A first-order differential equation with complex coefficients

수학 모음 (Maths collection) 전체보기

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? $$ \begin{align} (a_1 + ia_2) \frac{ \textrm{d} y }{ \textrm{d}x } + (b_1 + ib_2) y &= 0 \end{align} $$

 

Method 1. Using matrix notation: Let $$ \begin{align} y(x) = \psi(x) + i \chi(x) \end{align} $$ Substituting this into the differential equation gives $$ \begin{align} (a_1 + ia_2) \left( \psi' + i \chi' \right) + (b_1 + ib_2) \left( \psi + i \chi \right) &= 0 \\ \Big[ \left( a_1 \psi' + b_1 \psi \right) - \left( a_2 \chi' + b_2 \chi \right) \Big] + i \Big[ \left( a_2 \psi' + b_2 \psi \right) + \left( a_1 \chi' + b_1 \chi \right) \Big] &= 0 \end{align} $$ The real and imaginary parts yield a set of simultaneous equations $$ \begin{align} \left\{ \begin{array}{l} a_1 \psi' + b_1 \psi = a_2 \chi' + b_2 \chi \\ a_2 \psi' + b_2 \psi = - a_1 \chi' - b_1 \chi \end{array} \right. \qquad &\textrm{or} \qquad \left\{ \begin{array}{l} a_1 \psi' - a_2 \chi' = - b_1 \psi + b_2 \chi \\ a_2 \psi' + a_1 \chi' = - b_2 \psi - b_1 \chi \end{array} \right. \\ \\ \begin{pmatrix} a_1 & b_1 \\ a_2 & b_2 \end{pmatrix} \begin{pmatrix} \psi' \\ \psi \end{pmatrix} = \begin{pmatrix} a_2 & b_2 \\ - a_1 & - b_1 \end{pmatrix} \begin{pmatrix} \chi' \\ \chi \end{pmatrix} \qquad &\textrm{or} \qquad \begin{pmatrix} a_1 & - a_2 \\ a_2 & a_1 \end{pmatrix} \frac{ \textrm{d} }{ \textrm{d}x } \begin{pmatrix} \psi \\ \chi \end{pmatrix} = \begin{pmatrix} - b_1 & b_2 \\ - b_2 & - b_1 \end{pmatrix} \begin{pmatrix} \psi \\ \chi \end{pmatrix} \end{align} $$

