- #1
yungman
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This is not homework. This is actually a subset of proofing G(\vec{x},\vec{x_0}) = G(\vec{x_0},\vec{x}) where G is the Green's function. I don't want to present the whole thing, just the part I have question.
Let D be an open solid region with surface S. Let P \;=\; G(\vec{x},\vec{a}) \;\hbox{ and } P \;=\; G(\vec{x},\vec{b}) \; where both are green function at point a and b resp. inside D. This means Q is defined at point a ( harmonic at point a ) and P is defined at point b. Both P and Q are defined in D except at a and b resp. Both equal to zero on surface S.
Green function defined:
G(\vec{x},\vec{x_0}) \;=\; v + H \;\hbox { where } \;v=\; \frac{-1}{4\pi|\vec{x}-\vec{x_0|}} \;\hbox{ and }\; H \;\hbox { is a harmonic function in D and on S where }\; G(\vec{x},\vec{x_0}) \;=\; 0 \;\hbox { on D}.
In this proof, I need to make two spherical cutout each with radius =\epsilon with center at a and b. I call the spherical region of this two sphere A and B resp and the surface S_a \;&\; S_b resp. Then I let D_{\epsilon} = D -A-B so both P and Q are defined and harmonic in D_{\epsilon}.
Now come to the step I need to verify:
I want to prove:
^{lim}_{\epsilon\rightarrow 0} \int\int_{S_a} P\frac{\partial Q}{\partial n} \;-\; Q\frac{\partial P}{\partial n} \;dS \;=\; ^{lim}_{\epsilon\rightarrow 0} \int\int_{S_a} v\frac{1}{4\pi\epsilon^2} \;dS
This is my work:
^{lim}_{\epsilon\rightarrow 0} \int\int_{S_a} P\frac{\partial Q}{\partial n} \;-\; Q\frac{\partial P}{\partial n} \;dS \;=\; ^{lim}_{\epsilon\rightarrow 0} \int\int_{S_a} (-\frac{1}{4\pi r} + H)\frac{\partial Q}{\partial n} \;-\; Q\frac{\partial }{\partial n}(-\frac{1}{4\pi r} + H) \;dS (1)
Where:
^{lim}_{\epsilon\rightarrow 0}\; v\; =\; \frac{-1}{4\pi |\vec{x}-\vec{a}|} \;=\; ^{lim}_{\epsilon\rightarrow 0} \;\frac{-1}{4\pi r} \;. in sphere region A.
^{lim}_{\epsilon\rightarrow 0}( P=v+H )\;=\; ^{lim}_{\epsilon\rightarrow 0} (\frac{-1}{4\pi r } + H)
Form (1) I break into 3 parts:
^{lim}_{\epsilon\rightarrow 0} [ \int\int_{S_a} -\frac{1}{4\pi r}\frac{\partial Q}{\partial n} dS + \int\int_{S_a} (H\frac{\partial Q }{\partial n} \;-\; Q\frac{\partial H}{\partial n}) dS + \int\int_{S_a} Q \frac{\partial}{\partial n}(-\frac{1}{4\pi r}) \;dS]
^{lim}_{\epsilon\rightarrow 0} [ \int\int_{S_a} -\frac{1}{4\pi r}\frac{\partial Q}{\partial n} dS \;=\; -\frac{1}{4\pi \epsilon} \int\int_{S_a} \frac{\partial Q}{\partial n} dS \;=\; 0
Because Q is harmonic and \int\int_{S_a} \frac{\partial Q}{\partial n} dS \;=\; 0
From second identity:
\int\int_{S_a} (H\frac{\partial Q }{\partial n} \;-\; Q\frac{\partial H}{\partial n}) dS \;= \int\int\int_A (H\nabla^2 Q - Q\nabla^2 H) dV =0
because both H and Q are harmonic in A and on surface S_A.
Therefore.
^{lim}_{\epsilon\rightarrow 0} \int\int_{S_a} P\frac{\partial Q}{\partial n} \;-\; Q\frac{\partial P}{\partial n} \;dS \;=\; ^{lim}_{\epsilon\rightarrow 0}\int\int_{S_a} Q \frac{\partial}{\partial n}(-\frac{1}{4\pi r}) \;dS = \frac{1}{4\pi \epsilon^2} \int\int_{S_a} Q dS
The proof of the Strauss's book is very funky to put it politely. This is the way I proof it and please bare with the long explanation and tell me whether I am correct or not.
