Mixing
Laplace's equation: separation of variables
$begingroup$
Question:
Let $(r,theta)$ denote plane polar coordinates. Show that there are countably infinitely many $k in Bbb R$ for which
$$nabla^2 u=0 qquad 1≤r≤2 \ ku + frac{partial u}{partial r}=0 qquad r=1,2$$
has a non-trivial solution of the form $u(r,theta) = f(r)g(theta)$.
Attempt:
Writing out the Laplacian in plane polars and plugging in $u(r,theta) = f(r)g(theta)$, we get
$$nabla^2 u = frac{partial^2 u}{partial r^2} + frac 1r frac{partial u}{partial r}+ frac{1}{r^2} frac{partial u^2}{partial theta^2} = gfrac{d^2f}{dr^2} + frac gr frac{df}{dr} + frac{f}{r^2}frac{d^2g}{dtheta^2}$$
Thus
begin{align}
nabla^2 u = 0 & implies gfrac{d^2f}{dr^2} + frac gr frac{df}{dr} + frac{f}{r^2}frac{d^2g}{dtheta^2} = 0 \
& implies frac{r^2}{f}frac{d^2f}{dr^2} + frac rf frac{df}{dr} = - frac 1g frac{d^2g}{dtheta^2} = text{constant} : = lambda
end{align}
First assuming that $lambda neq 0$, I solved the $r$-equation:
$$frac{r^2}{f}frac{d^2f}{dr^2} + frac rf frac{df}{dr} = lambda implies r^2frac{d^2f}{dr^2} + rfrac{df}{dr}-lambda f = 0 implies f(r) = Ar^{sqrt lambda} + Br^{-sqrt lambda}$$
The boundary condition given is
$$ku + frac{partial u}{partial r} = 0 implies kfg + gfrac{df}{dr}=0 implies g(theta) big(k f(r) + f'(r)big)=0 qquad r=1,2 ; ; , ; ; theta in Bbb R$$
$g(theta) equiv 0$ gives only non-trivial solutions, thus we must have $kf(1)+f'(1) = kf(2)f'(2)=0$. That is:
begin{align*}
& k f(1)+f'(1)=0 implies k (A+B) + sqrtlambda(A-B)=0 \
& k f(2)+f'(2)=0 implies k big(A2^{sqrtlambda} + B2^{-sqrt{lambda}} big) + frac {sqrtlambda}{2} big(A2^{sqrtlambda} - B2^{-sqrt{lambda}} big)=0
end{align*}
In matrix form, this is
$$begin{pmatrix} k + sqrt lambda & k - sqrt lambda \ Big(k + frac {sqrtlambda}{2}Big) 2^{sqrt lambda} & Big(k - frac {sqrtlambda}{2}Big) 2^{-sqrt lambda} end{pmatrix} begin{pmatrix} A \ B end{pmatrix} = begin{pmatrix} 0 \ 0 end{pmatrix}$$
and we get a non-trivial solution iff the determinant of the matrix is zero.
However, it seems to me that for every $k in Bbb R$ there is a corresponding $lambda$ that gives a non-trivial solution. Have I done something wrong? Or am I misunderstanding something? Any help would be much appreciated.
pde laplacian
asked Jan 6 at 17:25
glowstonetreesglowstonetrees
2,336418
$endgroup$
add a comment |
$begingroup$
Question:
Let $(r,theta)$ denote plane polar coordinates. Show that there are countably infinitely many $k in Bbb R$ for which
$$nabla^2 u=0 qquad 1≤r≤2 \ ku + frac{partial u}{partial r}=0 qquad r=1,2$$
has a non-trivial solution of the form $u(r,theta) = f(r)g(theta)$.
