^2}\right)}\tag{11}$$

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There are 3 unknown parameters $\;a_1,a_2,\omega,\;$ in the given ODE $$y''+a_1y+a_2=Ae^{i\omega x},\tag1$$ which influence to the form of the particular solution.

The common solution is defined by the characteristic equation $$p(k)=k^2+a_1k+a_2=0,\tag2$$ which can be obtained by the substitution $\;y=e^{kx}\;$ to the homogenius ODE $(1)$. Then the root $\;k=i\omega\;$ can have multiplicities of $\;m=0,1,2.\;$

If $\;\color{brown}{\mathbf{p(i\omega)\not=0,\quad(m=0)}},\;$
then a particular solution has the form of $\;\varphi^\,_0(x)=c^\,_0e^{i\omega x}\;$ (see also the Arthur's answer).

In this case, $$c^\,_0\,p(i\omega)=A.\tag3$$ Using of the exponential form $$p(i\omega)=|p(i\omega)|e^{ia},\quad\text{where}\quad a=\arg p(i\omega),\tag4$$ allows to write $$c^\,_0 = \dfrac{Ae^{-ia}}{|p(i\omega)|,}$$ $$\color{green}{\mathbf{\varphi_0(x)=\dfrac A{|p(i\omega)|}\,e^{i(\omega x-a)}.\tag5}}$$

If $\;\color{brown}{\mathbf {\big(p(i\omega)=0\big)\wedge\big(p'(i\omega)\not=0\big),\quad (m=1)}},\;$
then the solution $\;(5)\;$ belongs to the common ones (the resonant case).

In this case, a particular solution has the form of $$\varphi_1(x) = c^\,_1 xe^{i\omega x},\tag6$$ wherein $$\varphi_1'(x) = c^\,_1(1+i\omega x)e^{i\omega x},$$ $$\varphi_1''(x)= i\omega c^\,_1(2+i\omega x)e^{i\omega x},$$ $$c^\,_1\big(2i\omega-\omega^2 x+a_1(1+i\omega x)+a_2 x\big)=A,$$ $$c^\,_1\big(2i\omega+a_1\big)=A.\tag7$$

Using of the exponential form $$p'(i\omega)=|p'(i\omega)|e^{ib},\quad\text{where}\quad b=\arg p'(i\omega),\tag8$$ allows to write $$c^\,_1 = \dfrac{Ae^{-ib}}{|p'(i\omega)|,}$$ $$\color{green}{\mathbf{\varphi_1(x)=\dfrac A{|p'(i\omega)|}\,xe^{i(\omega x-b)}.\tag9}}$$

If $\;\color{brown}{\mathbf {\big(p(i\omega)=0\big)\wedge\big(p'(i\omega)=0\big),\quad (m=2)}},\;$ then $\;a_1=-2i\omega,\;a_2=-\omega^2,\;$ $$y''-2i\omega y'-\omega^2y=Ae^{i\omega x},\tag{10}$$

wherein the solutions $\;(5),(9)\;$ belong to the common ones (the resonant case).

In this case, a particular solution takes the form of $$\varphi_2(x) = c^\,_2 x^2e^{i\omega x},\tag{11}$$ wherein $$\varphi_2'(x) = c^\,_2(2x+i\omega x^2)e^{i\omega x},$$ $$\varphi_2''(x)= c^\,_2(2+4i\omega x-\omega^2 x^2)e^{i\omega x},$$ $$c^\,_2\big(2+4i\omega x-\omega^2x^2-4i\omega x+2\omega^2 x^2-\omega^2x^2\big)=A,$$ $$c^\,_2=\dfrac A2,$$ $$\color{green}{\mathbf{\varphi_2(x)=\dfrac A2 x^2\,e^{i\omega x}.\tag{12}}}$$

$\color{brown}{\textbf{Results.}}$

The particular solution of $(1)$ is $\;y=\varphi_m(x),\;$ where $\;m\;$ is defined upper.

The particular solution in the form $(5)$ is correct if and only if $$p(iw)= -\omega^2+a_1\cdot i\omega+a_2\not=0.$$

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