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Week_08

# Week_08 - ACM 100c Methods of Applied Mathematics Week 8 1...

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ACM 100c - Methods of Applied Mathematics Week 8 May 17, 2009 1 Laplace equation in a semi-infinite strip In this section, we consider the solution of the Laplace equation in 2-D Cartesian coordi- nates, but in a semi-infinite region: 2 φ ( x, y ) = 0 0 x L 0 < y < + . The boundary conditions here are given by φ (0 , y ) = g 1 ( y ) , φ ( L, y ) = g 2 ( y ) , φ ( x, 0) = f ( x ) . We have to assume that the solution behaves well as y → ∞ We can think of the solution as the sum of the solutions of two sub-problems: φ = φ 1 + φ 2 where 2 φ 1 ( x, y ) = 0 with boundary conditions φ 1 ( x, 0) = 0 φ 1 (0 , y ) = g 1 ( y ) φ 1 ( L, y ) = g 2 ( y ) , and where 2 φ 2 ( x, y ) = 0 with boundary conditions φ 2 ( x, 0) = f ( x ) φ 2 (0 , y ) = 0 φ 2 ( L, y ) = g 2 ( y ) . We look at the solution for φ 1 and φ 2 in turn. For φ 2 , the solution is easily accomplished using finite transforms: φ 2 ( x, y ) = X n =1 A n ( y ) sin( nπx/L ) . 1

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The ODE’s for the coefficients A n are d 2 A n dy 2 - n 2 π 2 L 2 A n = 0 . From this, we can see that φ 2 ( x, y ) = X n =1 F n exp( - nπy/L ) sin( nπx/L ) , where F n = 2 L Z L 0 f ( x ) sin( nπx/L ) dx. In order to get φ 1 we use the Fourier sine transform since this is a useful approach when you have Dirichlet boundary conditions at one end of the semi-infinite interval. We write the solution as φ 1 ( x, y ) = Z 0 U S ( x, k ) sin( ky ) dy, U S ( x, k ) = 2 π Z 0 φ 2 ( x, y ) sin( dy ) dy. When we transform the equation we get the ODE d 2 U S dx 2 + 2 2 ( x, 0) π - k 2 U S ( x, k ) = 0 , with boundary conditions U S (0 , k ) = G 1 S ( k ) U S ( L, k ) = G 2 S ( k ) , where G 1 S and G 2 S are the sine transforms of the boundary conditions g 1 ( y ) and g 2 ( y ) . Note that the boundary value at y = 0 also appears as an inhomogeneous term in the ODE so in principle we really did not need to solve earlier for φ 2 . We can do it all at once with the sine transform. The solution is U S ( x, k ) = G 2 S ( k ) sinh( kL ) sinh( kx ) + G 1 S ( k ) sinh( kL ) sinh( k ( L - x )) At this point the solution is obtained by performing the inverse transform: φ 1 ( x, y ) = Z 0 U S ( x, k ) sin( ky ) dy, . At this point, the solution will be in the form of an integral. Note there is more than one approach to this problem. For example we could have written φ ( x, y ) = ψ ( x, y ) + g 1 ( y ) + ( g 2 - g 1 ) x/L. 2
If we plug this into the Laplace equation for φ we get a Poisson equation which now has homogeneous boundary conditions at x = 0 and x = L . We could then expand the solution as before using sines: ψ ( x, y ) = X n =1 C n ( y ) sin( nπx/L ) and solve the ODE’s for C n . All these formulas are simply different representations of the same solution φ ( x, y ) . In contrast to the solution of Laplace’s equation in a strip, the solution in a full half plane can be done in closed form using Fourier transforms. Consider the Laplace equation 2 φ ∂x 2 + 2 φ ∂y 2 = 0 in the half plane domain -∞ < x < and y > 0 . Here the boundary conditions are φ ( x, 0) = f ( x ) along with the implicit conditions that the solution behave well as | x | → ∞ and y → ∞ .

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