The mean value of the cube, Force Field Laplace equation

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SUMMARY

The discussion focuses on calculating the mean value of the function \( U \) defined as \( U=U_0+x \left(\frac{\partial U}{\partial x}\right)+y\left(\frac{\partial U}{\partial y}\right)+z \left(\frac{\partial U}{\partial z}\right)+\frac{1}{2}x^2\left(\frac{\partial^2 U}{\partial x^2}\right)+\frac{1}{2}y^2\left(\frac{\partial^2 U}{\partial y^2}\right)+\ldots \). The mean value is derived as \( \overline{U} \approx U_0 + \frac{a^2}{24}(\nabla^2 U) \). The integral limits for the calculations are from \(-\frac{a}{2}\) to \(\frac{a}{2}\). Key points include the necessity of evaluating derivatives at zero and correcting a typographical error in the equation regarding the power of \( y \).

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This discussion is beneficial for physics students, mathematicians, and engineers who are involved in fluid dynamics, thermodynamics, or any field requiring the application of Laplace equations and mean value calculations in three-dimensional space.

Arman777
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Homework Statement


I have a value of $$ U=U_0+x (∂U/∂x)+y(∂U/∂y)+z (∂U/∂z)+1/2x^2(∂^2U/∂x^2)+1/2y^(2∂^2U/∂y^2)+...$$

We need to find the mean value of the U. So the answer is

$$\overline{\rm U}\approx U_0+a^2/24(∇^2U)$$

Homework Equations



$$\overline{\rm U}=1/a^3 \int \int\int Udxdydz$$

The Attempt at a Solution



The problem I get is that I have to proof that,

$$K=1/a^3 \int \int\int x (∂U/∂x)+y(∂U/∂y)+z (∂U/∂z)dxdydz=0$$ but

$$L=1/a^3 \int \int\int 1/2x^2(∂^2U/∂x^2)+1/2y^(2∂^2U/∂y^2)+1/2z^(2∂^2U/∂z^2)=a^2/24(∇^2U)$$ but

I couldn't proceed why these are true.

The integral limits are from ##-a/2## to ##a/2##
 
Last edited:
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Arman777 said:

Homework Statement


I have a value of $$ U=U_0+x (∂U/∂x)+y(∂U/∂y)+z (∂U/∂z)+1/2x^2(∂^2U/∂x^2)+1/2y^(2∂^2U/∂y^2)+...$$
I think you have to be careful interpreting that. The derivatives should be evaluated at 0, so for the purposes of the integration they are constants.
I.e. ##\frac{\partial U}{\partial x}|_{x=0}## etc.
Also, you have dropped a power of 2 on the y in the last term above.
 

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