The latter appears to be more useful. $$ \begin{align} \begin{pmatrix} a_1 & - a_2 \\ a_2 & a_1 \end{pmatrix} \frac{ \textrm{d} }{ \textrm{d}x } \begin{pmatrix} \psi \\ \chi \end{pmatrix} &= - \begin{pmatrix} b_1 & - b_2 \\ b_2 & b_1 \end{pmatrix} \begin{pmatrix} \psi \\ \chi \end{pmatrix} \\ \\ \Rightarrow \qquad \frac{ \textrm{d} }{ \textrm{d}x } \begin{pmatrix} \psi \\ \chi \end{pmatrix} &= - \begin{pmatrix} a_1 & - a_2 \\ a_2 & a_1 \end{pmatrix}^{-1} \begin{pmatrix} b_1 & - b_2 \\ b_2 & b_1 \end{pmatrix} \begin{pmatrix} \psi \\ \chi \end{pmatrix} \\ &= - \frac1{ a_1^2 + a_2^2 } \begin{pmatrix} a_1 & a_2 \\ - a_2 & a_1 \end{pmatrix} \begin{pmatrix} b_1 & - b_2 \\ b_2 & b_1 \end{pmatrix} \begin{pmatrix} \psi \\ \chi \end{pmatrix} \\ &= - \frac1{ a_1^2 + a_2^2 } \begin{pmatrix} a_1 b_1 + a_2 b_2 & - ( a_1 b_2 - a_2 b_1 ) \\ a_1 b_2 - a_2 b_1 & a_1 b_1 + a_2 b_2 \end{pmatrix} \begin{pmatrix} \psi \\ \chi \end{pmatrix} \end{align} $$ Using the vector notation, $$ \begin{align} \textbf a = \begin{pmatrix} a_1 \\ a_2 \end{pmatrix} \qquad \textrm{and} \qquad \textbf b = \begin{pmatrix} b_1 \\ b_2 \end{pmatrix} \end{align} $$ we can express $$ \begin{align} \textbf a \cdot \textbf b &= a_1 b_1 + a_2 b_2 = p \\ (\textbf a \times \textbf b) \cdot \textbf k &= a_1 b_2 - a_2 b_1 = q \end{align} $$ Then the coupled differential equation reads $$ \begin{align} && \frac{ \textrm{d} }{ \textrm{d}x } \begin{pmatrix} \psi \\ \chi \end{pmatrix} &= - \frac1{ | \textbf a |^2 } \begin{pmatrix} \textbf a \cdot \textbf b & - (\textbf a \times \textbf b) \cdot \textbf k \\ (\textbf a \times \textbf b) \cdot \textbf k & \textbf a \cdot \textbf b \end{pmatrix} \begin{pmatrix} \psi \\ \chi \end{pmatrix} \\ &&&= - \frac1{ | \textbf a |^2 } \begin{pmatrix} p & - q \\ q & p \end{pmatrix} \begin{pmatrix} \psi \\ \chi \end{pmatrix} \\ &&&= - \frac1{ | \textbf a |^2 } \underbrace{ \begin{pmatrix} -\frac{i}{\sqrt{2}} & \frac{i}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \end{pmatrix} }_{ P^{-1} } \underbrace{ \begin{pmatrix} p - iq & 0 \\ 0 & p + iq \end{pmatrix} }_{D} \underbrace{ \begin{pmatrix} \frac{i}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ -\frac{i}{\sqrt{2}} & \frac{1}{\sqrt{2}} \end{pmatrix} }_{ P } \begin{pmatrix} \psi \\ \chi \end{pmatrix} \\ &&&= - \frac1{ | \textbf a |^2 } P^{-1}DP \begin{pmatrix} \psi \\ \chi \end{pmatrix} \\ \\ &\Rightarrow& P \begin{pmatrix} \psi' \\ \chi' \end{pmatrix} &= - \frac1{ | \textbf a |^2 } D P \begin{pmatrix} \psi \\ \chi \end{pmatrix} \\ &\Rightarrow& \begin{pmatrix} \frac{i}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ -\frac{i}{\sqrt{2}} & \frac{1}{\sqrt{2}} \end{pmatrix} \begin{pmatrix} \psi' \\ \chi' \end{pmatrix} &= - \frac1{ | \textbf a |^2 } \begin{pmatrix} p - iq & 0 \\ 0 & p + iq \end{pmatrix} \begin{pmatrix} \frac{i}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ -\frac{i}{\sqrt{2}} & \frac{1}{\sqrt{2}} \end{pmatrix} \begin{pmatrix} \psi \\ \chi \end{pmatrix} \\ \end{align} $$ Let $$ \begin{align} \Psi = \frac{ -i \psi + \chi }{ \sqrt{2} } \quad \left[ = -\frac{i}{ \sqrt{2} } (\psi + i \chi) = -\frac{i}{ \sqrt{2} }y \right] \end{align} $$ then the differential equation can be written as $$ \begin{align} && \frac{ \textrm{d} }{ \textrm{d}x } \begin{pmatrix} \Psi^* \\ \Psi \end{pmatrix} &=- \frac1{ | \textbf a |^2 } \begin{pmatrix} p - iq & 0 \\ 0 & p + iq \end{pmatrix} \begin{pmatrix} \Psi^* \\ \Psi \end{pmatrix} \\ \\ &\Rightarrow& \frac{ \textrm d \Psi }{ \textrm dx } &= - \frac{ p + i q }{ | \textbf a |^2 } \Psi \\ &\Rightarrow& \Psi &= A \textrm e^{ - \frac{ p + i q }{ | \textbf a |^2 } x } \\ &&&= A \textrm e^{ - \frac{ \textbf a \cdot \textbf b + i (\textbf a \times \textbf b) \cdot \textbf k }{ | \textbf a |^2 } x } \\ &&&= A \textrm e^{ - \frac{ \textbf a \cdot \textbf b }{ | \textbf a |^2 } x } \left[ \cos \left( \frac{ ( \textbf a \times \textbf b ) \cdot \textbf k }{ | \textbf a |^2 } x \right) - i \sin \left( \frac{ ( \textbf a \times \textbf b ) \cdot \textbf k }{ | \textbf a |^2 } x \right) \right] \end{align} $$ Finally, $$ \begin{align} y &\equiv \psi + i \chi \\ &= i \sqrt{2} \Psi \\ &= i \sqrt{2} A \textrm e^{ - \frac{ \textbf a \cdot \textbf b }{ | \textbf a |^2 } x } \left[ \cos \left( \frac{ ( \textbf a \times \textbf b ) \cdot \textbf k }{ | \textbf a |^2 } x \right) - i \sin \left( \frac{ ( \textbf a \times \textbf b ) \cdot \textbf k }{ | \textbf a |^2 } x \right) \right] \end{align} $$

