- #1
yungman
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I cannot get the answer of the problem as in the book but the book usually right. I did it in 2 totally different ways and I still get my own answer. Can anyone help me double check?
This is to find u(r,\theta,\phi) given:
\nabla^2 u(r,\theta,\phi) = -k u(r,\theta,\phi) = f(r)=1 \hbox { and boundary condition } u(1,\theta,\phi)=0
u(r,\theta,\phi) = \sum_{j=1}^{\infty}\sum_{n=0}^{\infty}\sum_{m=-n}^{n} B_{jnm} \; j_n(\lambda_{n,j}r) \; Y_{nm}(\theta\phi)
\Rightarrow \nabla^2 u(r,\theta,\phi) = -k u(r,\theta,\phi) = -\lambda^2_{n,j}u(r,\theta,\phi) = \sum_{j=1}^{\infty}\sum_{n=0}^{\infty}\sum_{m=-n}^{n} -\lambda^2_{n,j} \; B_{jnm} \; j_n(\lambda_{n,j}r) \; Y_{nm}(\theta,\phi) = 1(1)
We can just use simple spherical harmonics expansion on (1) where:
\int_0^1 \int_0^{2\pi} \int_0^{\pi} j_n^2(\lambda_{nj}r)|Y_{nm}(\theta,\phi)|^2 r^2 sin(\theta) d\theta d\phi dr = \frac{a^3}{2}j^2_{n+1}(\alpha_{(n+\frac{1}{2},j)})
(1) \Rightarrow -\lambda^2_{n,j} \; B_{jnm} \; \frac{a^3}{2}j^2_{n+1}(\alpha_{(n+\frac{1}{2},j)}) = \int_0^1 \int_0^{2\pi} \int_0^{\pi} \; j_n (\lambda_{nj}r) \; \overline{Y}_{nm}(\theta,\phi) \; r^2 \; sin(\theta) \; d\theta \; d\phi \; dr
As you can see, only n=m=0 produce non zero coefficients.
\Rightarrow -\lambda^2_{0,j} \; B_j \; \frac{1}{2}j^2_1(\alpha_{(\frac{1}{2},j)}) = \int_0^1 \; j_n (\lambda_{nj}r) \; r^2 \; dr \; \int_0^{2\pi} \; \frac {1}{2\sqrt{\pi}} \; d\phi \; \int_0^{\pi} P^0_0(cos\theta) \; sin(\theta) \; d\theta
Using boundary condition u(1,\theta,\phi)=0 \Rightarrow \; \lambda_{0,j}= j\pi
B_j = \frac{4\sqrt{\pi} (-1)^j}{j^2\pi^2}
u(r,\theta,\phi) = \sum_{j=1}^{\infty} B_j \; j_0(\lambda_{0,j}r) \; Y_{0,0}(\theta\phi) = 2\sum_{j=1}^{\infty} (-1)^j \; \frac{sin(j\pi r)}{j^3 \pi^3 r}
Where:
j_0(\lambda_{0,j}r) = \frac{sin(j\pi r)}{j \pi r} \hbox { and } Y_{0,0}(\theta\phi) = \frac{1}{2\sqrt{\pi}}
But the book's answer is:
u(r,\theta,\phi) = 2\sum_{j=1}^{\infty} (-1)^j \; \frac{sin(j\pi r)}{j^2 \pi^2 r}
I don't know what I did wrong. I did it the other way using spherical harmonics expansion of f(r)=1 and then use:
B_j=\frac{-A_j}{\lambda_{0,j}} and get the same result.
This is to find u(r,\theta,\phi) given:
\nabla^2 u(r,\theta,\phi) = -k u(r,\theta,\phi) = f(r)=1 \hbox { and boundary condition } u(1,\theta,\phi)=0
u(r,\theta,\phi) = \sum_{j=1}^{\infty}\sum_{n=0}^{\infty}\sum_{m=-n}^{n} B_{jnm} \; j_n(\lambda_{n,j}r) \; Y_{nm}(\theta\phi)
\Rightarrow \nabla^2 u(r,\theta,\phi) = -k u(r,\theta,\phi) = -\lambda^2_{n,j}u(r,\theta,\phi) = \sum_{j=1}^{\infty}\sum_{n=0}^{\infty}\sum_{m=-n}^{n} -\lambda^2_{n,j} \; B_{jnm} \; j_n(\lambda_{n,j}r) \; Y_{nm}(\theta,\phi) = 1(1)
We can just use simple spherical harmonics expansion on (1) where:
\int_0^1 \int_0^{2\pi} \int_0^{\pi} j_n^2(\lambda_{nj}r)|Y_{nm}(\theta,\phi)|^2 r^2 sin(\theta) d\theta d\phi dr = \frac{a^3}{2}j^2_{n+1}(\alpha_{(n+\frac{1}{2},j)})
(1) \Rightarrow -\lambda^2_{n,j} \; B_{jnm} \; \frac{a^3}{2}j^2_{n+1}(\alpha_{(n+\frac{1}{2},j)}) = \int_0^1 \int_0^{2\pi} \int_0^{\pi} \; j_n (\lambda_{nj}r) \; \overline{Y}_{nm}(\theta,\phi) \; r^2 \; sin(\theta) \; d\theta \; d\phi \; dr
As you can see, only n=m=0 produce non zero coefficients.
\Rightarrow -\lambda^2_{0,j} \; B_j \; \frac{1}{2}j^2_1(\alpha_{(\frac{1}{2},j)}) = \int_0^1 \; j_n (\lambda_{nj}r) \; r^2 \; dr \; \int_0^{2\pi} \; \frac {1}{2\sqrt{\pi}} \; d\phi \; \int_0^{\pi} P^0_0(cos\theta) \; sin(\theta) \; d\theta
Using boundary condition u(1,\theta,\phi)=0 \Rightarrow \; \lambda_{0,j}= j\pi
B_j = \frac{4\sqrt{\pi} (-1)^j}{j^2\pi^2}
u(r,\theta,\phi) = \sum_{j=1}^{\infty} B_j \; j_0(\lambda_{0,j}r) \; Y_{0,0}(\theta\phi) = 2\sum_{j=1}^{\infty} (-1)^j \; \frac{sin(j\pi r)}{j^3 \pi^3 r}
Where:
j_0(\lambda_{0,j}r) = \frac{sin(j\pi r)}{j \pi r} \hbox { and } Y_{0,0}(\theta\phi) = \frac{1}{2\sqrt{\pi}}
But the book's answer is:
u(r,\theta,\phi) = 2\sum_{j=1}^{\infty} (-1)^j \; \frac{sin(j\pi r)}{j^2 \pi^2 r}
I don't know what I did wrong. I did it the other way using spherical harmonics expansion of f(r)=1 and then use:
B_j=\frac{-A_j}{\lambda_{0,j}} and get the same result.
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