A first-order differential equation with complex coefficients

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This little investigation has been motivated by Exercise 7B, Challenge in CP2 §7.2 Second-order homogeneous differential equations.

 

Question 1. What happens with a first-order differential equation with complex constant coefficients? $$\begin{array}{rl}(a_1+i{a}_2)\frac{\text{d}y}{\text{d}x}+(b_1+i{b}_2)y& =0\end{array}$$

 

Method 1. Using matrix notation: Let $$\begin{array}{r}y(x)=\psi (x)+i\chi (x)\end{array}$$ Substituting this into the differential equation gives $$\begin{array}{rl}(a_1+i{a}_2)({\psi }^{\prime }+i{\chi }^{\prime })+(b_1+i{b}_2)(\psi +i\chi )& =0\\ [(a_1{\psi }^{\prime }+b_1\psi )-(a_2{\chi }^{\prime }+b_2\chi )]+i[(a_2{\psi }^{\prime }+b_2\psi )+(a_1{\chi }^{\prime }+b_1\chi )]& =0\end{array}$$ The real and imaginary parts yield a set of simultaneous equations $$\begin{array}{rl}\{\begin{array}{l}{a}_1{\psi }^{\prime }+b_1\psi =a_2{\chi }^{\prime }+b_2\chi \\ a_2{\psi }^{\prime }+b_2\psi =-a_1{\chi }^{\prime }-b_1\chi \end{array}& \text{or}\{\begin{array}{c}{a}_1{\psi }^{\prime }-a_2{\chi }^{\prime }=-b_1\psi +b_2\chi \\ a_2{\psi }^{\prime }+a_1{\chi }^{\prime }=-b_2\psi -b_1\chi \end{array}\\ \\ \left(\begin{array}{cc}{a}_1& b_1\\ a_2& b_2\end{array}\right)\left(\begin{array}{c}{\psi }^{\prime }\\ \psi \end{array}\right)=\left(\begin{array}{cc}{a}_2& b_2\\ -a_1& -b_1\end{array}\right)\left(\begin{array}{c}{\chi }^{\prime }\\ \chi \end{array}\right)& \text{or}\left(\begin{array}{cc}{a}_1& -a_2\\ a_2& a_1\end{array}\right)\frac{\text{d}}{\text{d}x}\left(\begin{array}{c}\psi \\ \chi \end{array}\right)=\left(\begin{array}{cc}-b_1& b_2\\ -b_2& -b_1\end{array}\right)\left(\begin{array}{c}\psi \\ \chi \end{array}\right)\end{array}$$

