Perturbation theory to solve diff eq?

[vibe3]

Hi all, I have a tricky problem in pertubation theory.

I have a function:

[tex]
f(\vec{r}) = P(\vec{r}) + \left( B(\vec{r}) + b(\vec{r}) \right)^2
[/tex]

where [tex] b(\vec{r}) [/tex] is a small perturbation and is equal to 0 when [tex] P(\vec{r}) = 0 [/tex]

Now, to solve the equation
[tex]
\nabla f(\vec{r}) = 0
[/tex]

for b(r) is fairly straightforward by noting that
[tex]
P + (B + b)^2 = C = B^2
[/tex]
using the fact that P = 0 ==> b = 0. And so,
[tex]
b(\vec{r}) = \frac{-P}{2B}
[/tex]
by expanding the above equation and neglecting the [tex]b^2[/tex] term. Now, my question is how do you solve the inhomogeneous equation:
[tex]
\nabla f(\vec{r}) = \vec{A}(\vec{r})
[/tex]
for b(r) where A is a known vector field with 0 curl?

 

[jvicens]

Hello there,

Thank you for sharing your problem with us. Perturbation theory can indeed be tricky, but I believe I can offer some insights that may help you solve this inhomogeneous equation.

Firstly, let's take a closer look at the equation you are trying to solve:

\nabla f(\vec{r}) = \vec{A}(\vec{r})

We can rewrite this equation as:

\nabla \left( P(\vec{r}) + \left( B(\vec{r}) + b(\vec{r}) \right)^2 \right) = \vec{A}(\vec{r})

Expanding the squared term, we get:

\nabla \left( P(\vec{r}) + B^2(\vec{r}) + 2B(\vec{r})b(\vec{r}) + b^2(\vec{r}) \right) = \vec{A}(\vec{r})

Now, we know that P(\vec{r}) = 0 when b(\vec{r}) = 0, so we can simplify the equation to:

\nabla \left( B^2(\vec{r}) + 2B(\vec{r})b(\vec{r}) + b^2(\vec{r}) \right) = \vec{A}(\vec{r})

Next, we can use the fact that A(\vec{r}) has 0 curl, which means that it can be written as the gradient of a scalar function, say \phi(\vec{r}).

\nabla \phi(\vec{r}) = \vec{A}(\vec{r})

Substituting this into our equation, we get:

\nabla \left( B^2(\vec{r}) + 2B(\vec{r})b(\vec{r}) + b^2(\vec{r}) \right) = \nabla \phi(\vec{r})

Now, we can use the identity:

\nabla (fg) = f\nabla g + g\nabla f

To rewrite our equation as:

B(\vec{r}) \nabla (2b(\vec{r})) + b(\vec{r}) \nabla (2B(\vec{r})) = \nabla \phi(\vec{r})

Since we know that B