A Converting a Vector Differential Equation in Fluid Mechanics

[Gribouille]

Hi guys,
I have encountered a problem in fluid mechanics that gives a three-dimensional vector differential equation
\begin{equation}
a \vec{f} + \nabla{a} + b \nabla{c} = \vec{0}
\end{equation}
where a, b, and c are unknown scalar functions of three-dimensional space and f is a known vector function of space. Do you have any ideas how to convert this into an equation for a (or b or c) alone or even solve it?

Thank you very much!

 

[phyzguy]

Well, you have three separate scalar equations
a f_x + \nabla a + b \nabla c = 0 ; a f_y + \nabla a + b \nabla c = 0 ; a f_z + \nabla a + b \nabla c = 0
If you subtract them pairwise, you get:
a(f_x - f_y) = 0 ; a(f_x - f_z) = 0 ; a(f_y - f_z) = 0
Unless f is a very special vector field (where fx = fy = fz everywhere), this seems to imply that a = 0. If a = 0, then your original equation becomes b \nabla c = 0, which of course has solutions independent of f, but probably isn't what you want.

 

[Gribouille]

Thanks for your reply, I highly appreciate it. The equations would read
\begin{eqnarray}
a f_x + \partial_x a + b \partial_x c &=& 0 \\
a f_y + \partial_y a + b \partial_y c &=& 0 \\
a f_z + \partial_z a + b \partial_z c &=& 0
\end{eqnarray}
since the gradient operator produces a vector.

To give you more background information, the equation I am trying to solve in its original form is
\begin{equation}
d \vec{x} \cdot (a \vec{f} + \nabla{a} + b \nabla{c}) = 0
\end{equation}

I will try to describe the progress I have made. If only x is variable and y = 0 and z = 0 are constant and zero, only the first of the three equations has to be satisfied and the differential equation becomes
\begin{equation}
a f_x + a^{\prime} = 0
\end{equation}
and b and c are constants. When x and y are variables and z is constant, b = 1 because two functions are enough to satisfy the set of equations. One can use the curl operator and arrives at
\begin{equation}
\nabla \times (a \vec{f}) = \vec{0}
\end{equation}
since the curl of the gradient is zero. This can be solved for a and c, too.

I have not managed to solve the equation when x, y, and z are variable.

Thanks for your time guys and I appreciate any kind of input.

 

[phyzguy]

Apologies for my stupid post. Let me think about it some more.

 

[Gribouille]

I think I found the solution. Divide by a and take the curl and you get
\begin{equation*}
\nabla \times \vec{f} + \nabla (b/a) \times \nabla c = \vec{0}
\end{equation*}
From this follows
\begin{eqnarray}
\nabla (b/a) \cdot \nabla \times \vec{f} = 0 \\
\nabla c \cdot \nabla \times \vec{f} = 0
\end{eqnarray}
Any suggestions if this approach seems correct?

 

[phyzguy]

I think I found the solution. Divide by a and take the curl and you get
\begin{equation*}
\nabla \times \vec{f} + \nabla (b/a) \times \nabla c = \vec{0}
\end{equation*}
From this follows
\begin{eqnarray}
\nabla (b/a) \cdot \nabla \times \vec{f} = 0 \\
\nabla c \cdot \nabla \times \vec{f} = 0
\end{eqnarray}
Any suggestions if this approach seems correct?

I agree with this so far. Where would you go next?

 

[Gribouille]

I would form the inner product of the original equation and the curl of f. That results in
\begin{equation*}
a ( \nabla \times \vec{f} ) \cdot \vec{f} + ( \nabla \times \vec{f} ) \cdot \nabla a + b ( \nabla \times \vec{f} ) \cdot \nabla c = 0
\end{equation*}
The last term on the left-hand side is zero and we have an equation of a alone.