Attempt:
Writing out the Laplacian in plane polars and plugging in $u(r,theta) = f(r)g(theta)$, we get
$$nabla^2 u = frac{partial^2 u}{partial r^2} + frac 1r frac{partial u}{partial r}+ frac{1}{r^2} frac{partial u^2}{partial theta^2} = gfrac{d^2f}{dr^2} + frac gr frac{df}{dr} + frac{f}{r^2}frac{d^2g}{dtheta^2}$$
Thus
begin{align}
nabla^2 u = 0 & implies gfrac{d^2f}{dr^2} + frac gr frac{df}{dr} + frac{f}{r^2}frac{d^2g}{dtheta^2} = 0 \
& implies frac{r^2}{f}frac{d^2f}{dr^2} + frac rf frac{df}{dr} = - frac 1g frac{d^2g}{dtheta^2} = text{constant} : = lambda
end{align}
First assuming that $lambda neq 0$, I solved the $r$-equation:
$$frac{r^2}{f}frac{d^2f}{dr^2} + frac rf frac{df}{dr} = lambda implies r^2frac{d^2f}{dr^2} + rfrac{df}{dr}-lambda f = 0 implies f(r) = Ar^{sqrt lambda} + Br^{-sqrt lambda}$$
The boundary condition given is
$$ku + frac{partial u}{partial r} = 0 implies kfg + gfrac{df}{dr}=0 implies g(theta) big(k f(r) + f'(r)big)=0 qquad r=1,2 ; ; , ; ; theta in Bbb R$$
$g(theta) equiv 0$ gives only non-trivial solutions, thus we must have $kf(1)+f'(1) = kf(2)f'(2)=0$. That is:
begin{align*}
& k f(1)+f'(1)=0 implies k (A+B) + sqrtlambda(A-B)=0 \
& k f(2)+f'(2)=0 implies k big(A2^{sqrtlambda} + B2^{-sqrt{lambda}} big) + frac {sqrtlambda}{2} big(A2^{sqrtlambda} - B2^{-sqrt{lambda}} big)=0
end{align*}
In matrix form, this is
$$begin{pmatrix} k + sqrt lambda & k - sqrt lambda \ Big(k + frac {sqrtlambda}{2}Big) 2^{sqrt lambda} & Big(k - frac {sqrtlambda}{2}Big) 2^{-sqrt lambda} end{pmatrix} begin{pmatrix} A \ B end{pmatrix} = begin{pmatrix} 0 \ 0 end{pmatrix}$$
and we get a non-trivial solution iff the determinant of the matrix is zero.
However, it seems to me that for every $k in Bbb R$ there is a corresponding $lambda$ that gives a non-trivial solution. Have I done something wrong? Or am I misunderstanding something? Any help would be much appreciated.
pde laplacian
asked Jan 6 at 17:25
glowstonetreesglowstonetrees
2,336418
$endgroup$
$begingroup$
I haven't done it myself, but i think the restriction on lambda will come from solving the equation for $g(theta)$
$endgroup$
– Calvin Khor
Jan 6 at 17:37
$begingroup$
Calvin is right: you need $g$ to be $2pi$ periodic and not identically zero, but your equation reads $g''+lambda g =0 $ which only has such a solution when $lambda$ takes on countably many particular values. Of course, you're not done: you still need to check that the boundary condition in $r$ can be satisfied for at least countably many of the pairs $(g_n,lambda_n)$.
$endgroup$
– Ian
Jan 6 at 17:40
$begingroup$
Ahhh I see. We have $$g''+lambda g =0 implies g(theta) = Ccos (sqrt lambda theta) + D sin (sqrt lambda theta)$$ which is $2pi$-periodic iff $lambda =lambda_n = n^2$ where $n in Bbb Z^+$
$endgroup$
– glowstonetrees
Jan 6 at 17:44
add a comment |
$begingroup$
Question:
Let $(r,theta)$ denote plane polar coordinates. Show that there are countably infinitely many $k in Bbb R$ for which
$$nabla^2 u=0 qquad 1≤r≤2 \ ku + frac{partial u}{partial r}=0 qquad r=1,2$$
has a non-trivial solution of the form $u(r,theta) = f(r)g(theta)$.
