Div(gradΦ): Solve 3y12+2y22-3y1y32

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In summary, Div(gradΦ) is a mathematical operation that represents the divergence of the gradient of a scalar field Φ and is commonly used in vector calculus. To solve the equation 3y12+2y22-3y1y32, you can use the method of separation of variables. The coefficients in this equation represent the rate of change of the variables and have various applications in science, such as describing the behavior of fluids and solving differential equations. Real-world examples that can be modeled using this equation include the motion of particles in fluids, electrical and magnetic fields, and heat transfer in materials.
  • #1
kezzstar
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Φ=3y12+2y22-3y1y32

Find: div(gradΦ).

I broke it down to find the partial derivative of all 3 coordinates y1,y2 and y3, so using the continuity equation:

y1=6y1-3y32
y2=4y2
y3=6y3y1

However, I'm not sure if this correct, would the first part give the gradient? Could someone confirm or solve this?

Regards
 
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  • #2
kezzstar said:
Find: div(gradΦ).

I broke it down to find the partial derivative of all 3 coordinates y1,y2 and y3, so using the continuity equation:

y1=6y1-3y32
y2=4y2
y3=6y3y1

However, I'm not sure if this correct, would the first part give the gradient? Could someone confirm or solve this?

what is the gradient of a function ? its the slope of the curve representing the function/

divergence also has a meaning in calculus.
 
  • #3
kezzstar said:
Φ=3y12+2y22-3y1y32

Find: div(gradΦ).

I broke it down to find the partial derivative of all 3 coordinates y1,y2 and y3, so using the continuity equation:

y1=6y1-3y32
y2=4y2
y3=6y3y1

However, I'm not sure if this correct, would the first part give the gradient? Could someone confirm or solve this?

Regards
The divergence of the gradient of a function is also known as the Laplacian of that function:

https://en.wikipedia.org/wiki/Laplace_operator

What you have calculated so far is:

$$\frac{∂Φ}{∂y_1} = 6y_1-3y_3^2$$
$$\frac{∂Φ}{∂y_2}=4y_2$$
$$\frac{∂Φ}{∂y_3}=-6y_3y_1$$

or

##∇Φ = (\frac{∂Φ}{∂y_1},\frac{∂Φ}{∂y_2},\frac{∂Φ}{∂y_3}) = (6y_1-3y_3^2,4y_2,-6y_1y_3)##

The final step is

##div(grad Φ) = ∇ ⋅ ∇Φ = \frac{∂^2Φ}{∂y_1^2}+\frac{∂^2Φ}{∂y_2^2}+\frac{∂^2Φ}{∂y_3^2}##

(Note: edited to fix sign error in third term.)
 
Last edited:
  • #4
Just to add, I believe you should get 10-6y1.

kezzstar said:
y3=6y3y1
I think you forgot a negative sign there.
 
  • #5
You wrote
##y_1=6y_1-3y_3^2##
when you should have written
##\frac{\partial\Psi}{\partial y_1}=6y_1-3y_3^2##.
I don't know if you made a mistake in understanding or used a lazy abuse of notation, but you really need to avoid misuse of the equality expression. If you want to write it as a semantic pairing of the left and right, but not an equality, you are better off using a colon or comma, or making a table.
 

What is Div(gradΦ)?

Div(gradΦ) is a mathematical operation that represents the divergence of the gradient of a scalar field Φ. It is also known as the Laplacian operator and is commonly used in vector calculus to describe the rate of change of a vector field.

How do I solve the equation 3y12+2y22-3y1y32?

To solve this equation, you can use the method of separation of variables. First, separate the equation into two parts: 3y12+2y22 and -3y1y32. Then, take the derivative of both sides with respect to y to obtain two separate equations. Finally, solve for y in each equation and plug the solutions back into the original equation to find the values of y.

What is the significance of the coefficients in the equation 3y12+2y22-3y1y32?

The coefficients in this equation represent the rate of change of the variables y1, y2, and y3. They determine the direction and magnitude of the change in the values of these variables. In this case, the coefficients indicate that the rate of change of y1 is 3, the rate of change of y2 is 2, and the rate of change of y3 is -3.

What are the applications of this equation in science?

This equation has various applications in fields such as physics, engineering, and mathematics. It is used to describe the behavior of fluids, electric and magnetic fields, and other physical phenomena. It is also used in optimization problems and in solving differential equations.

Are there any real-world examples that can be modeled using this equation?

Yes, there are many real-world examples that can be modeled using this equation. For instance, the motion of a particle in a fluid, the flow of electricity in a circuit, and the spread of pollutants in a body of water can all be described using this equation. It can also be used to model the heat transfer in a solid material or the diffusion of gases in the atmosphere.

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