Green's function

  • Thread starter rsq_a
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  • #1
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This appears on the bottom of p.279 of this book.

The author begins with Green's second identity:
[tex]
\int_V \alpha \nabla^2 \beta - \beta \nabla^2 \alpha \ dV =
\int_C \left( \alpha \frac{\partial \beta}{\partial n} - \beta \frac{\partial \alpha}{\partial n} \right) \ ds[/tex]

Here, C is a closed curve, s is the arc length for C and n is the outward unit normal. We then let [itex]\alpha[/itex] satisfy Laplace's equation and let [itex]g = 1/(4\pi)\log [(x-x^*)^2 + (y-y^*)^2][/itex], i.e. the free-space Green's function. Then he gets

[tex]
\alpha(x^*, y^*) = r\int_C \left( \alpha \frac{\partial \beta}{\partial n} - \beta \frac{\partial \alpha}{\partial n} \right) \ ds,[/tex]

where r = 1 if (x*,y*) is inside C, but r = 1/2 if (x*, y*) is on C.

I'm confused at this point. I thought that
[tex]
\int_V \alpha \nabla^2 \beta - \beta \nabla^2 \alpha \ dV =
\int_V \alpha \delta((x,y) - (x^*, y^*)) \ dV = \alpha(x^*, y^*)
[/tex]

Where is the factor of 1/2 or 1 coming in?

Moreover, the next equation, the factor of r = 1/2 has switched to the left-hand-side. I can't figure out how this is done (but perhaps if someone firsts helps me understand the above, this will be clear).
 

Answers and Replies

  • #2
98
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[tex]\alpha[/tex] is harmonic in V,but [tex]g[/tex] is harmonic in V except in [tex](x^*,y^*)[/tex],so (10.7)doesn't exist for [tex]g[/tex].

if [tex](x^*,y^*)[/tex] is inside C,we need to construct a ball [tex]B_\eplson[/tex] to cover [tex](x^*,y^*)[/tex],and the formula is wright for [tex]\int_{C \cup C_\varepsilon } {\left( {\alpha \frac{{\partial g}}{{\partial n}} - g\frac{{\partial \alpha }}{{\partial n}}} \right)} \;ds = 0[/tex],you can get the formula (10.8) from this.You need to do it yourself when the point is on C
 
  • #3
98
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[tex]C_\epsilon[/tex] is the boundry curve of B_\epsilon,you need to take limit for epsilon
 
  • #4
107
1
[tex]\alpha[/tex] is harmonic in V,but [tex]g[/tex] is harmonic in V except in [tex](x^*,y^*)[/tex],so (10.7)doesn't exist for [tex]g[/tex].

if [tex](x^*,y^*)[/tex] is inside C,we need to construct a ball [tex]B_\eplson[/tex] to cover [tex](x^*,y^*)[/tex],and the formula is wright for [tex]\int_{C \cup C_\varepsilon } {\left( {\alpha \frac{{\partial g}}{{\partial n}} - g\frac{{\partial \alpha }}{{\partial n}}} \right)} \;ds = 0[/tex],you can get the formula (10.8) from this.You need to do it yourself when the point is on C
Ah, I see. Can you verify that the formula (10.7) is indeed right, and the 'r' is supposed to be on the right-hand side?

I think it's supposed to be on the LHS. Then the 1/2 factor on the LHS of (10.9) would make sense.

I reason that for the case that (x*, y*) is inside C,
[tex]0 = \int_{C \cup C_\varepsilon } {\left( {\alpha \frac{{\partial g}}{{\partial n}} - g\frac{{\partial \alpha }}{{\partial n}}} \right)} \;ds = \int_{C} {\left( {\alpha \frac{{\partial g}}{{\partial n}} - g\frac{{\partial \alpha }}{{\partial n}}} \right)} \;ds -\alpha(x^*, y^*)[/tex]

because you end up with going around an entire circle for [itex]C_\epsilon[/itex], but if (x*, y*) is on the boundary, then
[tex]0 = \int_{C \cup C_\varepsilon } {\left( {\alpha \frac{{\partial g}}{{\partial n}} - g\frac{{\partial \alpha }}{{\partial n}}} \right)} \;ds = \int_{C} {\left( {\alpha \frac{{\partial g}}{{\partial n}} - g\frac{{\partial \alpha }}{{\partial n}}} \right)} \;ds - \frac{1}{2}\alpha(x^*, y^*)[/tex]

where the factor of 1/2 results from only going around half a circle.
 
