Strategy in solving vector equations involving grad, scalar product operators?

In summary, you can find \Lambda by solving for \Phi in terms of U and then using the directional derivative.
  • #1
jason12345
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What is the general strategy in solving vector equations involving grad and the scalar product?

In particular, I want to express [itex]\Lambda[/itex] in terms from [itex] \mathbf U \cdot \nabla\Lambda = \Phi[/itex] but it looks impossible, unless there is some vector identity I can use.

Thanks in advance.
 
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  • #2
Hi jason12345! :smile:

What you have there is a directional derivative.
See for instance wiki: http://en.wikipedia.org/wiki/Directional_derivative

If U is for instance a unit vector, then:
[tex]\vec U \cdot \vec\nabla \Lambda(\vec x) = \Phi(\vec x)[/tex]
is also written as:
[tex]\vec\nabla_{\vec U} \Lambda(\vec x) = \Phi(\vec x)[/tex]

You can find [itex]\Lambda[/itex] with for instance something like:
[tex]\Lambda(\vec x) = \int \Phi(\vec x + u \vec U) du[/tex]
 
  • #3
I like Serena said:
Hi jason12345! :smile:

What you have there is a directional derivative.
See for instance wiki: http://en.wikipedia.org/wiki/Directional_derivative

If U is for instance a unit vector, then:
[tex]\vec U \cdot \vec\nabla \Lambda(\vec x) = \Phi(\vec x)[/tex]
is also written as:
[tex]\vec\nabla_{\vec U} \Lambda(\vec x) = \Phi(\vec x)[/tex]

You can find [itex]\Lambda[/itex] with for instance something like:
[tex]\Lambda(\vec x) = \int \Phi(\vec x + u \vec U) du[/tex]

That's given me a lot to think about - thanks!

Maybe I could try for something simpler to start with so:

How would I find [itex]\nabla\Lambda[/itex] as the other terms with [itex]U[/itex] independent of (x,y,z)?

Perhaps I could use some vector identity?
 
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  • #4
I don't think it's possible to solve. There's an infinite number of solutions, even after we disregard the ones that are off by a constant.

To illustrate this, suppose we want to construct a Λ such that U⋅∇Λ=Φ is satisfied. The left hand side is a sum of three terms: Ux*dΛ/dx, Uy*dΛ/dy, and Uz*dΛ/dz. For any point, we literally have absolute power to set dΛ/dx to whatever we want. After that, we have absolute power to choose any value for dΛ/dy. There's an infinite number of choices for dΛ/dx, and for each of those choices, there's an infinite number of choices for the other two derivatives.

This is analogous to a normal dot product, say a*(0,0,1)=0.5. The only thing this equation tells you is that the x component of a is 0.5. The y and z components can be anything, so it's not possible to solve for a in the sense of writing a concise analytical expression that captures all solutions.
 
  • #5
In the example a*(1,0,0)=0.5 you can indeed say that the x-component is 0.5.
So the solution is a=(0.5,y,z) with 2 unknown variables y and z.


Similarly, if you know for instance that U=(1,0,0), the expression in the OP ([itex]\boldsymbol U \cdot \nabla \Lambda = \Phi[/itex]) reduces to:
[tex]\frac {\partial \Lambda} {\partial x} = \Phi[/tex]
So:
[tex]\Lambda = \int \Phi dx + C(y,z)[/tex]
And:
[tex]\nabla \Lambda = \begin{pmatrix} \Phi(x,y,z) \\ \frac {\partial} {\partial y} C(y,z) \\ \frac {\partial} {\partial z} C(y,z) \end{pmatrix}[/tex]

You can also see here that it is not simpler to solve for [itex]\nabla \Lambda[/itex].
You need to find [itex]\Lambda[/itex] first.


When U is not aligned with an axis, you will need to make a coordinate transformation in such a way that U is mapped to for instance (1,0,0).
Then you can use this same procedure.

As yet, I don't have a nice complete formula that includes the coordinate transformation though. That will take a bit of mind-benchpressing first. :smile:
 
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FAQ: Strategy in solving vector equations involving grad, scalar product operators?

1. What is the purpose of using vector equations in solving problems?

Vector equations are used to represent physical quantities that have both magnitude and direction, such as force, velocity, and acceleration. They allow us to mathematically describe these quantities and solve problems involving them.

2. What is the role of the grad operator in vector equations?

The grad operator, also known as the gradient, is used to determine the rate of change of a scalar field in a given direction. It is represented by the symbol ∇ and is often used in vector calculus to solve problems related to motion and forces.

3. How do I use the scalar product operator in solving vector equations?

The scalar product operator, also known as the dot product, is used to calculate the magnitude of the projection of one vector onto another. It is represented by the symbol · and is often used in solving problems involving work, energy, and projections.

4. What are some common techniques for solving vector equations involving grad and scalar product operators?

Some common techniques for solving vector equations include using the properties of vector operations, such as the distributive and associative properties, and applying the rules of calculus, such as the chain rule and product rule. It is also helpful to draw diagrams and use geometric interpretations of the equations.

5. Are there any real-world applications of solving vector equations involving grad and scalar product operators?

Yes, there are many real-world applications of vector equations. For example, the use of vector calculus is essential in understanding the motion of particles in fluids, electromagnetic fields, and other physical systems. It is also used in engineering, physics, and other fields to solve problems related to forces, motion, and energy.

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