# Soln of Leplace in Cartesian Coord

• Apteronotus
In summary, you are getting confused because you are trying to solve a problem that is not physically relevant.

#### Apteronotus

Hi

I'm getting confused solving the Laplace eqn in Cartesian coordinates.

The equation can be solved by solving each of
$\frac{X''(x)}{X(x)}=-k_x^2, \qquad\qquad \frac{Y''(y)}{Y(y)}=-k_y^2, \qquad\qquad \frac{Z''(z)}{Z(z)}=k_z^2$

and then substituting into the equation
$\Phi(\vec{x})=X(x)Y(y)Z(z)$

The solution to the (general) ODE's is
$X(x)=A_1e^{ik_xx}+A_2e^{-ik_xx}, \qquad\qquad Y(y)=B_1e^{ik_yy}+B_2e^{-ik_yy}, \qquad\qquad Z(z)=C_1e^{k_zz}+C_2e^{k_zz}$

But the solution when solving for the electric potential is given by
$\Phi_{k_x,k_y,\pm}(\vec{x})=exp\left(ik_xx+ik_yy \pm\sqrt{k_x^2+k_y^2}\right)$

Where am I going wrong?
Why have they let the coefficients $A_{1,2}, B_{1,2}, C_{1,2}=1$?

Thanks

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The values for the coefficients in the general solution can be found by applying the boundary conditions of the problem. Nobody can answer your question without more information about what the boundary conditions were. (You need to give us the whole problem.) Depending on the geometry of the problem, certain conditions will force certain coefficients to have particular values, which will usually simplify your solution to something with a manageable form. You will normally end up doing something like Fourier's trick to determine the value of the coefficient determined by the last boundary condition, and you will end up with an expression for the potential. Sometimes, things will simplify really nicely.

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So then is the general solution always

$\Phi(\vec{x})=\left[A_1e^{ik_xx}+A_2e^{ik_xx}\right]\cdot\left[B_1e^{ik_yy}+B_2e^{ik_yy}\right]\cdot\left[C_1e^{k_zz}+C_2e^{-k_zz}\right]$

and depending on the boundary condition we solve for the $A_{1,2}, B_{1,2}, C_{1,2}, k_1 \quad\mbox{and}\quad k_2$?

The problem I have is solving the electric potential of a uniform field, when a dielectric cube is placed within it. I can assume that the field lines are parallel to two of the cube's sides.

If you know how I can approach this problem, I would be grateful if you could guide me.

## 1. What is the Laplace equation in Cartesian coordinates?

The Laplace equation in Cartesian coordinates is a second-order partial differential equation that describes the distribution of a scalar quantity (such as temperature or electric potential) in a 2D or 3D space. It is written as ∇²Φ = 0, where Φ is the scalar quantity and ∇² is the Laplace operator.

## 2. How is the Laplace equation solved in Cartesian coordinates?

The Laplace equation in Cartesian coordinates can be solved using separation of variables. This involves assuming a solution of the form Φ(x,y,z) = X(x)Y(y)Z(z), plugging it into the equation, and solving for the functions X, Y, and Z. Then, the general solution can be found by taking the sum of all possible solutions.

## 3. What are the boundary conditions for the Laplace equation in Cartesian coordinates?

The boundary conditions for the Laplace equation in Cartesian coordinates are typically specified as either Dirichlet or Neumann boundary conditions. Dirichlet boundary conditions specify the value of the scalar quantity at the boundary, while Neumann boundary conditions specify the derivative of the scalar quantity at the boundary.

## 4. What are some applications of the Laplace equation in Cartesian coordinates?

The Laplace equation in Cartesian coordinates has many applications in physics and engineering. It is commonly used to model steady-state heat conduction, electrostatics, and fluid flow. It is also used in image processing and computer graphics for smoothing and edge-preserving filtering.

## 5. Are there any limitations to using the Laplace equation in Cartesian coordinates?

While the Laplace equation in Cartesian coordinates is a powerful tool for solving many physical and mathematical problems, it does have some limitations. It can only be used for steady-state problems, where the scalar quantity does not change with time. It also assumes a linear relationship between the scalar quantity and its sources, which may not hold in all situations.