What is the meaning of having imaginary solutions to a differential equation

Question

I am trying to solve this equation

$$x+e^{xt}=0$$

for different values of x and see when there is no real solution. Plotting this on a graph and assuming x is real gives me the different values of t. I want to find the value of t when the above equation has no real solution. To do this, I first differentiate the equation and set that value equal to $0$. This gives me the relation for the minimum value of $$x+e^{-xt}$$. When the value of t is greater than this the curve doesnt cut the x axis and there is no real solution. But now assuming x is imaginary I can still solve to get a solution which consists of complex terms. But I dont understand what this means in the real domain. If there is no real value of $x$ for which there is a solution, then what does allowing imaginary values of $x$ do? How does a system like this work in real life?

Answer 1

I don't know about "real life", but one place your equation can come up is in solving a PDE such as

$$ \dfrac{\partial u}{\partial t} = \dfrac{\partial^2 u}{\partial x^2} $$

with boundary condition

$$ \dfrac{\partial u}{\partial x}(0,t) + u(L,t) = 0$$

If you assume a solution of the form $u(x,t) = \exp(r x + r^2 t)$ you get from the boundary condition $$ r + e^{rL} = 0 \tag{1}$$

You may only be interested in real solutions of your PDE, but note that the real and imaginary parts of a solution are solutions. Thus if $r = \alpha + i \beta$ is a complex solution of (1), you get real solutions of your PDE of the forms

$$ \eqalign{\text{Re}(\exp(r x + r^2 t)) &= \exp(\alpha x + (\alpha^2-\beta^2)t) \cos(\beta x + 2 \alpha \beta t)\cr \text{Im}(\exp(r x + r^2 t)) &= \exp(\alpha x + (\alpha^2-\beta^2)t) \sin(\beta x + 2 \alpha \beta t)\cr}$$

Answer 2

In cases where the result represent an impedance, the real part is the ohmic resistance, independent of frequency. The imaginary part is either capacitive or inductive depending of the sign. Complex numbers are used for systems that can store then release energy after some time.

Answer 3

I know this was a question asked years ago. New members like me may come across this so I still post my understanding. An imaginary part is a result of real problems described in fewer degrees of freedom than actual degrees of freedom.

For example, a circular motion in $x-y$ plane can be described in $x$ degree of freedom only with an imaginary part, $z= \cos x + i \sin x$ where the imaginary part for $y$ projection. Likewise, for vibration $m y” + c y’ + k y=0$, you will get real stiffness $k$. However, if you describe it as $m y”+ k y =0$, the damping part not shown up in the equation will entail an imaginary stiffness $k$, Which accounts for both stiffness and damping. Hope this helps.