The latter appears to be more useful. $$\begin{array}{rl}\left(\begin{array}{cc}{a}_1& -a_2\\ a_2& a_1\end{array}\right)\frac{\text{d}}{\text{d}x}\left(\begin{array}{c}\psi \\ \chi \end{array}\right)& =-\left(\begin{array}{cc}{b}_1& -b_2\\ b_2& b_1\end{array}\right)\left(\begin{array}{c}\psi \\ \chi \end{array}\right)\\ \\ \Rightarrow \frac{\text{d}}{\text{d}x}\left(\begin{array}{c}\psi \\ \chi \end{array}\right)& =-{\left(\begin{array}{cc}{a}_1& -a_2\\ a_2& a_1\end{array}\right)}^{-1}\left(\begin{array}{cc}{b}_1& -b_2\\ b_2& b_1\end{array}\right)\left(\begin{array}{c}\psi \\ \chi \end{array}\right)\\ & =-\frac{1}{a_1^2+a_2^2}\left(\begin{array}{cc}{a}_1& a_2\\ -a_2& a_1\end{array}\right)\left(\begin{array}{cc}{b}_1& -b_2\\ b_2& b_1\end{array}\right)\left(\begin{array}{c}\psi \\ \chi \end{array}\right)\\ & =-\frac{1}{a_1^2+a_2^2}\left(\begin{array}{cc}{a}_1{b}_1+a_2{b}_2& -(a_1{b}_2-a_2{b}_1)\\ a_1{b}_2-a_2{b}_1& a_1{b}_1+a_2{b}_2\end{array}\right)\left(\begin{array}{c}\psi \\ \chi \end{array}\right)\end{array}$$ Using the vector notation, $$\begin{array}{r}\mathbf{\text{a}}=\left(\begin{array}{c}{a}_1\\ a_2\end{array}\right)\text{and}\mathbf{\text{b}}=\left(\begin{array}{c}{b}_1\\ b_2\end{array}\right)\end{array}$$ we can express $$\begin{array}{rl}\mathbf{\text{a}}\cdot \mathbf{\text{b}}& =a_1{b}_1+a_2{b}_2=p\\ (\mathbf{\text{a}}\times \mathbf{\text{b}})\cdot \mathbf{\text{k}}& =a_1{b}_2-a_2{b}_1=q\end{array}$$ Then the coupled differential equation reads $$\begin{array}{rlrl}& & \frac{\text{d}}{\text{d}x}\left(\begin{array}{c}\psi \\ \chi \end{array}\right)& =-\frac{1}{|\mathbf{\text{a}}{|}^2}\left(\begin{array}{cc}\mathbf{\text{a}}\cdot \mathbf{\text{b}}& -(\mathbf{\text{a}}\times \mathbf{\text{b}})\cdot \mathbf{\text{k}}\\ (\mathbf{\text{a}}\times \mathbf{\text{b}})\cdot \mathbf{\text{k}}& \mathbf{\text{a}}\cdot \mathbf{\text{b}}\end{array}\right)\left(\begin{array}{c}\psi \\ \chi \end{array}\right)\\ & & & =-\frac{1}{|\mathbf{\text{a}}{|}^2}\left(\begin{array}{cc}p& -q\\ q& p\end{array}\right)\left(\begin{array}{c}\psi \\ \chi \end{array}\right)\\ & & & =-\frac{1}{|\mathbf{\text{a}}{|}^2}\underset{P^{-1}}{\underbrace{\left(\begin{array}{cc}-\frac{i}{\sqrt{2}}& \frac{i}{\sqrt{2}}\\ \frac{1}{\sqrt{2}}& \frac{1}{\sqrt{2}}\end{array}\right)}}\underset{D}{\underbrace{\left(\begin{array}{cc}p-iq& 0\\ 0& p+iq\end{array}\right)}}\underset{P}{\underbrace{\left(\begin{array}{cc}\frac{i}{\sqrt{2}}& \frac{1}{\sqrt{2}}\\ -\frac{i}{\sqrt{2}}& \frac{1}{\sqrt{2}}\end{array}\right)}}\left(\begin{array}{c}\psi \\ \chi \end{array}\right)\\ & & & =-\frac{1}{|\mathbf{\text{a}}{|}^2}{P}^{-1}DP\left(\begin{array}{c}\psi \\ \chi \end{array}\right)\\ \\ & \Rightarrow & P\left(\begin{array}{c}{\psi }^{\prime }\\ {\chi }^{\prime }\end{array}\right)& =-\frac{1}{|\mathbf{\text{a}}{|}^2}DP\left(\begin{array}{c}\psi \\ \chi \end{array}\right)\\ & \Rightarrow & \left(\begin{array}{cc}\frac{i}{\sqrt{2}}& \frac{1}{\sqrt{2}}\\ -\frac{i}{\sqrt{2}}& \frac{1}{\sqrt{2}}\end{array}\right)\left(\begin{array}{c}{\psi }^{\prime }\\ {\chi }^{\prime }\end{array}\right)& =-\frac{1}{|\mathbf{\text{a}}{|}^2}\left(\begin{array}{cc}p-iq& 0\\ 0& p+iq\end{array}\right)\left(\begin{array}{cc}\frac{i}{\sqrt{2}}& \frac{1}{\sqrt{2}}\\ -\frac{i}{\sqrt{2}}& \frac{1}{\sqrt{2}}\end{array}\right)\left(\begin{array}{c}\psi \\ \chi \end{array}\right)\end{array}$$ Let $$\begin{array}{r}\mathrm{\Psi }=\frac{-i\psi +\chi }{\sqrt{2}}[=-\frac{i}{\sqrt{2}}(\psi +i\chi )=-\frac{i}{\sqrt{2}}y]\end{array}$$ then the differential equation can be written as $$\begin{array}{rlrl}& & \frac{\text{d}}{\text{d}x}\left(\begin{array}{c}{\mathrm{\Psi }}^{\ast }\\ \mathrm{\Psi }\end{array}\right)& =-\frac{1}{|\mathbf{\text{a}}{|}^2}\left(\begin{array}{cc}p-iq& 0\\ 0& p+iq\end{array}\right)\left(\begin{array}{c}{\mathrm{\Psi }}^{\ast }\\ \mathrm{\Psi }\end{array}\right)\\ \\ & \Rightarrow & \frac{\text{d}\mathrm{\Psi }}{\text{d}x}& =-\frac{p+iq}{|\mathbf{\text{a}}{|}^2}\mathrm{\Psi }\\ & \Rightarrow & \mathrm{\Psi }& =A{\text{e}}^{-\frac{p+iq}{|\mathbf{\text{a}}{|}^2}x}\\ & & & =A{\text{e}}^{-\frac{\mathbf{\text{a}}\cdot \mathbf{\text{b}}+i(\mathbf{\text{a}}\times \mathbf{\text{b}})\cdot \mathbf{\text{k}}}{|\mathbf{\text{a}}{|}^2}x}\\ & & & =A{\text{e}}^{-\frac{\mathbf{\text{a}}\cdot \mathbf{\text{b}}}{|\mathbf{\text{a}}{|}^2}x}[\cos \left(\frac{(\mathbf{\text{a}}\times \mathbf{\text{b}})\cdot \mathbf{\text{k}}}{|\mathbf{\text{a}}{|}^2}x\right)-i\sin \left(\frac{(\mathbf{\text{a}}\times \mathbf{\text{b}})\cdot \mathbf{\text{k}}}{|\mathbf{\text{a}}{|}^2}x\right)]\end{array}$$ Finally, $$\begin{array}{rl}y& \equiv \psi +i\chi \\ & =i\sqrt{2}\mathrm{\Psi }\\ & =i\sqrt{2}A{\text{e}}^{-\frac{\mathbf{\text{a}}\cdot \mathbf{\text{b}}}{|\mathbf{\text{a}}{|}^2}x}[\cos \left(\frac{(\mathbf{\text{a}}\times \mathbf{\text{b}})\cdot \mathbf{\text{k}}}{|\mathbf{\text{a}}{|}^2}x\right)-i\sin \left(\frac{(\mathbf{\text{a}}\times \mathbf{\text{b}})\cdot \mathbf{\text{k}}}{|\mathbf{\text{a}}{|}^2}x\right)]\end{array}$$

