Neumann problem for a circle.
$begingroup$
We consider the interior neumann problem : $$nabla^{2} u =0:::, ::r<R$$
$$ dfrac{partial u}{partial n}=dfrac{partial u}{partial r}=f(theta)::,::r=R::,:: 0<theta<2pi$$
We first establish the fact which as the solution says is the compatibility condition : $$int_B fds=0$$
which is established using Green's second formula. Now the solution proceeds as follows :
(1) The compatibility condition obtained above can be written as $$R int_0^{2pi} f(theta) dtheta=0$$
(2) In the case of the dirichlet problem of the circle , the solution of the laplace equation is given as : $$u(r,theta) = dfrac{a_0}{2} + sum_{k=1}^{infty}r^{k}(a_k cosktheta + b_k sinktheta)$$
(3) Differentiating with respect to r and applying the boundary conditions we obtain : $$dfrac{partial u}{partial r}(R,theta) = sum_{k=1}^{infty}kR^{k-1}(a_k cosktheta + b_k sinktheta) = f(theta)$$
(4) Now the expression of $f(theta)$ in a series of the form (3) is possible only by the virtue of the compatibility condition since : $$a_0 = dfrac{1}{pi} int_0^{2pi} f(tau) d tau = 0$$
The rest of the proof was okay , can anyone explain the points (1) and (4) ?
Like in (1) are we changing the variable into $theta$ ? If yes , then how is $R$ there ?
And I have no idea about (4). Kindly help !
Thanks in advance !
ordinary-differential-equations
asked Mar 7 '16 at 19:33
User9523User9523
98511026
$endgroup$
add a comment |
$begingroup$
We consider the interior neumann problem : $$nabla^{2} u =0:::, ::r<R$$
$$ dfrac{partial u}{partial n}=dfrac{partial u}{partial r}=f(theta)::,::r=R::,:: 0<theta<2pi$$
We first establish the fact which as the solution says is the compatibility condition : $$int_B fds=0$$
which is established using Green's second formula. Now the solution proceeds as follows :
(1) The compatibility condition obtained above can be written as $$R int_0^{2pi} f(theta) dtheta=0$$
(2) In the case of the dirichlet problem of the circle , the solution of the laplace equation is given as : $$u(r,theta) = dfrac{a_0}{2} + sum_{k=1}^{infty}r^{k}(a_k cosktheta + b_k sinktheta)$$
(3) Differentiating with respect to r and applying the boundary conditions we obtain : $$dfrac{partial u}{partial r}(R,theta) = sum_{k=1}^{infty}kR^{k-1}(a_k cosktheta + b_k sinktheta) = f(theta)$$
(4) Now the expression of $f(theta)$ in a series of the form (3) is possible only by the virtue of the compatibility condition since : $$a_0 = dfrac{1}{pi} int_0^{2pi} f(tau) d tau = 0$$
The rest of the proof was okay , can anyone explain the points (1) and (4) ?
Like in (1) are we changing the variable into $theta$ ? If yes , then how is $R$ there ?
And I have no idea about (4). Kindly help !
Thanks in advance !
ordinary-differential-equations
asked Mar 7 '16 at 19:33
User9523User9523
98511026
$endgroup$
add a comment |
$begingroup$
We consider the interior neumann problem : $$nabla^{2} u =0:::, ::r<R$$
$$ dfrac{partial u}{partial n}=dfrac{partial u}{partial r}=f(theta)::,::r=R::,:: 0<theta<2pi$$
We first establish the fact which as the solution says is the compatibility condition : $$int_B fds=0$$
which is established using Green's second formula. Now the solution proceeds as follows :
(1) The compatibility condition obtained above can be written as $$R int_0^{2pi} f(theta) dtheta=0$$
(2) In the case of the dirichlet problem of the circle , the solution of the laplace equation is given as : $$u(r,theta) = dfrac{a_0}{2} + sum_{k=1}^{infty}r^{k}(a_k cosktheta + b_k sinktheta)$$
(3) Differentiating with respect to r and applying the boundary conditions we obtain : $$dfrac{partial u}{partial r}(R,theta) = sum_{k=1}^{infty}kR^{k-1}(a_k cosktheta + b_k sinktheta) = f(theta)$$
(4) Now the expression of $f(theta)$ in a series of the form (3) is possible only by the virtue of the compatibility condition since : $$a_0 = dfrac{1}{pi} int_0^{2pi} f(tau) d tau = 0$$
The rest of the proof was okay , can anyone explain the points (1) and (4) ?
Like in (1) are we changing the variable into $theta$ ? If yes , then how is $R$ there ?
And I have no idea about (4). Kindly help !