Attempt:
Writing out the Laplacian in plane polars and plugging in $u(r,theta) = f(r)g(theta)$, we get
$$nabla^2 u = frac{partial^2 u}{partial r^2} + frac 1r frac{partial u}{partial r}+ frac{1}{r^2} frac{partial u^2}{partial theta^2} = gfrac{d^2f}{dr^2} + frac gr frac{df}{dr} + frac{f}{r^2}frac{d^2g}{dtheta^2}$$
Thus
begin{align}
nabla^2 u = 0 & implies gfrac{d^2f}{dr^2} + frac gr frac{df}{dr} + frac{f}{r^2}frac{d^2g}{dtheta^2} = 0 \
& implies frac{r^2}{f}frac{d^2f}{dr^2} + frac rf frac{df}{dr} = - frac 1g frac{d^2g}{dtheta^2} = text{constant} : = lambda
end{align}
First assuming that $lambda neq 0$, I solved the $r$-equation:
$$frac{r^2}{f}frac{d^2f}{dr^2} + frac rf frac{df}{dr} = lambda implies r^2frac{d^2f}{dr^2} + rfrac{df}{dr}-lambda f = 0 implies f(r) = Ar^{sqrt lambda} + Br^{-sqrt lambda}$$
The boundary condition given is
$$ku + frac{partial u}{partial r} = 0 implies kfg + gfrac{df}{dr}=0 implies g(theta) big(k f(r) + f'(r)big)=0 qquad r=1,2 ; ; , ; ; theta in Bbb R$$
$g(theta) equiv 0$ gives only non-trivial solutions, thus we must have $kf(1)+f'(1) = kf(2)f'(2)=0$. That is:
begin{align*}
& k f(1)+f'(1)=0 implies k (A+B) + sqrtlambda(A-B)=0 \
& k f(2)+f'(2)=0 implies k big(A2^{sqrtlambda} + B2^{-sqrt{lambda}} big) + frac {sqrtlambda}{2} big(A2^{sqrtlambda} - B2^{-sqrt{lambda}} big)=0
end{align*}
In matrix form, this is
$$begin{pmatrix} k + sqrt lambda & k - sqrt lambda \ Big(k + frac {sqrtlambda}{2}Big) 2^{sqrt lambda} & Big(k - frac {sqrtlambda}{2}Big) 2^{-sqrt lambda} end{pmatrix} begin{pmatrix} A \ B end{pmatrix} = begin{pmatrix} 0 \ 0 end{pmatrix}$$
and we get a non-trivial solution iff the determinant of the matrix is zero.
However, it seems to me that for every $k in Bbb R$ there is a corresponding $lambda$ that gives a non-trivial solution. Have I done something wrong? Or am I misunderstanding something? Any help would be much appreciated.
pde laplacian
asked Jan 6 at 17:25
glowstonetreesglowstonetrees
2,336418
$endgroup$
Question:
Let $(r,theta)$ denote plane polar coordinates. Show that there are countably infinitely many $k in Bbb R$ for which
$$nabla^2 u=0 qquad 1≤r≤2 \ ku + frac{partial u}{partial r}=0 qquad r=1,2$$
has a non-trivial solution of the form $u(r,theta) = f(r)g(theta)$.
Attempt:
Writing out the Laplacian in plane polars and plugging in $u(r,theta) = f(r)g(theta)$, we get
$$nabla^2 u = frac{partial^2 u}{partial r^2} + frac 1r frac{partial u}{partial r}+ frac{1}{r^2} frac{partial u^2}{partial theta^2} = gfrac{d^2f}{dr^2} + frac gr frac{df}{dr} + frac{f}{r^2}frac{d^2g}{dtheta^2}$$
Thus
begin{align}
nabla^2 u = 0 & implies gfrac{d^2f}{dr^2} + frac gr frac{df}{dr} + frac{f}{r^2}frac{d^2g}{dtheta^2} = 0 \
& implies frac{r^2}{f}frac{d^2f}{dr^2} + frac rf frac{df}{dr} = - frac 1g frac{d^2g}{dtheta^2} = text{constant} : = lambda
end{align}
First assuming that $lambda neq 0$, I solved the $r$-equation:
$$frac{r^2}{f}frac{d^2f}{dr^2} + frac rf frac{df}{dr} = lambda implies r^2frac{d^2f}{dr^2} + rfrac{df}{dr}-lambda f = 0 implies f(r) = Ar^{sqrt lambda} + Br^{-sqrt lambda}$$
The boundary condition given is
$$ku + frac{partial u}{partial r} = 0 implies kfg + gfrac{df}{dr}=0 implies g(theta) big(k f(r) + f'(r)big)=0 qquad r=1,2 ; ; , ; ; theta in Bbb R$$
$g(theta) equiv 0$ gives only non-trivial solutions, thus we must have $kf(1)+f'(1) = kf(2)f'(2)=0$. That is:
begin{align*}
& k f(1)+f'(1)=0 implies k (A+B) + sqrtlambda(A-B)=0 \
