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

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Unformatted text preview: 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 ≤ x ≤ L < 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 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 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 ∞ U S ( x,k ) sin( ky ) dy, U S ( x,k ) = 2 π Z ∞ φ 2 ( x,y ) sin( dy ) dy. When we transform the equation we get the ODE d 2 U S dx 2 + 2 kφ 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 ∞ 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 ) ....
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This note was uploaded on 12/12/2009 for the course ACM 95c taught by Professor Nilesa.pierce during the Spring '09 term at Caltech.

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

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