- #1
jmb
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Trying to verify the general integral solution to Poisson's equation I reach the following contradiction, what am I missing/doing wrong?
A solution [tex]u(\mathbf{x})[/tex] to Poisson's equation satisfies:
[tex]\nabla^2 u(\mathbf{x}) = - \frac{\rho(\mathbf{x})}{\epsilon_0}[/tex]
we can find such a solution in a given domain by evaluating:
[tex]u(\mathbf{x}) = \int_{V} \frac{\rho(\mathbf{x'})}{4 \pi \epsilon_0 |\mathbf{x}-\mathbf{x'}|} d\mathbf{x'}[/tex]
or equivalently (if we are only interested in the electric field):
[tex]\mathbf{E} = -\nabla u = - \int_{V} \frac{\rho(\mathbf{x'})}{4 \pi \epsilon_0} \frac{\mathbf{x}-\mathbf{x'}}{|\mathbf{x}-\mathbf{x'}|^3} d\mathbf{x'}[/tex]
We can think of these solutions as either the Green's function solution or, physically, as the convolution of the point source solution with the source density [tex]\rho(\mathbf{x})[/tex].
However if I take the Laplacian (or the divergence in the case of the electric field solution) of either of the above I get zero instead of [tex]-\frac{\rho}{\epsilon_0}[/tex]. What am I missing??
To evaluate the Laplacian/divergence of the above I make use of the fact that [tex]\mathbf{x'}[/tex] is a 'dummy variable', which allows me to switch the order of integration and differentiation when applying [tex]\nabla^2[/tex] to the integral on the RHS and also means the components of [tex]\mathbf{x'}[/tex] all have derivative zero. With these assumptions the operation is fairly straightforward to carry out, but I have anyway verified my answer with Mathematica.
A solution [tex]u(\mathbf{x})[/tex] to Poisson's equation satisfies:
[tex]\nabla^2 u(\mathbf{x}) = - \frac{\rho(\mathbf{x})}{\epsilon_0}[/tex]
we can find such a solution in a given domain by evaluating:
[tex]u(\mathbf{x}) = \int_{V} \frac{\rho(\mathbf{x'})}{4 \pi \epsilon_0 |\mathbf{x}-\mathbf{x'}|} d\mathbf{x'}[/tex]
or equivalently (if we are only interested in the electric field):
[tex]\mathbf{E} = -\nabla u = - \int_{V} \frac{\rho(\mathbf{x'})}{4 \pi \epsilon_0} \frac{\mathbf{x}-\mathbf{x'}}{|\mathbf{x}-\mathbf{x'}|^3} d\mathbf{x'}[/tex]
We can think of these solutions as either the Green's function solution or, physically, as the convolution of the point source solution with the source density [tex]\rho(\mathbf{x})[/tex].
However if I take the Laplacian (or the divergence in the case of the electric field solution) of either of the above I get zero instead of [tex]-\frac{\rho}{\epsilon_0}[/tex]. What am I missing??
To evaluate the Laplacian/divergence of the above I make use of the fact that [tex]\mathbf{x'}[/tex] is a 'dummy variable', which allows me to switch the order of integration and differentiation when applying [tex]\nabla^2[/tex] to the integral on the RHS and also means the components of [tex]\mathbf{x'}[/tex] all have derivative zero. With these assumptions the operation is fairly straightforward to carry out, but I have anyway verified my answer with Mathematica.