& k f(2)+f'(2)=0 implies k big(A2^{sqrtlambda} + B2^{-sqrt{lambda}} big) + frac {sqrtlambda}{2} big(A2^{sqrtlambda} - B2^{-sqrt{lambda}} big)=0
end{align*}
In matrix form, this is
$$begin{pmatrix} k + sqrt lambda & k - sqrt lambda \ Big(k + frac {sqrtlambda}{2}Big) 2^{sqrt lambda} & Big(k - frac {sqrtlambda}{2}Big) 2^{-sqrt lambda} end{pmatrix} begin{pmatrix} A \ B end{pmatrix} = begin{pmatrix} 0 \ 0 end{pmatrix}$$
and we get a non-trivial solution iff the determinant of the matrix is zero.
However, it seems to me that for every $k in Bbb R$ there is a corresponding $lambda$ that gives a non-trivial solution. Have I done something wrong? Or am I misunderstanding something? Any help would be much appreciated.
pde laplacian
pde laplacian
asked Jan 6 at 17:25
glowstonetreesglowstonetrees
2,336418
asked Jan 6 at 17:25
glowstonetreesglowstonetrees
2,336418
edited Jan 6 at 17:35
glowstonetrees
asked Jan 6 at 17:25
glowstonetreesglowstonetrees
2,336418
asked Jan 6 at 17:25
glowstonetreesglowstonetrees
2,336418
asked Jan 6 at 17:25
glowstonetreesglowstonetrees
2,336418
2,336418
$begingroup$
I haven't done it myself, but i think the restriction on lambda will come from solving the equation for $g(theta)$
$endgroup$
– Calvin Khor
Jan 6 at 17:37
$begingroup$
Calvin is right: you need $g$ to be $2pi$ periodic and not identically zero, but your equation reads $g''+lambda g =0 $ which only has such a solution when $lambda$ takes on countably many particular values. Of course, you're not done: you still need to check that the boundary condition in $r$ can be satisfied for at least countably many of the pairs $(g_n,lambda_n)$.
$endgroup$
– Ian
Jan 6 at 17:40
$begingroup$
Ahhh I see. We have $$g''+lambda g =0 implies g(theta) = Ccos (sqrt lambda theta) + D sin (sqrt lambda theta)$$ which is $2pi$-periodic iff $lambda =lambda_n = n^2$ where $n in Bbb Z^+$
$endgroup$
– glowstonetrees
Jan 6 at 17:44
add a comment |
$begingroup$
I haven't done it myself, but i think the restriction on lambda will come from solving the equation for $g(theta)$
$endgroup$
– Calvin Khor
Jan 6 at 17:37
$begingroup$
Calvin is right: you need $g$ to be $2pi$ periodic and not identically zero, but your equation reads $g''+lambda g =0 $ which only has such a solution when $lambda$ takes on countably many particular values. Of course, you're not done: you still need to check that the boundary condition in $r$ can be satisfied for at least countably many of the pairs $(g_n,lambda_n)$.
$endgroup$
– Ian
Jan 6 at 17:40
$begingroup$
Ahhh I see. We have $$g''+lambda g =0 implies g(theta) = Ccos (sqrt lambda theta) + D sin (sqrt lambda theta)$$ which is $2pi$-periodic iff $lambda =lambda_n = n^2$ where $n in Bbb Z^+$
$endgroup$
– glowstonetrees
Jan 6 at 17:44
$begingroup$
I haven't done it myself, but i think the restriction on lambda will come from solving the equation for $g(theta)$
$endgroup$
– Calvin Khor
Jan 6 at 17:37
$begingroup$
I haven't done it myself, but i think the restriction on lambda will come from solving the equation for $g(theta)$
$endgroup$
– Calvin Khor
Jan 6 at 17:37
$begingroup$
Calvin is right: you need $g$ to be $2pi$ periodic and not identically zero, but your equation reads $g''+lambda g =0 $ which only has such a solution when $lambda$ takes on countably many particular values. Of course, you're not done: you still need to check that the boundary condition in $r$ can be satisfied for at least countably many of the pairs $(g_n,lambda_n)$.
