I How can I verify the solution for Poisson equation with forgotten factor?

[baptiste]

Hello

I have a physics article who solve Poisson equation of the form:

L² ∂²f/∂x²=sinh(f)

The proposed solution is:

tanh(f/4)=exp(x/L) tanh(f0/4)

with f0 a constant

I suspect an error, something like a forgotten factor.

How can I verify? (I tried but I failed)

I forgot the limit conditions:

x belong to [-∞,0] and f and its first dérivative are null at -∞Thanks

 

[Mark44]

Hello

I have a physics article who solve Poisson equation of the form:

L² ∂²f/∂x²=sinh(f)

Is there some reason this equation involves partial derivatives? From the proposed solution f appears to be a function of x alone.
If so, the above should be $L^2\frac{d^2 f}{dx^2} = \sinh(f)$

The proposed solution is:

tanh(f/4)=exp(x/L) tanh(f0/4)

with f0 a constant

I suspect an error, something like a forgotten factor.

How can I verify? (I tried but I failed)

Please show us what you tried.

I forgot the limit conditions:

x belong to [-∞,0] and f and its first dérivative are null at -∞Thanks

 

[baptiste]

Yes you are right, f is a single variable function, so the equation is
L²d²f/dx²=sinh(f)
Sorry for the error.

So I tried to derivate the expression tanh(f/4)=exp(x/L) tanh(f0/4) or f=4*atanh(exp(x/L) tanh(f0/4)) two times (both forms).
That leads to huge expressions which have nothing to do with the supposed result 1/L² sinh(f)

Thanks for your help

 

[player100]

Multiple both sides by $\frac{df}{dx}$. Then you get $$L^2 \frac{d^2f}{dx^2} \frac{df}{dx} = sinh(f) \frac{df}{dx}.$$ Notice that $\frac{d^2f}{dx^2} \frac{df}{dx} = \frac{1}{2} \frac{d}{dx} (\frac{df}{dx})^2$. Integrate both sides with respect to x and you get $$L^2(\frac{df}{dx})^2 = 2\int sinh(f) df = 2 cosh(f) + C.$$

So, $$L\frac{df}{dx} = \pm \sqrt{2cosh(f) + C}.$$ I think this is as simplified as it gets...

 

[baptiste]

Yes OK I see
If we take the given solution, we have Ldf/dx=1/L (cosh(f/4)) sinh(f/4)
then by some trigonometric transformation we get
L df/dx=+/- sqrt(2 cosh(f)-2).
So the solution is true.

I will read the article one more time to find where is the bug in my simulations.

And I will try to learn matjax next time

Thank you