Thanks
Alan
Let D be an open solid region with surface S. Let P \;=\; G(\vec{x},\vec{a}) \;\hbox{ and } P \;=\; G(\vec{x},\vec{b}) \; where both are green function at point a and b resp. inside D. This means Q is defined at point a ( harmonic at point a ) and P is defined at point b. Both P and Q are defined in D except at a and b resp. Both equal to zero on surface S.
Green function defined:
G(\vec{x},\vec{x_0}) \;=\; v + H \;\hbox { where } \;v=\; \frac{-1}{4\pi|\vec{x}-\vec{x_0|}} \;\hbox{ and }\; H \;\hbox { is a harmonic function in D and on S where }\; G(\vec{x},\vec{x_0}) \;=\; 0 \;\hbox { on D}.
In this proof, I need to make two spherical cutout each with radius =\epsilon with center at a and b. I call the spherical region of this two sphere A and B resp and the surface S_a \;&\; S_b resp. Then I let D_{\epsilon} = D -A-B so both P and Q are defined and harmonic in D_{\epsilon}.
Now come to the step I need to verify:
I want to prove:
^{lim}_{\epsilon\rightarrow 0} \int\int_{S_a} P\frac{\partial Q}{\partial n} \;-\; Q\frac{\partial P}{\partial n} \;dS \;=\; ^{lim}_{\epsilon\rightarrow 0} \int\int_{S_a} v\frac{1}{4\pi\epsilon^2} \;dS
This is my work:
^{lim}_{\epsilon\rightarrow 0} \int\int_{S_a} P\frac{\partial Q}{\partial n} \;-\; Q\frac{\partial P}{\partial n} \;dS \;=\; ^{lim}_{\epsilon\rightarrow 0} \int\int_{S_a} (-\frac{1}{4\pi r} + H)\frac{\partial Q}{\partial n} \;-\; Q\frac{\partial }{\partial n}(-\frac{1}{4\pi r} + H) \;dS (1)
Where:
^{lim}_{\epsilon\rightarrow 0}\; v\; =\; \frac{-1}{4\pi |\vec{x}-\vec{a}|} \;=\; ^{lim}_{\epsilon\rightarrow 0} \;\frac{-1}{4\pi r} \;. in sphere region A.
^{lim}_{\epsilon\rightarrow 0}( P=v+H )\;=\; ^{lim}_{\epsilon\rightarrow 0} (\frac{-1}{4\pi r } + H)
Form (1) I break into 3 parts:
^{lim}_{\epsilon\rightarrow 0} [ \int\int_{S_a} -\frac{1}{4\pi r}\frac{\partial Q}{\partial n} dS + \int\int_{S_a} (H\frac{\partial Q }{\partial n} \;-\; Q\frac{\partial H}{\partial n}) dS + \int\int_{S_a} Q \frac{\partial}{\partial n}(-\frac{1}{4\pi r}) \;dS]
^{lim}_{\epsilon\rightarrow 0} [ \int\int_{S_a} -\frac{1}{4\pi r}\frac{\partial Q}{\partial n} dS \;=\; -\frac{1}{4\pi \epsilon} \int\int_{S_a} \frac{\partial Q}{\partial n} dS \;=\; 0
Because Q is harmonic and \int\int_{S_a} \frac{\partial Q}{\partial n} dS \;=\; 0
From second identity:
\int\int_{S_a} (H\frac{\partial Q }{\partial n} \;-\; Q\frac{\partial H}{\partial n}) dS \;= \int\int\int_A (H\nabla^2 Q - Q\nabla^2 H) dV =0
because both H and Q are harmonic in A and on surface S_A.
Therefore.
^{lim}_{\epsilon\rightarrow 0} \int\int_{S_a} P\frac{\partial Q}{\partial n} \;-\; Q\frac{\partial P}{\partial n} \;dS \;=\; ^{lim}_{\epsilon\rightarrow 0}\int\int_{S_a} Q \frac{\partial}{\partial n}(-\frac{1}{4\pi r}) \;dS = \frac{1}{4\pi \epsilon^2} \int\int_{S_a} Q dS
The proof of the Strauss's book is very funky to put it politely. This is the way I proof it and please bare with the long explanation and tell me whether I am correct or not.
Thanks
Alan
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