$endgroup$
– Ian
Jan 6 at 17:40
$begingroup$
Calvin is right: you need $g$ to be $2pi$ periodic and not identically zero, but your equation reads $g''+lambda g =0 $ which only has such a solution when $lambda$ takes on countably many particular values. Of course, you're not done: you still need to check that the boundary condition in $r$ can be satisfied for at least countably many of the pairs $(g_n,lambda_n)$.
$endgroup$
– Ian
Jan 6 at 17:40
$begingroup$
Ahhh I see. We have $$g''+lambda g =0 implies g(theta) = Ccos (sqrt lambda theta) + D sin (sqrt lambda theta)$$ which is $2pi$-periodic iff $lambda =lambda_n = n^2$ where $n in Bbb Z^+$
$endgroup$
– glowstonetrees
Jan 6 at 17:44
$begingroup$
Ahhh I see. We have $$g''+lambda g =0 implies g(theta) = Ccos (sqrt lambda theta) + D sin (sqrt lambda theta)$$ which is $2pi$-periodic iff $lambda =lambda_n = n^2$ where $n in Bbb Z^+$
$endgroup$
– glowstonetrees
Jan 6 at 17:44
add a comment |
1 Answer
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From the periodicity condition, we get a general solution
$$ u_n(r,theta) = f(r)g(theta) = (Ar^n + Br^{-n})(Ccos ntheta + Dsin n theta) $$
where $n=1,2,3,dots$
Since the boundary conditions are radially symmetric, it follows that $ kf(r) +f'(r) =0 $ on $r=1,2$. Plugging this in
begin{align}
k(A+B) + n(A-B) = 0 \
k(A2^n + B2^{-n}) +n(A2^{n-1} -B2^{-n-1})=0
end{align}
Converting into a matrix equation
$$ begin{pmatrix} k+n & k-n \ 2^nk + 2^{n-1}n & 2^{-n}k - 2^{-n-1}n end{pmatrix}begin{pmatrix} A \ B end{pmatrix} = 0 $$
There's a non-trivial solution for $(A, B)$ if the determinant of the coefficient matrix is zero
$$ (k+n)(2k-n) - 2^{2n}(k-n)(2k+n) = 0 $$
In quadratic form this is
$$ 2(2^{2n}-1)k^2 - n(2^{2n}+1)k - (2^{2n}-1)n^2 = 0 $$
where $Delta = b^2-4ac = n^2(2^{2n}+1)^2 + 8n^2(2^{2n}-1)^2 > 0$
So there exists two values of $k$ for each corresponding positive integer $n$, thus "countably infinite"
edited Jan 7 at 16:17
Ian
67.8k25388
answered Jan 7 at 10:06
DylanDylan
12.5k31026
$endgroup$
$begingroup$
Just as an aside, it sufficed to just note that $a$ and $c$ in the quadratic had different signs. You didn't need to write out the whole discriminant.