 

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

 

Method 2. As an ordinary DE with complex-valued function: For $y \in \mathbb C$, $$ \begin{align} a \frac{ \textrm{d} y }{ \textrm{d}x } + b y &= 0 \qquad \Rightarrow \qquad y = C \textrm e^{ - \frac{b}{a} x } \end{align} $$ where $$ \begin{align} a&= a_1 + i a_2 \\ b&= b_1 + i b_2 \end{align} $$ For the exponent, we find: $$ \begin{align} \frac{b}{a} &= \frac{ a^*b }{ a^*a } \\ &= \frac{ ( a_1 - i a_2 ) ( b_1 + i b_2 ) }{ a_1^2 + a_2^2 } \\ &= \frac{ ( a_1 b_1 + a_2 b_2 ) + i ( a_1 b_2 - a_2 b_1 ) }{ a_1^2 + a_2^2 } \\ &= \frac{ ( \textbf a \cdot \textbf b ) + i ( \textbf a \times \textbf b ) \cdot \textbf k }{ | \textbf a |^2 } \end{align} $$ which gives $$ \begin{align} \Rightarrow \qquad y &= C \textrm e^{ - \frac{ ( \textbf a \cdot \textbf b ) + i ( \textbf a \times \textbf b ) \cdot \textbf k }{ | \textbf a |^2 } x } \\ &= C \textrm e^{ - \frac{ \textbf a \cdot \textbf b }{ | \textbf a |^2 } x } \left[ \cos \left( \frac{ ( \textbf a \times \textbf b ) \cdot \textbf k }{ | \textbf a |^2 } x \right) - i \sin \left( \frac{ ( \textbf a \times \textbf b ) \cdot \textbf k }{ | \textbf a |^2 } x \right) \right] \end{align} $$ 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?

 

$$ \begin{align} (a_1 + ia_2) \frac{ \textrm{d}^2 y }{ \textrm{d}x^2 } + (b_1 + ib_2) \frac{ \textrm{d} y }{ \textrm{d}x } + (c_1 + ic_2) y &= 0 \end{align} $$ Then we would expect to have a complex-valued solution, i.e. $$ \begin{align} y(x) = \psi(x) + i \chi(x) \end{align} $$ Substituting these into the differential equation gives $$ \begin{align} && (a_1 + ia_2) \left( \psi'' + i \chi'' \right) + (b_1 + ib_2) \left( \psi' + i \chi' \right) + (c_1 + ic_2) \left( \psi + i \chi \right) &= 0 \\ &\Rightarrow& \Big[ \left( a_1 \psi'' + b_1 \psi' + c_1 \psi \right) - \left( a_2 \chi'' + b_2 \chi' + c_2 \chi \right) \Big] + i \Big[ \left( a_2 \psi'' + b_2 \psi' + c_2 \psi \right) + \left( a_1 \chi'' + b_1 \chi' + c_1 \chi \right) \Big] &= 0 \end{align} $$ so it gives a coupled second-order differential equations.