- #1

Safinaz

- 241

- 7

## \ddot \sigma - p e^\sigma - q e^{2\sigma} =0 ##

Or

## \frac{d^2 \sigma}{dt^2} - p e^\sigma - q e^{2\sigma} =0 ##

Where p and q are constants.

Thanks.

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- Thread starter Safinaz
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- #1

Safinaz

- 241

- 7

## \ddot \sigma - p e^\sigma - q e^{2\sigma} =0 ##

Or

## \frac{d^2 \sigma}{dt^2} - p e^\sigma - q e^{2\sigma} =0 ##

Where p and q are constants.

Thanks.

- #2

pasmith

Homework Helper

- 2,427

- 1,033

\frac12 \left(\dot\sigma^2(t) - \dot\sigma^2(0)\right) = \int_{\sigma(0)}^{\sigma(t)} f(s)\,ds.[/tex] I haven't checked whether the resulting ODE for [itex]\dot \sigma[/itex] can be solved analytically.

- #3

Mark44

Mentor

- 36,436

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The equation is second order, all right, but it is not a partial differential equation, as you initially wrote in the thread title. It's a second order ordinary differential equation (ODE).

Last edited:

- #4

- 13,256

- 1,274

Here's how things go in the era of AI of the 21st century: You "feed" this ODE to Mathematica on the Wolframalpha website. In a matter of seconds, you will see if it has a solution in terms of known elementary or special functions. If such a solution is shown by the software, then you can ask people how this solution is computed by the program. If the AI can't break it, neither can people, then a series-expansion is needed (i.e. an approximation for t very close to 0).

## \ddot \sigma - p e^\sigma - q e^{2\sigma} =0 ##

Or

## \frac{d^2 \sigma}{dt^2} - p e^\sigma - q e^{2\sigma} =0 ##

Where p and q are constants.

Thanks.

- #5

S.G. Janssens

Science Advisor

Education Advisor

- 1,223

- 818

No.Here's how things go in the era of AI of the 21st century: You "feed" this ODE to Mathematica on the Wolframalpha website. In a matter of seconds, you will see if it has a solution in terms of known elementary or special functions. If such a solution is shown by the software, then you can ask people how this solution is computed by the program. If the AI can't break it, neither can people, then a series-expansion is needed (i.e. an approximation for t very close to 0).

- #6

- 13,256

- 1,274

I am sorry, but if the OP has no "feeling" that his highly nonlinear ODE may be solved by special functions, he can't just pretend that AI did not exist, like in the time of Gradsheyn and Rytzhik doing complicated integrals by hand. You may think it is cheating, but I say this is only being being realistic.No.

- #7

Ssnow

Gold Member

- 571

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Ssnow

- #8

anuttarasammyak

Gold Member

- 1,789

- 894

We can interpret it as equation of motion

[tex]\ddot{x}=-\frac{\partial U}{\partial x}[/tex]

where

[tex]U=-pe^x-\frac{q}{2}e^{2x}[/tex]

We can investigate shape of the potential energy including

[tex]U(-\infty)=0, |U(+\infty)|=\infty [/tex]

Say p,q>0

total energy E<0 would tell the region where the particle cannot go. For E>0 the particle would go infinite.

[tex]\ddot{x}=-\frac{\partial U}{\partial x}[/tex]

where

[tex]U=-pe^x-\frac{q}{2}e^{2x}[/tex]

We can investigate shape of the potential energy including

[tex]U(-\infty)=0, |U(+\infty)|=\infty [/tex]

Say p,q>0

total energy E<0 would tell the region where the particle cannot go. For E>0 the particle would go infinite.

Last edited:

- #9

aheight

- 320

- 108

(* edited: noticed DE in p easily solved by separation of variables *)

I suggest start by writing it as

$$

y''-a e^y-b e^{2y}=0

$$

Then let ##y'=p## as the standard notation. Then ##y''=p\frac{dp}{dy}## leaving

$$

p\frac{dp}{dy}-a e^y-b e^{2 y}=0

$$

or

$$

pdp=(ae^y+be^{2y})dy

$$

It's at least reduced to first-order and can be solved by integrating:

$$

p_1(y)=-\sqrt{2 a e^y+b e^{2 y}+2 c_1}

$$

$$

p_2(y)=\sqrt{2 a e^y+b e^{2 y}+2 c_1}

$$

Then need to solve:

$$

y'=-\sqrt{2 a e^y+b e^{2 y}+2 c_1}

$$

$$

y'=\sqrt{2 a e^y+b e^{2 y}+2 c_1}

$$

which Mathematical gives solutions in terms of radicals, tanh, and log expressions:

I suggest start by writing it as

$$

y''-a e^y-b e^{2y}=0

$$

Then let ##y'=p## as the standard notation. Then ##y''=p\frac{dp}{dy}## leaving

$$

p\frac{dp}{dy}-a e^y-b e^{2 y}=0

$$

or

$$

pdp=(ae^y+be^{2y})dy

$$

It's at least reduced to first-order and can be solved by integrating:

$$

p_1(y)=-\sqrt{2 a e^y+b e^{2 y}+2 c_1}

$$

$$

p_2(y)=\sqrt{2 a e^y+b e^{2 y}+2 c_1}

$$

Then need to solve:

$$

y'=-\sqrt{2 a e^y+b e^{2 y}+2 c_1}

$$

$$

y'=\sqrt{2 a e^y+b e^{2 y}+2 c_1}

$$

which Mathematical gives solutions in terms of radicals, tanh, and log expressions:

Last edited:

- #10

anuttarasammyak

Gold Member

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- 894

As said in post #8 these solutions come from energy conservation law. c_1 is total energy for m=1.would be interesting to figure out how these are the solutions.

Last edited:

- #11

aheight

- 320

- 108

I'd like to continue a bit to solve this somewhat manually. We have the general expression:

$$

\frac{dy}{dx}=\sqrt{a e^y+b e^{2y}+c}

$$

Separating variables again and integrating (I did the integration with Mathematica):

$$

\begin{align*}

\int\frac{dy}{\sqrt{a e^y+b e^{2y}+c}}&=x+c_2 \\

-\frac{\tanh ^{-1}\left(\frac{a e^y+2 c}{2 \sqrt{c} \sqrt{a e^y+b e^{2 y}+c}}\right)}{\sqrt{c}}&=x+c_2 \\

\frac{a e^y+2 c}{2 \sqrt{c} \sqrt{a e^y+b e^{2 y}+c}}&=\tanh(\sqrt{c}(x+c_2))

\end{align*}

$$

So basically now reduced to an algebraic problem solving for y in the expression:

$$

ae^y+2c=k\sqrt{ae^y+be^{2y}+c}

$$

$$

\frac{dy}{dx}=\sqrt{a e^y+b e^{2y}+c}

$$

Separating variables again and integrating (I did the integration with Mathematica):

$$

\begin{align*}

\int\frac{dy}{\sqrt{a e^y+b e^{2y}+c}}&=x+c_2 \\

-\frac{\tanh ^{-1}\left(\frac{a e^y+2 c}{2 \sqrt{c} \sqrt{a e^y+b e^{2 y}+c}}\right)}{\sqrt{c}}&=x+c_2 \\

\frac{a e^y+2 c}{2 \sqrt{c} \sqrt{a e^y+b e^{2 y}+c}}&=\tanh(\sqrt{c}(x+c_2))

\end{align*}

$$

So basically now reduced to an algebraic problem solving for y in the expression:

$$

ae^y+2c=k\sqrt{ae^y+be^{2y}+c}

$$

Last edited:

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