# Can a real ODE have a complex solution?

By a real ODE I mean an ordinary differential equation with only real coefficients and the resulting function is a function of a real argument. If such a solution exists, can you give an example?

Edit: To add to this, is it still possible if the initial conditions must also be real?

• You generally shouldn't change a question after asking it unless you're asked to clarify. Such things get called "chameleon" questions and people tend to run away from them, because the askers are never satisfied. If an answer to your question leads you to new ideas, ask another question (or think about the new ideas yourself for a little while). – John Hughes Feb 21 '17 at 20:54
• Do you mean that you have a vector field on $\Bbb C^n$ so that its restriction to the real part $\Bbb R^n$ is completely real? And then ask if there may be solutions that start in the real part but leave the real subspace? Do you assume that this vector field is continuous? Even complex differentiable? -- If yes, this should probably asked as a new question. Think about if you want to exclude $\dot z=f(x+iy)=1+i\sqrt{|y|}$. – LutzL Feb 22 '17 at 6:59

$$f(x)^2 + 1 = 0$$ That's a differential equation where the "derivative" coefficient is zero; as it happens, the solution is one of the constant functions $f(x) = \pm i$.

If you want one with a derivative term, consider $$f(x)^2 + 1 = xf'(x)$$ At $x = 0$, any solution must have $f(0) = \pm i$.

• Consider also the simple harmonic motion equation, $\frac{d^2x}{dt^2} = -kx$ – Devsman Feb 21 '17 at 21:35
• See BobaFret's answer. – John Hughes Feb 21 '17 at 21:36

$y'' + y = 0$ has solutions $e^{ix}$ and $e^{-ix}$.

Update: $y = ix$ is a solution to $(y')^2 + 1 = 0$ with $y(0)=0$

• yes, but those can be combined to give an equally-good generating set that's read (namely sine and cosine), and with real initial conditions, you'll always get real solutions, so they may not persuade OP. – John Hughes Feb 21 '17 at 20:51
• ...and as we see, the OP has changed to question to rule out this case. Sigh. – John Hughes Feb 21 '17 at 20:54

Consider $y(x)=(1-x^2)^{3/2}$.

Then $$y'(x)=-\frac322x(1-x^2)^{1/2}=-3x\sqrt[3]{y(x)},$$ with the initial condition $y(0)=1$.

But $y(\sqrt2)=i.$

• very nice example. It's worth commenting that what happens here is that existence and uniqueness theorems break down. – Joel Feb 22 '17 at 0:32

There is always a basis of real solutions for a real linear ODE with constant coefficients having degree greater than 1, thus ruling out the weird examples like a polynomial equation having no derivatives at all. See Theorem 4.1 of http://www.math.uconn.edu/~kconrad/blurbs/linmultialg/diffeqdim.pdf.

Try

$$\left[\frac{dy}{dx}\right]^2 = -(y^2 + 1)$$

This equation can have no real solution at all. Proof by contradiction: assume $y(x)$ is a real valued solution. Then $\left[\frac{dy}{dx}\right]^2$ is real as well, but that implies $\sqrt{-(y(x)^2 + 1)}$ is real, yet $y(x)^2 + 1 > 0$ always no matter what real function $y(x)$ is, thus $-(y(x)^2 + 1) < 0$ always and so this square root can never be real. Contradiction.

EDIT: I just throw it into Wolfram, and it looks like all solutions may be real valued at some isolated points -- but isolated points is not a real-differentiable function of a real variable!