Theoretical Question: how can complex solutions be allowed?

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SUMMARY

This discussion centers on the treatment of complex solutions in homogeneous linear differential equations. The participant questions the validity of referring to complex solutions, such as y_1 = e^{z_1 x} and y_2 = e^{z_2 x}, as mere "rewriting" when deriving real solutions Y_1 = e^{a_1 x} cos(b_1x) and Y_2 = e^{a_2 x} sin(b_2x). They argue that while complex solutions are instrumental in obtaining real results, they should not be dismissed as simple transformations. The conversation highlights the importance of understanding the role of complex numbers in the context of real vector spaces and their implications in theoretical frameworks.

PREREQUISITES
  • Understanding of homogeneous linear differential equations
  • Knowledge of complex numbers and their properties
  • Familiarity with characteristic polynomials
  • Basic concepts of vector spaces in mathematics
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  • Study the derivation of solutions for homogeneous linear differential equations using complex roots
  • Explore the implications of complex solutions in quantum mechanics
  • Learn about the relationship between complex and real vector spaces
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This discussion is beneficial for mathematics students, particularly those studying differential equations, theoretical physicists dealing with quantum mechanics, and educators seeking to clarify the role of complex solutions in real contexts.

nonequilibrium
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Hello, I have my first course in differential equations.

So to solve a homogeneous linear differential equation of order n, you solve the characteristic polynome. Now say it has two complex roots z_1 = a_1 + b_1i and z_2 = a_2 - b_2i. These two roots give two independent solutions, namely y_1 = e^{z_1 x} and y_2 = e^{z_2 x}. My course notes go on -as my professor literally said in class- to rewrite these solutions by taking (y_1 + y_2)/2 and (y_1 - y_2)/2i so we get the two real (and independent) solutions Y_1 = e^{a_1 x} \cos(b_1x) and Y_2 =e^{a_2 x} \sin(b_2x).

But can we really just call this rewriting? This cleary implies that y_1 and y_2 are just as well solutions, but that we simply prefer to rewrite them like that. I'd be more inclined to think that we only rely on the complex roots to lead us to the right answers, but that y_1 and y_2 form no solutions in their own right. Okay if we view our differential equation as a complex differential equation, then I suppose it's a valid solution, but it seems really wrong to me that one would mix real and complex cases. For example the reason we were looking for Y_1 and Y_2 in the first place was to use them as two vectors for our base for the vectorspace describing all the solutions to the differential equation. In that context using something with i in it seems absurd, doesn't it? In the whole chapter we've used R as our evident field and proven existence and unicity assuming the structure of R. (and R looks differently when viewed in the bigger picture of C, because according to R any constant in C is worth two constants in R)

I hope someone gets what is bothering me? If the professor had just said "we don't actually see these complex functions as solutions but only rely on them to get to our answer, an answer which can then exist and be justified in its own right" then all would be good, but he repeatedly stated it as "merely a rewriting", and the complex numbers keep returning (now with systems of DE's) and him stating the same thing... Just wondering if my view is correct or if I'm missing something big!

Please note that my question is purely theoretical and not practical.
 
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I could perfectly well write the general solution to my problem as

y(x) = A_1 y_1(x) + A_2 y_2(x)

for some y(0) = y_0 \in \mathbb{R} and y'(0) = y'_0 \in \mathbb{R}. In working out what A_1 and A_2 are, I would find they're complex. Any answer I get out for y(x), however, is going to end up being real, as it will happen that the imaginary component cancels out. My y(x) is still a perfectly good solution, it just happens that it only gives me real results for real initial conditions and real inputs. Because of that, it's convenient to take a linear combination of the solutions y_1 and y_2 to get real solutions. (Remember that any linear combination of solutions is still a solution for a linear DE). You could do the problem without using the complex solutions, but it's not obvious that Y_1 = e^{a_1 x} \cos(b_1x) and Y_2 = e^{a_2 x} \sin(b_2x) are solutions, whereas the complex solutions are pretty easy to find.

For quantum mechanical problems, factors of i appear right in the DE itself, so the solutions you get will have imaginary parts that don't cancel out for the given boundary conditions, so the complex solutions are very much "real" solutions.
 

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