Can the General Solution for this Exponential Equation be Found?

[Heimdall]

Hello,


I'm trying to solve this equation :

\frac{d^2f\left(x\right)}{dx^2}= A\exp\left(f\left(x\right)/B\right)

(which comes from plasma equilibrium)

I can't find out what's the general solution... is there a method to solve with this kind of second member

 

[arildno]

Well, you can make some headway as follows:
\frac{d^{2}f}{dx^{2}}f'(x)=Ae^{\frac{f}{B}}f'(x)\to\frac{1}{2}((f'(x))^{2}-(f'(x_{0}))^{2})=BA(e^{\frac{f(x)}{B}}-e^{\frac{f(x_{0})}{B}
Whereby it follows that:
f'(x)^{2}=Ce^{\frac{f}{B}}+K
where C=2AB, and K depends on the values of the function and its derivative at x_{0}[/tex]<br /> <br /> There is no particular reason why this should have a simple exact solution.

 

[Heimdall]

lets say that f\left(x_0=0\right)=f&#039;\left(0\right)=0

so that we have :

f&#039;^2 = 2BA\left(\exp\left(f/B\right)-1\right)

or f&#039;^2 = 4AB\exp\left(f/2B\right)sinsh\left(f/2B\right)

I know there's a trick somewhere because the solution should be something like

log\left(cosh\left(\right)\right)

 

[arildno]

Okay, let's see:

Let:
u^{2}=e^{\frac{f}{B}}\to{2u}\frac{du}{dx}=u\frac{f&#039;}{B}\to{2Bdu}=f&#039;dx
This substitution seems very promising..
We get:
\frac{f&#039;}{\sqrt{2AB}\sqrt{e^{\frac{f}{B}}-1}}=\pm{1}
which leads to:
\sqrt{\frac{2B}{A}}\int\frac{du}{\sqrt{u^{2}-1}}=\pm{x}+C
We set u=Cosh(v), yielding:
\sqrt{\frac{2B}{A}}sgn(Sinh(v))v=\pm{x}+C
essentially yielding the result for f you mentioned; just get the signs right.

 

[Heimdall]

Hum,

2u\frac{du}{dx} = \frac{f&#039;}{B}e^{f/B} = \frac{f&#039;}{B}u^2

which means that

2B\frac{du}{u} = f&#039;dx

now the integral is :

\int\frac{2Bdu}{u\sqrt{u^2-1}}

and with your u = ch(v) you get

\int \frac{dv}{ch\left(v\right)}

 

[arildno]

Okay, made a mistake there.

 

[Heimdall]

I've looked on the web and

\int \frac{dv}{ch\left(v\right)} = gd\left(v\right)

where gd(v) is the Gudermannian function :

gd\left(v\right) = 2\arctan\left(th\left(\frac{v}{2}\right)\right)which doesn't look nice if you come back to f

 

[Heimdall]

\frac{f&#039;}{\sqrt{2AB}\sqrt{e^{\frac{f}{B}}-1}}=\pm{1}

I haven't seen it the first time but you can't even do this if x=0 (because f(0)=0)

 

[arildno]

That need not be destructive, since f' is zero also.
You have an improper integral.

 

[arildno]

Hum,

2u\frac{du}{dx} = \frac{f&#039;}{B}e^{f/B} = \frac{f&#039;}{B}u^2

which means that

2B\frac{du}{u} = f&#039;dx

now the integral is :

\int\frac{2Bdu}{u\sqrt{u^2-1}}

and with your u = ch(v) you get

\int \frac{dv}{ch\left(v\right)}

So, then we continue!

We set y=tanh(v/2)\frac{dy}{dv}=\frac{1}{2}\frac{1}{Cosh^{2}(\frac{v}{2})}=\frac{1}{2}(1-y^{2})\to{dv}=\frac{2dy}{1-y^{2}}
Thereby, we get:
\int\frac{1*dv}{Cosh(v)}=\int\frac{Cosh^{2}(\frac{v}{2})-Sinh^{2}(\frac{v}{2})}{Cosh^{2}(\frac{v}{2})+Sinh^{2}(\frac{v}{2})}dv=\int\frac{1-y^{2}}{1+y^{2}}\frac{2dy}{1-y^{2}}=\int\frac{2dy}{1+y^{2}}=2arctan(y)+C
And we are essentially done.

 

[arildno]

I've looked on the web and

\int \frac{dv}{ch\left(v\right)} = gd\left(v\right)

where gd(v) is the Gudermannian function :

gd\left(v\right) = 2\arctan\left(th\left(\frac{v}{2}\right)\right)


which doesn't look nice if you come back to f

I didn't see this post. Sorry about that.

 

[Heimdall]

Hi :)

In the mean time I got another solution :

I = \int\frac{f&#039;dx}{\sqrt{1-\exp{\left(f/B\right)}}}

let u^2 = 1-\exp{\left(f/B\right)}

so we have :

-2Budu\frac{1}{1-u^2} = f&#039;dx

and we get :

I = -2B\int \frac{du}{1-u^2} = -2Bargth\left(u\right)

I = -2Bargth\left(\sqrt{1-\exp{\left(f/K\right)}}\right) = -\sqrt{2AB}x

\sqrt{1-\exp{\left(f/B\right)}} = th\left(\sqrt{\frac{A}{2B}x}\right)

f = -2B\ Ln\left(ch\left(\sqrt{\frac{A}{2B}}x\right)\right) :)

 

[arildno]

Note that in this case, you have 1-exp beneath the square root sign; not exp-1.

That will naturally yield a totally different solution, valid whenever AB<0, K>0.