  • #5
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I'm sorry I don't know how to derive (10.8) from (10.7),(10.7) is right,but my calculation don't get (10.8).
this is my calculation:

[tex]
\left. {\frac{{\partial g}}{{\partial n}}} \right|_{C_\varepsilon } = \left. { - \frac{1}{{2\pi r}}} \right|_{C_\varepsilon } = - \frac{1}{{2\pi \varepsilon }}[/tex],,,,,,,,,,,,,,,,,,,,,,,,,,,
[tex]\left. g \right|_{C_\varepsilon } = \frac{1}{{2\pi }}\ln \varepsilon[/tex],,,,,,,,,,,,,,,,,
so [tex]\int_{C_\varepsilon } {\left( {\alpha \frac{{\partial g}}{{\partial n}} - g\frac{{\partial \alpha }}{{\partial n}}} \right)} \:ds = - \frac{1}{{2\pi \varepsilon }}\int_{C_\varepsilon } \alpha \:ds - \frac{1}{{2\pi }}\ln \varepsilon \int_{C_\varepsilon } {\left( {\frac{{\partial \alpha }}{{\partial n}}} \right)} \:ds = - \int_C {\left( {\alpha \frac{{\partial g}}{{\partial n}} - g\frac{{\partial \alpha }}{{\partial n}}} \right)} \:ds[/tex],,,,,,,,,,,,,,,,,,,,,,
then [tex]
\frac{1}{{2\pi \varepsilon }}\int_{C_\varepsilon } \alpha \:ds + \frac{1}{{2\pi }}\ln \varepsilon \int_{C_\varepsilon } {\left( {\frac{{\partial \alpha }}{{\partial n}}} \right)} \:ds = \int_C {\left( {\alpha \frac{{\partial g}}{{\partial n}} - g\frac{{\partial \alpha }}{{\partial n}}} \right)} \:ds[/tex],
but I don't know how to eliminate the second term,and I also don't know where does the r come from.

Sorry
 
Last edited:
  • #6
107
1
I'm sorry I don't know how to derive (10.8) from (10.7),(10.7) is right,but my calculation don't get (10.8).
this is my calculation:

[tex]
\left. {\frac{{\partial g}}{{\partial n}}} \right|_{C_\varepsilon } = \left. { - \frac{1}{{2\pi r}}} \right|_{C_\varepsilon } = - \frac{1}{{2\pi \varepsilon }}[/tex],,,,,,,,,,,,,,,,,,,,,,,,,,,
[tex]\left. g \right|_{C_\varepsilon } = \frac{1}{{2\pi }}\ln \varepsilon[/tex],,,,,,,,,,,,,,,,,
so [tex]\int_{C_\varepsilon } {\left( {\alpha \frac{{\partial g}}{{\partial n}} - g\frac{{\partial \alpha }}{{\partial n}}} \right)} \:ds = - \frac{1}{{2\pi \varepsilon }}\int_{C_\varepsilon } \alpha \:ds - \frac{1}{{2\pi }}\ln \varepsilon \int_{C_\varepsilon } {\left( {\frac{{\partial \alpha }}{{\partial n}}} \right)} \:ds = - \int_C {\left( {\alpha \frac{{\partial g}}{{\partial n}} - g\frac{{\partial \alpha }}{{\partial n}}} \right)} \:ds[/tex],,,,,,,,,,,,,,,,,,,,,,
then [tex]
\frac{1}{{2\pi \varepsilon }}\int_{C_\varepsilon } \alpha \:ds + \frac{1}{{2\pi }}\ln \varepsilon \int_{C_\varepsilon } {\left( {\frac{{\partial \alpha }}{{\partial n}}} \right)} \:ds = \int_C {\left( {\alpha \frac{{\partial g}}{{\partial n}} - g\frac{{\partial \alpha }}{{\partial n}}} \right)} \:ds[/tex],
but I don't know how to eliminate the second term,and I also don't know where does the r come from.

Sorry
The second term ends up being lower-order than the first as [itex]\epsilon \to 0[\itex] (you'll get epsilon*log(epsilon)). And thus we are left with

[tex]
\frac{1}{{2\pi \varepsilon }}\int_{C_\varepsilon } \alpha \:ds = \int_C {\left( {\alpha \frac{{\partial g}}{{\partial n}} - g\frac{{\partial \alpha }}{{\partial n}}} \right)} \:ds[/tex]

The LHS will either give you [itex]alpha[/itex] (not on the boundary) or [itex]\alpha/2[/itex] (on the boundary). I'm not sure how that gives you (10.7). As I said in my post above, I think the 'r' should be on the other side in the formula.
 
  • #7
98
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we can get this equation [tex]
\int_V {\alpha \nabla ^2 \beta } dV + \int_V {\frac{{\partial \alpha }}{{\partial x}}\frac{{\partial \beta }}{{\partial x}}} + \frac{{\partial \alpha }}{{\partial y}}\frac{{\partial \beta }}{{\partial y}}dV = \int_C {\left( {\alpha \frac{{\partial \beta }}{{\partial n}}} \right)} ds[/tex]by Green Theorem,the same one[tex]\int_V {\beta \nabla ^2 \alpha } dV + \int_V {\frac{{\partial \beta }}{{\partial x}}\frac{{\partial \alpha }}{{\partial x}}} + \frac{{\partial \beta }}{{\partial y}}\frac{{\partial \alpha }}{{\partial y}}dV = \int_C {\left( {\beta \frac{{\partial \alpha }}{{\partial n}}} \right)} ds[/tex],then you can ge (10.7) by subtracting the first one from the second one.

and the r is on the right side,not left.But I can't see clear of your last post,the formula can't be displayed well.

How can we eliminate the second term?when eplison tends to 0,ln(eplison) tend to be minus infinity......
 

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