Green's Function: Understanding Laplace's 2nd Identity

[rsq_a]

This appears on the bottom of p.279 of this book.

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

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

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

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
$$\int_V \alpha \nabla^2 \beta - \beta \nabla^2 \alpha \ dV = \int_V \alpha \delta((x,y) - (x^*, y^*)) \ dV = \alpha(x^*, y^*)$$

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).

 

[kakarotyjn]

$$\alpha$$ is harmonic in V,but $$g$$ is harmonic in V except in $$(x^*,y^*)$$,so (10.7)doesn't exist for $$g$$.

if $$(x^*,y^*)$$ is inside C,we need to construct a ball $$B_\eplson$$ to cover $$(x^*,y^*)$$,and the formula is wright for $$\int_{C \cup C_\varepsilon } {\left( {\alpha \frac{{\partial g}}{{\partial n}} - g\frac{{\partial \alpha }}{{\partial n}}} \right)} \;ds = 0$$,you can get the formula (10.8) from this.You need to do it yourself when the point is on C

 

[kakarotyjn]

$$C_\epsilon$$ is the boundry curve of B_\epsilon,you need to take limit for epsilon

 

[rsq_a]

$$\alpha$$ is harmonic in V,but $$g$$ is harmonic in V except in $$(x^*,y^*)$$,so (10.7)doesn't exist for $$g$$.

if $$(x^*,y^*)$$ is inside C,we need to construct a ball $$B_\eplson$$ to cover $$(x^*,y^*)$$,and the formula is wright for $$\int_{C \cup C_\varepsilon } {\left( {\alpha \frac{{\partial g}}{{\partial n}} - g\frac{{\partial \alpha }}{{\partial n}}} \right)} \;ds = 0$$,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,
$$0 = \int_{C \cup C_\varepsilon } {\left( {\alpha \frac{{\partial g}}{{\partial n}} - g\frac{{\partial \alpha }}{{\partial n}}} \right)} \;ds = \int_{C} {\left( {\alpha \frac{{\partial g}}{{\partial n}} - g\frac{{\partial \alpha }}{{\partial n}}} \right)} \;ds -\alpha(x^*, y^*)$$

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

where the factor of 1/2 results from only going around half a circle.

 

[kakarotyjn]

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:

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

Sorry

 

[rsq_a]

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:

$$\left. {\frac{{\partial g}}{{\partial n}}} \right|_{C_\varepsilon } = \left. { - \frac{1}{{2\pi r}}} \right|_{C_\varepsilon } = - \frac{1}{{2\pi \varepsilon }}$$,,,,,,,,,,,,,,,,,,,,,,,,,,,
$$\left. g \right|_{C_\varepsilon } = \frac{1}{{2\pi }}\ln \varepsilon$$,,,,,,,,,,,,,,,,,
so $$\int_{C_\varepsilon } {\left( {\alpha \frac{{\partial g}}{{\partial n}} - g\frac{{\partial \alpha }}{{\partial n}}} \right)} \:ds = - \frac{1}{{2\pi \varepsilon }}\int_{C_\varepsilon } \alpha \:ds - \frac{1}{{2\pi }}\ln \varepsilon \int_{C_\varepsilon } {\left( {\frac{{\partial \alpha }}{{\partial n}}} \right)} \:ds = - \int_C {\left( {\alpha \frac{{\partial g}}{{\partial n}} - g\frac{{\partial \alpha }}{{\partial n}}} \right)} \:ds$$,,,,,,,,,,,,,,,,,,,,,,
then $$\frac{1}{{2\pi \varepsilon }}\int_{C_\varepsilon } \alpha \:ds + \frac{1}{{2\pi }}\ln \varepsilon \int_{C_\varepsilon } {\left( {\frac{{\partial \alpha }}{{\partial n}}} \right)} \:ds = \int_C {\left( {\alpha \frac{{\partial g}}{{\partial n}} - g\frac{{\partial \alpha }}{{\partial n}}} \right)} \:ds$$,
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

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

The LHS will either give you $alpha$ (not on the boundary) or $\alpha/2$ (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.

 

[kakarotyjn]

we can get this equation $$\int_V {\alpha \nabla ^2 \beta } dV + \int_V {\frac{{\partial \alpha }}{{\partial x}}\frac{{\partial \beta }}{{\partial x}}} + \frac{{\partial \alpha }}{{\partial y}}\frac{{\partial \beta }}{{\partial y}}dV = \int_C {\left( {\alpha \frac{{\partial \beta }}{{\partial n}}} \right)} ds$$by Green Theorem,the same one$$\int_V {\beta \nabla ^2 \alpha } dV + \int_V {\frac{{\partial \beta }}{{\partial x}}\frac{{\partial \alpha }}{{\partial x}}} + \frac{{\partial \beta }}{{\partial y}}\frac{{\partial \alpha }}{{\partial y}}dV = \int_C {\left( {\beta \frac{{\partial \alpha }}{{\partial n}}} \right)} ds$$,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...