Thanks in advance !
ordinary-differential-equations
asked Mar 7 '16 at 19:33
User9523User9523
98511026
$endgroup$
We consider the interior neumann problem : $$nabla^{2} u =0:::, ::r<R$$
$$ dfrac{partial u}{partial n}=dfrac{partial u}{partial r}=f(theta)::,::r=R::,:: 0<theta<2pi$$
We first establish the fact which as the solution says is the compatibility condition : $$int_B fds=0$$
which is established using Green's second formula. Now the solution proceeds as follows :
(1) The compatibility condition obtained above can be written as $$R int_0^{2pi} f(theta) dtheta=0$$
(2) In the case of the dirichlet problem of the circle , the solution of the laplace equation is given as : $$u(r,theta) = dfrac{a_0}{2} + sum_{k=1}^{infty}r^{k}(a_k cosktheta + b_k sinktheta)$$
(3) Differentiating with respect to r and applying the boundary conditions we obtain : $$dfrac{partial u}{partial r}(R,theta) = sum_{k=1}^{infty}kR^{k-1}(a_k cosktheta + b_k sinktheta) = f(theta)$$
(4) Now the expression of $f(theta)$ in a series of the form (3) is possible only by the virtue of the compatibility condition since : $$a_0 = dfrac{1}{pi} int_0^{2pi} f(tau) d tau = 0$$
The rest of the proof was okay , can anyone explain the points (1) and (4) ?
Like in (1) are we changing the variable into $theta$ ? If yes , then how is $R$ there ?
And I have no idea about (4). Kindly help !
Thanks in advance !
ordinary-differential-equations
ordinary-differential-equations
asked Mar 7 '16 at 19:33
User9523User9523
98511026
asked Mar 7 '16 at 19:33
User9523User9523
98511026
edited Mar 8 '16 at 12:20
User9523
asked Mar 7 '16 at 19:33
User9523User9523
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asked Mar 7 '16 at 19:33
User9523User9523
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asked Mar 7 '16 at 19:33
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1 Answer
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The integral in (1) is transformed into polar co-ordinates by considering the " arc-length " relation. $ds$ here is the arc length. $R$ is the radius of the region. So , using $ theta = dfrac{l}{r}$ , where $l$ is the arc length and $r$ is the radius . Thus we get $ theta = dfrac{ds}{R}$ . Substituting this gives the result.
(4) actually tells the value of the constant $a_0$. That's all.
answered Mar 12 '16 at 10:03
User9523User9523
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1 Answer
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1 Answer
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$begingroup$
The integral in (1) is transformed into polar co-ordinates by considering the " arc-length " relation. $ds$ here is the arc length. $R$ is the radius of the region. So , using $ theta = dfrac{l}{r}$ , where $l$ is the arc length and $r$ is the radius . Thus we get $ theta = dfrac{ds}{R}$ . Substituting this gives the result.
(4) actually tells the value of the constant $a_0$. That's all.
answered Mar 12 '16 at 10:03
User9523User9523
98511026
$endgroup$
add a comment |
$begingroup$
The integral in (1) is transformed into polar co-ordinates by considering the " arc-length " relation. $ds$ here is the arc length. $R$ is the radius of the region. So , using $ theta = dfrac{l}{r}$ , where $l$ is the arc length and $r$ is the radius . Thus we get $ theta = dfrac{ds}{R}$ . Substituting this gives the result.
(4) actually tells the value of the constant $a_0$. That's all.
answered Mar 12 '16 at 10:03
User9523User9523
98511026
$endgroup$
add a comment |
$begingroup$
The integral in (1) is transformed into polar co-ordinates by considering the " arc-length " relation. $ds$ here is the arc length. $R$ is the radius of the region. So , using $ theta = dfrac{l}{r}$ , where $l$ is the arc length and $r$ is the radius . Thus we get $ theta = dfrac{ds}{R}$ . Substituting this gives the result.
(4) actually tells the value of the constant $a_0$. That's all.
answered Mar 12 '16 at 10:03
User9523User9523
98511026
$endgroup$
The integral in (1) is transformed into polar co-ordinates by considering the " arc-length " relation. $ds$ here is the arc length. $R$ is the radius of the region. So , using $ theta = dfrac{l}{r}$ , where $l$ is the arc length and $r$ is the radius . Thus we get $ theta = dfrac{ds}{R}$ . Substituting this gives the result.
(4) actually tells the value of the constant $a_0$. That's all.
answered Mar 12 '16 at 10:03
User9523User9523
98511026
answered Mar 12 '16 at 10:03
User9523User9523
98511026
answered Mar 12 '16 at 10:03
User9523User9523
98511026
answered Mar 12 '16 at 10:03
User9523User9523
98511026
98511026
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