Undergrad Does this ODE have any real solutions?

[Radlor]

TL;DR

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.

 

[jedishrfu]

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]

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

 

[Radlor]

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 separated 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 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.

 

[fresh_42]

I may well have missed something obvious, cheers.

Yes, all local solutions.

 

[Radlor]

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]

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.

 

[Radlor]

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 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.

 

[fresh_42]

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]

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 <br /> y&#039;^2 + 2m^2y^2 = f(x) \\<br /> z&#039;^2 + 2m^2z^2 = f(x)<br /> 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 $\omega^2 = 2m^2$.

 

[Fred Wright]

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