$endgroup$
– Ian
Jan 7 at 16:18
add a comment |
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$begingroup$
From the periodicity condition, we get a general solution
$$ u_n(r,theta) = f(r)g(theta) = (Ar^n + Br^{-n})(Ccos ntheta + Dsin n theta) $$
where $n=1,2,3,dots$
Since the boundary conditions are radially symmetric, it follows that $ kf(r) +f'(r) =0 $ on $r=1,2$. Plugging this in
begin{align}
k(A+B) + n(A-B) = 0 \
k(A2^n + B2^{-n}) +n(A2^{n-1} -B2^{-n-1})=0
end{align}
Converting into a matrix equation
$$ begin{pmatrix} k+n & k-n \ 2^nk + 2^{n-1}n & 2^{-n}k - 2^{-n-1}n end{pmatrix}begin{pmatrix} A \ B end{pmatrix} = 0 $$
There's a non-trivial solution for $(A, B)$ if the determinant of the coefficient matrix is zero
$$ (k+n)(2k-n) - 2^{2n}(k-n)(2k+n) = 0 $$
In quadratic form this is
$$ 2(2^{2n}-1)k^2 - n(2^{2n}+1)k - (2^{2n}-1)n^2 = 0 $$
where $Delta = b^2-4ac = n^2(2^{2n}+1)^2 + 8n^2(2^{2n}-1)^2 > 0$
So there exists two values of $k$ for each corresponding positive integer $n$, thus "countably infinite"
edited Jan 7 at 16:17
Ian
67.8k25388
answered Jan 7 at 10:06
DylanDylan
12.5k31026
$endgroup$
$begingroup$
Just as an aside, it sufficed to just note that $a$ and $c$ in the quadratic had different signs. You didn't need to write out the whole discriminant.
$endgroup$
– Ian
Jan 7 at 16:18
add a comment |
$begingroup$
From the periodicity condition, we get a general solution
$$ u_n(r,theta) = f(r)g(theta) = (Ar^n + Br^{-n})(Ccos ntheta + Dsin n theta) $$
where $n=1,2,3,dots$
Since the boundary conditions are radially symmetric, it follows that $ kf(r) +f'(r) =0 $ on $r=1,2$. Plugging this in
begin{align}
k(A+B) + n(A-B) = 0 \
k(A2^n + B2^{-n}) +n(A2^{n-1} -B2^{-n-1})=0
end{align}
Converting into a matrix equation
$$ begin{pmatrix} k+n & k-n \ 2^nk + 2^{n-1}n & 2^{-n}k - 2^{-n-1}n end{pmatrix}begin{pmatrix} A \ B end{pmatrix} = 0 $$
There's a non-trivial solution for $(A, B)$ if the determinant of the coefficient matrix is zero
$$ (k+n)(2k-n) - 2^{2n}(k-n)(2k+n) = 0 $$
In quadratic form this is
$$ 2(2^{2n}-1)k^2 - n(2^{2n}+1)k - (2^{2n}-1)n^2 = 0 $$
where $Delta = b^2-4ac = n^2(2^{2n}+1)^2 + 8n^2(2^{2n}-1)^2 > 0$
So there exists two values of $k$ for each corresponding positive integer $n$, thus "countably infinite"
edited Jan 7 at 16:17
Ian
67.8k25388
answered Jan 7 at 10:06
DylanDylan
12.5k31026
$endgroup$
$begingroup$
Just as an aside, it sufficed to just note that $a$ and $c$ in the quadratic had different signs. You didn't need to write out the whole discriminant.
$endgroup$
– Ian
Jan 7 at 16:18
add a comment |
$begingroup$
From the periodicity condition, we get a general solution
$$ u_n(r,theta) = f(r)g(theta) = (Ar^n + Br^{-n})(Ccos ntheta + Dsin n theta) $$
where $n=1,2,3,dots$
Since the boundary conditions are radially symmetric, it follows that $ kf(r) +f'(r) =0 $ on $r=1,2$. Plugging this in
begin{align}
k(A+B) + n(A-B) = 0 \
k(A2^n + B2^{-n}) +n(A2^{n-1} -B2^{-n-1})=0
end{align}
Converting into a matrix equation
$$ begin{pmatrix} k+n & k-n \ 2^nk + 2^{n-1}n & 2^{-n}k - 2^{-n-1}n end{pmatrix}begin{pmatrix} A \ B end{pmatrix} = 0 $$
There's a non-trivial solution for $(A, B)$ if the determinant of the coefficient matrix is zero
$$ (k+n)(2k-n) - 2^{2n}(k-n)(2k+n) = 0 $$
In quadratic form this is
