Method of Images - Green's Function

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SUMMARY

The discussion focuses on using the method of images to derive a Green's function for a specified boundary value problem. The equation presented is the Laplace operator applied to the Green's function, expressed as \(\nabla^2 G = \delta(\underline{x} - \underline{x}_0)\), with boundary conditions defined by \(\frac{\partial G(x,0)}{\partial y} = 0\) for \(y \geq 0\). The solution derived is \(G(r) = \frac{1}{2 \pi} \ln |\underline{x} - \underline{x}_0|\), although the contributor expresses uncertainty about the correctness of applying the method of images in this context.

PREREQUISITES
  • Understanding of Green's functions in partial differential equations
  • Familiarity with the method of images in solving boundary value problems
  • Knowledge of the Laplace operator and its properties
  • Basic calculus, particularly in relation to multivariable functions
NEXT STEPS
  • Study the derivation of Green's functions for different boundary conditions
  • Explore the method of images in electrostatics and fluid dynamics
  • Learn about the properties and applications of the Laplace operator
  • Investigate the implications of singularities in Green's functions
USEFUL FOR

Mathematicians, physicists, and engineers working with partial differential equations, particularly those interested in boundary value problems and Green's functions.

OliviaB
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Use the method of images to find a Green's function for the problem in the attached image.

Demonstrate the functions satisfies the homogenous boundary condition.
 

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It's taken me ages, and I really am struggling to understand this, so I am not sure if this is correct but here goes...

[tex]\bigtriangledown^2 G = \delta(\underline{x} - \underline{x}_0)[/tex] [tex]\frac{\partial G(x,0)}{\partial y} = 0[/tex] for [tex]y \geq 0 \ - \infty < x < \infty[/tex]

Let [tex]r = |\underline{x} - \underline{x}_0 | = \sqrt{(x - x_0)^2 + (y - y_0)^2}[/tex]

be the distance between [tex]\underline{x}[/tex] and [tex]\underline{x}_0[/tex]

Then [tex]\bigtriangledown^2 G = \delta(\underline{x} - \underline{x}_0)[/tex] becomes

[tex]\bigtriangledown^2 G = \frac{1}{r} \frac{\partial}{\partial r} \Big( r \frac{\partial}{\partial r} \Big) = 0[/tex]

everywhere (although not at [tex]r = 0[/tex]) and subject to

[tex]\iint\limits_{\infty} \bigtriangledown G dV = \iint\limits_{\infty} \delta (0) dV = 1[/tex]

which gives

[tex]G(r) = A \ln r + B[/tex]

[tex]A = \frac{1}{2 \pi}[/tex]

Choosing [tex]B = 0[/tex]

[tex]G = \frac{1}{2 \pi} \ln r = \frac{1}{2 \pi} \ln |\underline{x} - \underline{x}_0|[/tex]

I don't think this is the method of images though...(Headbang)
 

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