Resolving Ecuations with $\Psi$, $\vec{v}$, and $\nabla$

[Raparicio]

Dear Friends,

Another questions about mathmatics.

How can be resolved a ecuation like this?

$${ \frac{ \partial{(m \vec {v} - \Psi \vec {v})}}{ \partial {t} } = \nabla (\Psi \vec {v}^2 )$$

Asuming that v could be any vector, m a constant, and psi a wave function. It's not the similar that a wave ecuation.

And more:

Is this ok? $$\Psi (\vec {v} \nabla) \vec {v} = \Psi \vec {v} (\nabla \vec {v})$$

 

[dextercioby]

Dear Friends,

Another questions about mathmatics.

How can be resolved a ecuation like this?

$${ \frac{ \partial{(m \vec {v} - \Psi \vec {v})}}{ \partial {t} } = \nabla (\Psi \vec {v}^2 )$$

Asuming that v could be any vector, m a constant, and psi a wave function. It's not the similar that a wave ecuation.

So i should understand that the unknown from your equation would be $\Psi$ ??That is to say,all other quantites are known...
So your equation should be looking like that
$$-\vec{v}\frac{\partial\Psi}{\partial t}=\vec{v}^{2}\nabla\Psi+\Psi\nabla(\vec{v}^{2})-m\frac{\partial\vec{v}}{\partial t}$$

I'lm afraind your equation is not 'balanced'.It has only one unknown and three equations.It's actually a system of 3 differential eq.with partial derivatives,but still only one function.Now,if 'v' is an unknown vector function as well,then the eq.(the system of eq.is not 'balanced' again).This time it would 4 unknowns,but only three quations.

My guess,it's something fishy with your eq.Double check it.

And more:

Is this ok? $$\Psi (\vec {v} \nabla) \vec {v} = \Psi \vec {v} (\nabla \vec {v})$$

Write it in tensor notation.Simplify through the (assumed nonzero) scalar function $\Psi$ and u'll get
$$v_{i}\frac{\partial v_{j}}{\partial x_{i}} \vec{e}_{j}$$
is the LHS.
$$v_{j}\frac{\partial v_{i}}{\partial x_{i}} \vec{e}_{j}$$
is the RHS.
You can see they're different.The first is a contracted tensor product between the vector 'v' and its gradient (which is a second rank tensor),while the second is the product between the vector and its divergence (which is a scalar).Though 'balanced' wrt to tensor rank,the two sides are different.
Therefore,u don't have an equality.

Daniel.

 

[wais]

To resolve the equation { \frac{ \partial{(m \vec {v} - \Psi \vec {v})}}{ \partial {t} } = \nabla (\Psi \vec {v}^2 ), we can use the chain rule and product rule for differentiation. First, we can rewrite the equation as { \frac{ \partial{(m - \Psi) \vec {v}}}{ \partial {t} } = \nabla (\Psi \vec {v} \cdot \vec {v} ). Then, using the chain rule, we can take the partial derivative of the left side with respect to time and the partial derivative of the right side with respect to position. This will give us the following equation:

{ (m - \Psi) \frac{\partial \vec{v}}{\partial t} + \vec{v} \frac{\partial \Psi}{\partial t} = 2\Psi \vec{v} \cdot \nabla \vec{v} + \Psi \vec{v} \cdot \nabla \vec{v} }

Next, we can simplify the equation by using the product rule for differentiation on the right side. This will give us:

{ (m - \Psi) \frac{\partial \vec{v}}{\partial t} + \vec{v} \frac{\partial \Psi}{\partial t} = 3\Psi \vec{v} \cdot \nabla \vec{v} }

Finally, we can solve for the partial derivative of velocity with respect to time by dividing both sides of the equation by (m - \Psi). This will give us the final equation:

{ \frac{\partial \vec{v}}{\partial t} = \frac{3\Psi}{m - \Psi} \vec{v} \cdot \nabla \vec{v} - \frac{\vec{v}}{m - \Psi} \frac{\partial \Psi}{\partial t} }

As for the second question, it is not clear what is meant by \Psi (\vec {v} \nabla) \vec {v}. The notation is not standard and it is not clear what operation is being performed. It is also not clear how this relates to the original equation given. Therefore, it is difficult to say if it is correct or not.