$$ 2(2^{2n}-1)k^2 - n(2^{2n}+1)k - (2^{2n}-1)n^2 = 0 $$
where $Delta = b^2-4ac = n^2(2^{2n}+1)^2 + 8n^2(2^{2n}-1)^2 > 0$
So there exists two values of $k$ for each corresponding positive integer $n$, thus "countably infinite"
edited Jan 7 at 16:17
Ian
67.8k25388
answered Jan 7 at 10:06
DylanDylan
12.5k31026
$endgroup$
From the periodicity condition, we get a general solution
$$ u_n(r,theta) = f(r)g(theta) = (Ar^n + Br^{-n})(Ccos ntheta + Dsin n theta) $$
where $n=1,2,3,dots$
Since the boundary conditions are radially symmetric, it follows that $ kf(r) +f'(r) =0 $ on $r=1,2$. Plugging this in
begin{align}
k(A+B) + n(A-B) = 0 \
k(A2^n + B2^{-n}) +n(A2^{n-1} -B2^{-n-1})=0
end{align}
Converting into a matrix equation
$$ begin{pmatrix} k+n & k-n \ 2^nk + 2^{n-1}n & 2^{-n}k - 2^{-n-1}n end{pmatrix}begin{pmatrix} A \ B end{pmatrix} = 0 $$
There's a non-trivial solution for $(A, B)$ if the determinant of the coefficient matrix is zero
$$ (k+n)(2k-n) - 2^{2n}(k-n)(2k+n) = 0 $$
In quadratic form this is
$$ 2(2^{2n}-1)k^2 - n(2^{2n}+1)k - (2^{2n}-1)n^2 = 0 $$
where $Delta = b^2-4ac = n^2(2^{2n}+1)^2 + 8n^2(2^{2n}-1)^2 > 0$
So there exists two values of $k$ for each corresponding positive integer $n$, thus "countably infinite"
edited Jan 7 at 16:17
Ian
67.8k25388
answered Jan 7 at 10:06
DylanDylan
12.5k31026
edited Jan 7 at 16:17
Ian
67.8k25388
edited Jan 7 at 16:17
Ian
67.8k25388
edited Jan 7 at 16:17
Ian
67.8k25388
67.8k25388
answered Jan 7 at 10:06
DylanDylan
12.5k31026
answered Jan 7 at 10:06
DylanDylan
12.5k31026
answered Jan 7 at 10:06
DylanDylan
12.5k31026
12.5k31026
$begingroup$
Just as an aside, it sufficed to just note that $a$ and $c$ in the quadratic had different signs. You didn't need to write out the whole discriminant.
$endgroup$
– Ian
Jan 7 at 16:18
add a comment |
$begingroup$
Just as an aside, it sufficed to just note that $a$ and $c$ in the quadratic had different signs. You didn't need to write out the whole discriminant.
$endgroup$
– Ian
Jan 7 at 16:18
$begingroup$
Just as an aside, it sufficed to just note that $a$ and $c$ in the quadratic had different signs. You didn't need to write out the whole discriminant.
$endgroup$
– Ian
Jan 7 at 16:18
$begingroup$
Just as an aside, it sufficed to just note that $a$ and $c$ in the quadratic had different signs. You didn't need to write out the whole discriminant.
$endgroup$
– Ian
Jan 7 at 16:18
add a comment |
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$begingroup$
I haven't done it myself, but i think the restriction on lambda will come from solving the equation for $g(theta)$
$endgroup$
– Calvin Khor
Jan 6 at 17:37
$begingroup$
Calvin is right: you need $g$ to be $2pi$ periodic and not identically zero, but your equation reads $g''+lambda g =0 $ which only has such a solution when $lambda$ takes on countably many particular values. Of course, you're not done: you still need to check that the boundary condition in $r$ can be satisfied for at least countably many of the pairs $(g_n,lambda_n)$.
$endgroup$
– Ian
Jan 6 at 17:40
$begingroup$
Ahhh I see. We have $$g''+lambda g =0 implies g(theta) = Ccos (sqrt lambda theta) + D sin (sqrt lambda theta)$$ which is $2pi$-periodic iff $lambda =lambda_n = n^2$ where $n in Bbb Z^+$
$endgroup$
– glowstonetrees
Jan 6 at 17:44