 

In addition, if we let $$\begin{array}{r}A=\frac{\lambda -i\mu }{\sqrt{2}}\in \mathbb{C}\end{array}$$ and it gives $$\begin{array}{rlrl}& \Rightarrow & y& =(i\lambda +\mu ){\text{e}}^{-\frac{\mathbf{\text{a}}\cdot \mathbf{\text{b}}}{|\mathbf{\text{a}}{|}^2}x}[\cos \left(\frac{(\mathbf{\text{a}}\times \mathbf{\text{b}})\cdot \mathbf{\text{k}}}{|\mathbf{\text{a}}|}x\right)-i\sin \left(\frac{(\mathbf{\text{a}}\times \mathbf{\text{b}})\cdot \mathbf{\text{k}}}{|\mathbf{\text{a}}{|}^2}x\right)]\\ \\ & \Rightarrow & \psi & =\text{Re}(y)={\text{e}}^{-\frac{\mathbf{\text{a}}\cdot \mathbf{\text{b}}}{|\mathbf{\text{a}}{|}^2}x}[\mu \cos \left(\frac{(\mathbf{\text{a}}\times \mathbf{\text{b}})\cdot \mathbf{\text{k}}}{|\mathbf{\text{a}}{|}^2}x\right)+\lambda \sin \left(\frac{(\mathbf{\text{a}}\times \mathbf{\text{b}})\cdot \mathbf{\text{k}}}{|\mathbf{\text{a}}{|}^2}x\right)]\\ & & \chi & =\text{Im}(y)={\text{e}}^{-\frac{\mathbf{\text{a}}\cdot \mathbf{\text{b}}}{|\mathbf{\text{a}}{|}^2}x}[\lambda \cos \left(\frac{(\mathbf{\text{a}}\times \mathbf{\text{b}})\cdot \mathbf{\text{k}}}{|\mathbf{\text{a}}{|}^2}x\right)-\mu \sin \left(\frac{(\mathbf{\text{a}}\times \mathbf{\text{b}})\cdot \mathbf{\text{k}}}{|\mathbf{\text{a}}{|}^2}x\right)]\end{array}$$

