ATH: Solving Differential Equations for B(x) with Constraints and a Constant K

[Mentz114]

This is not homework, these equations are constraints I've encountered in some GR work.

But I'm hopeless with differential equations and I have two I need to solve. Even after reading some instructive texts, I still can't do it.


$$\frac{dB(x)}{dx} = \frac{B(x)}{x} - \frac{1}{x}$$

$$\frac{dB(x)}{dx} = \frac{B(x)}{x} - \frac{1}{x} + \frac{K}{B(x)}$$

K is a constant. I know the first one has a solution but the second one might not.

M

 

[Rainbow Child]

This is not homework, these equations are constraints I've encountered in some GR work.

But I'm hopeless with differential equations and I have two I need to solve. Even after reading some instructive texts, I still can't do it.


$$\frac{dB(x)}{dx} = \frac{B(x)}{x} - \frac{1}{x}$$

$$\frac{dB(x)}{dx} = \frac{B(x)}{x} - \frac{1}{x} + \frac{K}{B(x)}$$

K is a constant. I know the first one has a solution but the second one might not.

M

For the 1st one

$$B'(x) = \frac{B(x)}{x} - \frac{1}{x}\Rightarrow \frac{B'(x)}{x}-\frac{B(x)}{x^2}=-\frac{1}{x^2}\Rightarrow (\frac{B(x)}{x})'=(\frac{1}{x})'\Rightarrow \frac{B(x)}{x}=\frac{1}{x}+c\Rightarrow B(x)=c\,x+1$$

For the 2nd one, let $$B(x)=K\,x\,f(x)$$, then

$$f'(x)=\frac{-f(x)+1}{K\,x^2\,f(x)}\Rightarrow\frac{f(x)\,d\,f(x)}{-f(x)+1}=\frac{d\,x}{K\,x^2}\Rightarrow f(x)+\log(f(x)-1)=\frac{1}{K\,x}+c\Rightarrow \frac{B(x)}{K\,x}+\log(\frac{B(x)}{K\,x}-1)=\frac{1}{K\,x}+c$$

which is the general implicit solution and can be written with in terms of the product log function.

 

[Mentz114]

Hi Rainbow,

thanks for the help, much appreciated. It looks like witchcraft to me. Unfortunately I transposed the signs and your solution depends crucially on the sign. So I'm back to square one.

Which serves me right for being a pratt.
M

 

[coomast]

For the 2nd one, let $$B(x)=K\,x\,f(x)$$, then

$$f'(x)=\frac{-f(x)+1}{K\,x^2\,f(x)}\Rightarrow\frac{f(x)\,d\,f(x)}{-f(x)+1}=\frac{d\,x}{K\,x^2}\Rightarrow f(x)+\log(f(x)-1)=\frac{1}{K\,x}+c\Rightarrow \frac{B(x)}{K\,x}+\log(\frac{B(x)}{K\,x}-1)=\frac{1}{K\,x}+c$$

which is the general implicit solution and can be written with in terms of the product log function.

I get a slightly different solution:

$$\frac{B(x)}{Kx}+\ln\left|1-\frac{B(x)}{Kx}\right|=\frac{1}{Kx}+c$$

There seems to be something with the absolute value.
Mmm, I use ln, sorry for that, it makes it easier for me to see what is meant.

@Rainbow Child: What do you exactly mean by "product log function"

 

[Rainbow Child]

I get a slightly different solution:

$$\frac{B(x)}{Kx}+\ln\left|1-\frac{B(x)}{Kx}\right|=\frac{1}{Kx}+c$$

There seems to be something with the absolute value.
Mmm, I use ln, sorry for that, it makes it easier for me to see what is meant.

@Rainbow Child: What do you exactly mean by "product log function"

I was working on the complex plane, that's why I didn't put the absolute value.

The product log function is the multivalued function $$w(z)$$ defined by

$$z=w(z)\,e^{w(z)}$$

Thus for the solution at hand

$$\frac{B(x)}{K\,x}+\log(\frac{B(x)}{K\,x}-1)=\frac{1}{K\,x}+c$$

we have

$$(\frac{B(x)}{K\,x}-1)\,\exp(\frac{B(x)}{K\,x})=\exp(\frac{1}{K\,x}+c)\Rightarrow (\frac{B(x)}{K\,x}-1)\,\exp(\frac{B(x)}{K\,x}-1)=\exp(\frac{1}{K\,x}+c-1)\Rightarrow \frac{B(x)}{K\,x}-1=w\left(\exp(\frac{1}{K\,x}+c-1)\right)$$

yielding to

$$B(x)=K\,x\,\left(1+w(C\,e^{\frac{1}{K\,x}})\right), \quad C=\exp(c-1)$$

 

[coomast]

I was under the assumption that the differential equation of the original post was to have a real valued function as solution. Therefore the use of ln instead of log. In case I assume that it is to be real, then I think that the solution I gave was correct. That leaves us with a minus sign difficulty or am I making a mistake?

I never used the product log function (or Lambert W function, google). Always good to learn something new. I looked in the books of Erdelyi, Magnus, Oberhettinger, Tricomi and the one of Abramowitz, Stegun, but couldn't find it. It seems to be a rarely used one. Very nice.

 

[Rainbow Child]

I wrote $$\log$$ because I am used with this notation for the natural logarithm. For the real domain your solution is the correct one.

The product log function $$w(z)$$ often appears when you are dealing with GR, i.e. in maximally symmetric two-dimensional surfaces. But in GR we are always allowed to make a coordinate transformation, and get rid of the product log function.

 

[Mentz114]

Thanks to both of you. I may have learned enough to solve the correct equations.