# Does this ODE have any real solutions?

• I
In summary, the given ODE has complex solutions and a trivial case where both y(x) and z(x) are equal to 0. However, it does not have any non-trivial real solutions. The ODE can be separated into two identical equations, suggesting that there may be a missing term that couples y(x) and z(x). The solutions can also be written in terms of a complex scalar function, with the requirement that the real and imaginary parts of the function are both real. It is also possible to construct an infinite number of trivial solutions by choosing a holomorphic function and setting y(x) equal to z(x).
TL;DR Summary
Are there real solutions to this ODE.
The ODE is:

(y'(x)^2 - z'(x)^2) + 2m^2( y(x)^2 - z(x)^2) = 0

Where y(x) and z(x) are real unknown functions of x, m is a constant.

I believe there are complex solutions, as well as the trivial case z(x) = y(x) = 0 , but I cannot find any real solutions. Are there any, and if so how is best to find them?

Thanks for any help.

Where did you find this ODE?

It looks like you can separate it out into two equations:

y'(x)^2 + 2m^2 * y(x)^2 = 0

and

z'(x)^2 + 2m^2 * z(x)^2 = 0

which look identical to me.

Did I miss something? or is there a missing term that couples Y and Z ?

What do you mean by solution? Does it have to be on the entire real line, or do local solutions count?

jedishrfu said:
Where did you find this ODE?

It looks like you can separate it out into two equations:

y'(x)^2 + 2m^2 * y(x)^2 = 0

and

z'(x)^2 + 2m^2 * z(x)^2 = 0

which look identical to me.

Did I miss something? or is there a missing term that couples Y and Z ?

Hi, yeah there is no coupling, hence I also separated the two. In which case, the only solutions would be of the form

y = z = K e^{\pm i m x}

And therefore no real solutions?

As to where it's from, I am trying to prove an equation involving a complex scalar field implies that the field is trivial. After seperating into real and imaginiary terms, the complex terms canceled and left me with the equation above (z and y could be seen as the real and imaginiary functions of the complex function respectively).

I may well have missed something obvious, cheers.

jedishrfu
I may well have missed something obvious, cheers.
Yes, all local solutions.

jedishrfu
fresh_42 said:
Yes, all local solutions.

To expand on that, I understand that around x=0, the solutions would be real, or for a constant value of y and z. That's fine in my case, since then my complex field is trivial within my theory and I am happy.

If you mean something else, that real solutions exist that aren't constant, then could you expand on what you meant please.

To expand on that, I understand that around x=0, the solutions would be real, or for a constant value of y and z. That's fine in my case, since then my complex field is trivial within my theory and I am happy.

If you mean something else, that real solutions exist that aren't constant, then could you expand on what you meant please.
I haven't worked it out, but there could be regions where ##\sinh## and ##\cosh## could work.

fresh_42 said:
I haven't worked it out, but there could be regions where ##\sinh## and ##\cosh## could work.

From what I can work out, it seems the global solution would be something like

y= z = c_1\cosh(\sqrt{2}mi x) \ \ \text{or} \ \ y= z = c_2\sinh(\sqrt{2}mi x)

Or some combination of such. I don't see any way a real solution is defined across a continuous region . In which case it looks to me that the same conditions regarding locality hold as before, and that y and z are either constant and real, or complex.

In which case if y and z are the component functions of a complex scalar function, i.e. $$f(\zeta) = y(x) + i z(x)$$, then since f(y) and f(z) must be real functions, the complex function doesn't exist continuously across ##(-\infty, \infty)##, or is real only for a constant complex value of ##f(\zeta)##.

Thanks for your help, appreciate bit of a bizarre problem.

At least for ##m=1## we get a solution ##y=\cosh(x)\, , \,z=\sinh(x)##.

The equation can be written as ##\dfrac{u'}{u}\cdot \dfrac{v'}{v}=-2m^2##. Why don't you choose real numbers ##a,b## such that ##ab+2m^2=0## and solve ##\dfrac{u'}{u}=a\, , \,\dfrac{v'}{v}=b\, , ## resp.?

jedishrfu said:
Where did you find this ODE?

It looks like you can separate it out into two equations:

y'(x)^2 + 2m^2 * y(x)^2 = 0

and

z'(x)^2 + 2m^2 * z(x)^2 = 0

which look identical to me.

Did I miss something? or is there a missing term that couples Y and Z ?

More generally, you can have $$y'^2 + 2m^2y^2 = f(x) \\ z'^2 + 2m^2z^2 = f(x)$$ and you might be able to choose $f$ to keep $y$ and $z$ real.

I believe there are complex solutions, as well as the trivial case z(x) = y(x) = 0 , but I cannot find any real solutions. Are there any, and if so how is best to find them?

There are an infinite number of trivial solutions to this ODE.

Supose ##f:\Omega\rightarrow\mathbb{C}## is a holomorphic function, then ##\big(y(x), z(x)\big) = \big(f(x),\pm f(x)\big)## are trivial solutions to your original ODE $$(y^\prime)^2 - (z^\prime)^2 + 2m^2(y^2 - z^2) = 0.$$
So to answer your question "are there real solutions to this ODE"; yes there are an infinite number! Simply pick any once differentiable real functions ##y(x)## and let ##y(x) = \pm z(x)##.

EDIT: A non-trivial real valued solution is ##y(x) = \sin(\omega x)## and ##z(x) = \cos(\omega x)##, where ##\omega^2 = 2m^2##.

Last edited:
Delta2
What's wrong with (by guessing): ##y=cosh(\sqrt {2}mx) ## ## z=sinh(\sqrt {2}mx)##?

Last edited:

## 1. What is an ODE?

An ODE (ordinary differential equation) is a mathematical equation that relates a function to its derivatives. It is commonly used to model physical systems in science and engineering.

## 2. What are real solutions?

Real solutions refer to the values of the variable in an equation that make the equation true when substituted in. In the case of an ODE, real solutions would be the values of the function that satisfy the equation.

## 3. How do you determine if an ODE has real solutions?

To determine if an ODE has real solutions, you can use various methods such as substitution, separation of variables, or using an integrating factor. These methods involve solving the equation to find the values of the function that satisfy the equation.

## 4. Are there cases where an ODE does not have real solutions?

Yes, there are cases where an ODE does not have real solutions. This can happen when the equation is not well-defined or when the values of the function that satisfy the equation are complex numbers instead of real numbers.

## 5. Why is it important to know if an ODE has real solutions?

Knowing if an ODE has real solutions is important in order to understand and model physical systems accurately. It also helps in finding the behavior of the system over time and predicting future outcomes.

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