 

Method 2. As an ordinary DE with complex-valued function: For $y\in \mathbb{C}$, $$\begin{array}{rl}a\frac{\text{d}y}{\text{d}x}+by& =0\Rightarrow y=C{\text{e}}^{-\frac{b}{a}x}\end{array}$$ where $$\begin{array}{rl}a& =a_1+i{a}_2\\ b& =b_1+i{b}_2\end{array}$$ For the exponent, we find: $$\begin{array}{rl}\frac{b}{a}& =\frac{a^{\ast }b}{a^{\ast }a}\\ & =\frac{(a_1-i{a}_2)(b_1+i{b}_2)}{a_1^2+a_2^2}\\ & =\frac{(a_1{b}_1+a_2{b}_2)+i(a_1{b}_2-a_2{b}_1)}{a_1^2+a_2^2}\\ & =\frac{(\mathbf{\text{a}}\cdot \mathbf{\text{b}})+i(\mathbf{\text{a}}\times \mathbf{\text{b}})\cdot \mathbf{\text{k}}}{|\mathbf{\text{a}}{|}^2}\end{array}$$ which gives $$\begin{array}{rl}\Rightarrow y& =C{\text{e}}^{-\frac{(\mathbf{\text{a}}\cdot \mathbf{\text{b}})+i(\mathbf{\text{a}}\times \mathbf{\text{b}})\cdot \mathbf{\text{k}}}{|\mathbf{\text{a}}{|}^2}x}\\ & =C{\text{e}}^{-\frac{\mathbf{\text{a}}\cdot \mathbf{\text{b}}}{|\mathbf{\text{a}}{|}^2}x}[\cos \left(\frac{(\mathbf{\text{a}}\times \mathbf{\text{b}})\cdot \mathbf{\text{k}}}{|\mathbf{\text{a}}{|}^2}x\right)-i\sin \left(\frac{(\mathbf{\text{a}}\times \mathbf{\text{b}})\cdot \mathbf{\text{k}}}{|\mathbf{\text{a}}{|}^2}x\right)]\end{array}$$ If we identify $C=i\sqrt{2}A\in \mathbb{C}$, it agrees with the result obtained above.

 

Question 2. What about the second-order differential equations with complex coefficients?

 

$$\begin{array}{rl}(a_1+i{a}_2)\frac{{\text{d}}^{2}y}{\text{d}{x}^2}+(b_1+i{b}_2)\frac{\text{d}y}{\text{d}x}+(c_1+i{c}_2)y& =0\end{array}$$ Then we would expect to have a complex-valued solution, i.e. $$\begin{array}{r}y(x)=\psi (x)+i\chi (x)\end{array}$$ Substituting these into the differential equation gives $$\begin{array}{rlrl}& & (a_1+i{a}_2)({\psi }^{″}+i{\chi }^{″})+(b_1+i{b}_2)({\psi }^{\prime }+i{\chi }^{\prime })+(c_1+i{c}_2)(\psi +i\chi )& =0\\ & \Rightarrow & [(a_1{\psi }^{″}+b_1{\psi }^{\prime }+c_1\psi )-(a_2{\chi }^{″}+b_2{\chi }^{\prime }+c_2\chi )]+i[(a_2{\psi }^{″}+b_2{\psi }^{\prime }+c_2\psi )+(a_1{\chi }^{″}+b_1{\chi }^{\prime }+c_1\chi )]& =0\end{array}$$ so it gives a coupled second-order differential equations.