# Second order differential equation with complex coefficient

I have some doubts about this kind of second - order differential equation, which is used a lot in physics and for which there are many topics (but in this case the situation is a bit different because k is in general complex number):

I have many doubts about the solutions of this equation, because I have seen different expressions for them in several cases (all applied to electromagnetic problems):

1)I have seen this kind of solution (it is the expression used to describe voltage or current along a transmission line):

where T1 and T2 are complex values.

So, from this kind of analysis, I'll say that:

The solution is a complex linear combination of exponential functions with arguments kx and -kx, with k complex (because in general we have supposed k complex from the beginning).

So, my first question is: is this true for any case? Or may the solution be different depending on k?

2) In other situations (for instance analysis of rectangular and circular waveguides) I have seen different solutions:

with T0 complex value.

Second question: is this solution equivalent to that seen in 1)? And is it true for any value of k?

3) I have seen also another kind of solution:

So my third question is: is this solution equivalent to that seen in 1) and 2)? And is it true for any value of k?

Then, I have a last question: I have seen that these solutions have been used in different situations, by specifing the domain of x. If x was defined in a bounded domain, I have usually seen solutions shown in 1), while for unbounded domains, I have usually seen 3). I wanted to know if it is just a reason of convenience for those specific applications, or if it is a strict math rule.

First note that

$$sinh(kx) = \frac{e^{kx}-e^{-kx}}{2i}$$ and $$cosh(kx) = \frac{e^{kx}+e^{-kx}}{2}$$

this is a defintion of these functions.

So we can see that these definitions nearly look like

$$T_1 \cdot e^{kx} + T_2 \cdot e^{-kx}$$

we just have to choose $$T_1$$ and $$T_2$$ so that we get multiples of $$cosh(kx)$$ and $$sinh(kx)$$, getting your third solution from the first. I'll leave that to you to try and figure out :)

Your second solution looks like it might be slightly off(I may be wrong and hopefully someone can correct me if I am, but I'll work through it).

We note that

$$sinh(kx) = -isin(ikx)$$ and $$cosh(kx) = cos(ikx)$$,

this just comes from the definitions of the Hyperbolic and Trigonometric functions.

$$T_2 \cdot cos(ikx) - T_1 \cdot i \cdot sin(ikx)$$

Now all we need to do is apply what is called a Harmonic Identity, this one in particular says that,

$$Acos(x) - Bsin(x) = Rcos(x+\alpha)$$

where $$R = \sqrt{A^2 + B^2}$$ and $$\alpha = arctan(\frac{b}{a})$$.

So now you apply that identity to what we have above with $$A = T_2$$, $$B = T_1 \cdot i$$ and $$x = ikx$$ and your second answer should pop out!

If what I've done is right it should be of the form $$T_0cos(ikx+\alpha)$$ instead.

I hope this helps :)

• Thank you for your answer. Now the equivalence between 1 and 3 is clear. About 2, is it possible for this expression to be true only with a bounded domain of x, and the other ones in unbounded one? Commented Apr 15, 2020 at 9:00
• What would you suppose the bound on $x$ might be? I can't see a reason it should be bounded but I might be missing something. Commented Apr 15, 2020 at 12:56
• Correct, thank you very much for your deep and accurate analysis Commented Apr 26, 2020 at 20:29