# I Does this ODE have any real solutions?

Summary
Are there real solutions to this ODE.
The ODE is:
\begin{equation}
(y'(x)^2 - z'(x)^2) + 2m^2( y(x)^2 - z(x)^2) = 0
\end{equation}

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.

Related Differential Equations News on Phys.org

#### jedishrfu

Mentor
Where did you find this ODE?

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

\begin{equation}
y'(x)^2 + 2m^2 * y(x)^2 = 0
\end{equation}

and

\begin{equation}
z'(x)^2 + 2m^2 * z(x)^2 = 0
\end{equation}

which look identical to me.

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

#### fresh_42

Mentor
2018 Award
What do you mean by solution? Does it have to be on the entire real line, or do local solutions count?

Where did you find this ODE?

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

\begin{equation}
y'(x)^2 + 2m^2 * y(x)^2 = 0
\end{equation}

and

\begin{equation}
z'(x)^2 + 2m^2 * z(x)^2 = 0
\end{equation}

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 seperated the two. In which case, the only solutions would be of the form

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

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 cancelled 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

#### fresh_42

Mentor
2018 Award
I may well have missed something obvious, cheers.
Yes, all local solutions.

• jedishrfu

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.

#### fresh_42

Mentor
2018 Award
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.

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

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

Or some combination of such. I don't see any way a real solution is defined across a continous 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.

#### fresh_42

Mentor
2018 Award
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.?

• #### pasmith

Homework Helper
Where did you find this ODE?

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

\begin{equation}
y'(x)^2 + 2m^2 * y(x)^2 = 0
\end{equation}

and

\begin{equation}
z'(x)^2 + 2m^2 * z(x)^2 = 0
\end{equation}

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.

• #### William Crawford

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

#### Fred Wright

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

Last edited:

### Want to reply to this thread?

"Does this ODE have any real solutions?"

### Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving