I Does this ODE have any real solutions?

4
1
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.
 
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4,359
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

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2018 Award
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What do you mean by solution? Does it have to be on the entire real line, or do local solutions count?
 
4
1
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.
 
4
1
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

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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.
 
4
1
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

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2018 Award
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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
1,734
408
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 [tex]
y'^2 + 2m^2y^2 = f(x) \\
z'^2 + 2m^2z^2 = f(x)
[/tex] and you might be able to choose [itex]f[/itex] to keep [itex]y[/itex] and [itex]z[/itex] 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:
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What's wrong with (by guessing): ##y=cosh(\sqrt {2}mx) ## ## z=sinh(\sqrt {2}mx)##?
 
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