8 Laplaces Equation in 2D 8.1 Simple idea SLIDE 11 u is given on surface out fdex-*)+(ry Does not match boundary conditions! Note 8 Simple idea for solving Laplace's equation Here is a simple idea for computing the solution of La s equation o u(r, y)=alog V(a-xo)2+(y-yo where ro, yo is a point inside the square. Clearly such a u will satisfy Vu=0 outside the square, but u may not match the boundary conditions. By adjusting a, it is possible to make sure to match the boundary condition at at least one b Exercise 2 Suppose the potential on the surface of the square is a constant Can you match that constant potential everywhere on the perimeter of the square by judiciously selecting a. I 8.1.1 More points SLIDE 12 u is given on surface a2u du 0 out (x,y)● (., y.) Letu=∑11lg(√(-)2+(-)=∑aG(x-r;y-) Pick the ais to match the boundary conditions8 Laplace’s Equation in 2D 8.1 Simple idea Slide 11 2 2 2 2 + = 0 outs u u x y ∂ ∂ ∂ ∂ Surface ( ) x y 0 0 , 2 2 2 2 + = 0 outside u u x y ∂ ∂ ∂ ∂ Problem Solved u is given on surface ( ) ( )( ) 2 2 Let log u xx yy = − +− 0 0 Does not match boundary conditions! Note 8 Simple idea for solving Laplace’s equation in 2D Here is a simple idea for computing the solution of Laplace’s equation outside the square. Simply let u(x, y) = α log (x − x0)2 + (y − y0)2 where x0, y0 is a point inside the square. Clearly such a u will satisfy ∇2u = 0 outside the square, but u may not match the boundary conditions. By adjusting α, it is possible to make sure to match the boundary condition at at least one point.  Exercise 2 Suppose the potential on the surface of the square is a constant. Can you match that constant potential everywhere on the perimeter of the square by judiciously selecting α. 8.1.1 "More points" Slide 12 2 2 2 2 + = 0 out u u x y ∂ ∂ ∂ ∂ ( ) x y 1 1 , ( ) x2 2 ,y u is given on surface ( ) x y n n , Let u = n i=1 αi log (x − xi)2 + (y − yi)2  = n i=1 αiG(x − xi, y − yi) Pick the αi’s to match the boundary conditions! 8