I'm currently reading a paper about the two/three body center problem. In this paper there is this equation, which i'm not sure how to come up with. My problem is, that the functuon evolved radial aswell as vectorial dependence and I'm not sure how to apply the nabla operator to this function.
To sum up I have two questions:
1) What ist the relation of (28) to the two body coloumb problem?
2) How can I proove that (29) is really a solution of (28)? It would be very nice if someone could help me with this. Thanks in advance!
I'm going to answer one of your questions which I understood as "how to prove that (29) is really a solution to (28)?"
First, let us assume $\mathbf{x}=\{x,y,z\}$ is a radius vector in 3D Cartesian coordinate system. The following will be valid for any number of dimensions though.
Further: $\mathbf{a}=\{a_x,a_y,a_z\}$ is a constant vector.
We will use the absolute values of both vectors which are:
$$|\mathbf{x}| =\sqrt{x^2+y^2+z^2}$$
$$|\mathbf{a}| =a=\sqrt{a_x^2+a_y^2+a_z^2}$$
And the dot product:
$$\mathbf{ax}=a_xx+a_yy+a_zz$$
Nabla operator is represented in this coordinate system as:
$$\nabla=\mathbf{i}\frac{\partial}{\partial x}+\mathbf{j}\frac{\partial}{\partial y}+\mathbf{k}\frac{\partial}{\partial z}$$
And:
$$\mathbf{a}\nabla=a_x\frac{\partial}{\partial x}+a_y\frac{\partial}{\partial y}+a_z\frac{\partial}{\partial z}$$
Our equation is:
$$\mathbf{a} \nabla f(\mathbf{x})= - \frac{1}{\sqrt{x^2+y^2+z^2}} \tag{28}$$
The proposed solution is:
$$f(\mathbf{x})^{\pm}= \pm \frac{1}{a} \ln \left( a\sqrt{x^2+y^2+z^2} \mp (a_xx+a_yy+a_zz)\right) \tag{29}$$
Let us choose one of the signs to be more clear and compute the partial derivatives:
$$a_x\frac{\partial f}{\partial x} = \frac{1}{a} \frac{1}{a\sqrt{x^2+y^2+z^2} - (a_xx+a_yy+a_zz)} \left(a \frac{a_xx}{\sqrt{x^2+y^2+z^2}} - a_x^2 \right)$$
$$a_y \frac{\partial f}{\partial y} = \frac{1}{a} \frac{1}{a\sqrt{x^2+y^2+z^2} - (a_xx+a_yy+a_zz)} \left(a \frac{a_yy}{\sqrt{x^2+y^2+z^2}} - a_y^2 \right)$$
$$a_z\frac{\partial f}{\partial z} = \frac{1}{a} \frac{1}{a\sqrt{x^2+y^2+z^2} - (a_xx+a_yy+a_zz)} \left(a \frac{a_zz}{\sqrt{x^2+y^2+z^2}} - a_z^2 \right)$$
Now we can finally write:
$$\mathbf{a} \nabla f(\mathbf{x})=\frac{1}{a} \frac{1}{a\sqrt{x^2+y^2+z^2} - (a_xx+a_yy+a_zz)} \left(a \frac{a_xx+a_yy+a_zz}{\sqrt{x^2+y^2+z^2}} - a^2 \right)$$
$$\mathbf{a} \nabla f(\mathbf{x})=\frac{1}{a\sqrt{x^2+y^2+z^2} - (a_xx+a_yy+a_zz)} \left(\frac{a_xx+a_yy+a_zz}{\sqrt{x^2+y^2+z^2}} - a \right)$$
$$\mathbf{a} \nabla f(\mathbf{x})=-\frac{1}{\sqrt{x^2+y^2+z^2}}$$
I think it is obvious that (29) really is a solution to (28)