In mathematics, Laplace equation is one of the most interesting topics in second order partial differential equations. The two dimensional heat equations in Cartesian form in unsteady state is
`(delu)/(delt)` = `alpha^2 [(del^2u)/(delx^2) + (del^2u)/(dely^2)]`
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For the steady state `(delu)/(delt)` = 0, then the above equation becomes
`[(del^2u)/(delx^2) + (del^2u)/(dely^2)]` = 0
(or)
`grad^2 u` = 0
This is the two dimensional heat equation in unsteady state (or) Laplace equation. In this article, we shall discuss about how to solve the Laplace equation. The following are the example problem in solve Laplace equation.
Solve Laplace equation - Laplace equation:
Finite difference solution for the two dimensional heat equation in steady state (or) Laplace equation is
`grad^2 u` = 0
(or)
`(del^2u)/(delx^2)` + `(del^2u)/(dely^2)` = 0
Here we know the values of `(del^2u)/(delx^2)` , `(del^2u/dely^2)`
`(del^2u)/(delx^2)` = `(u_(i+1, j) + u_(i-1, j)-2u_(ij))/(h^2)`
`(del^2u)/(dely^2)` = `(u_(i, j+1) + u_(i, j-1)-2u_(ij))/(h^2)`
substitute the values of `(del^2u)/(delx^2)` , `(del^2u)/(dely^2)` in the Laplace equation, then we get
`4u_(ij)` = `(u_(i+1, j) + u_(i-1, j) + u_(i, j+1) + u_(i, j-1))`
`u_(ij)` = (`1/4` ) `(u_(i+1, j) + u_(i-1, j) + u_(i, j+1) + u_(i, j-1))`
This is the finite difference solution for the Laplace equation, which is also called as the standard five-point formula.
Solve Laplace equation
Solve Laplace equation - Example problem:
Example 1:
Solve the laplace equation for the given region
Solve Laplace equation - Example
Solution:
The finite difference sheme for the Laplace equation is
Solve Laplace equation - Example
u(ij) = (`1/4` ) [`(u_(i+1, j) + u_(i-1, j) + u_(i, j+1) + u_(i, j-1))` ]
Using this we can solve this,
u1 = `1/4` [200 + 50 + u2 + u4]
= `1/4` [250 + u2 + u4] → (1)
u2 = `1/4` [u1 + 0 + 0 + u3]
= `1/4` [u1 + u3] → (2)
u3 = `1/4` [u4 + u2 + 100 + 50]
= `1/4` [150 + u2 + u4] → (3)
u4 = `1/4` [100 + 200 + u1 + u3]
= `1/4` [300 + u1 + u3] → (4)
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Solve the equation by using the gauss seidal method
u1 = `1/4` [250 + u2 + u4]
u2 = `1/4` [u1 + u3]
u3 = `1/4` [150 + u2 + u4]
u4 = `1/4` [300 + u1 + u3]
Let the intial approximation is