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Solving Laplace's equation in cylindrical coordinates (ODE)

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1

$begingroup$

I know this looks like a physics thing, but it's purely the math that I'm stuck on. I'm attempting to write Laplace's equation, $nabla^2V=0$, in cylindrical coordinates for a potential, $V(r,phi,z)$, independent of $z$.
By definition (in cylindrical),
$$
nabla^2V=frac{1}{r}frac{partial}{partial r}left(rfrac{partial V}{partial r}right)+frac{1}{r^2}frac{partial^2 V}{partial phi^2}+frac{partial^2 V}{partial z^2}
$$
in cylindrical coordinates. $frac{partial^2 V}{partial z^2}=0$ and $V(r,phi) = R(r)Phi(phi)$. Plugging in $V(r,phi)$ and using separation of variables:
$$
frac{r}{R}frac{d}{dr}left(rfrac{dR}{dr}right)+frac{1}{Phi}frac{d^2Phi}{dphi^2}=0
$$
Now I have two ordinary differential equations. Using a separation constant $k^2$, I can solve for $Phi(phi)$ easily:
$$
frac{d^2Phi}{dphi^2}=-Phi k^2 quad Rightarrow quad Phi(phi)=Asin(kphi)+Bcos(kphi)
$$
Here's where I'm stuck, solving for $R(r)$
$$
frac{r}{R}frac{d}{dr}left(rfrac{dR}{dr}right)=k^2
$$
I tried applying the product rule and writing it out like this:
$$
rR''+R'-frac{k^2}{r}R=0
$$
but I don't see how to get past this. I'm eventually supposed to end up with something in this form:
$$
V(r, phi)=A_0+B_0ln(r)+sum^infty_{k=1}[r^k(A_kcos(kphi)+B_ksin(kphi))+r^{-k}(C_kcos(kphi)+D_ksin(kphi)]
$$
Some of the constants might be different but this is what I'm going for. A Fourier thing happened somewhere.

ordinary-differential-equations

share|cite|improve this question

edited 1 hour ago

griffin175

asked 6 hours ago

griffin175griffin175

84

$endgroup$

add a comment | 

1

$begingroup$

I know this looks like a physics thing, but it's purely the math that I'm stuck on. I'm attempting to write Laplace's equation, $nabla^2V=0$, in cylindrical coordinates for a potential, $V(r,phi,z)$, independent of $z$.
By definition (in cylindrical),
$$
nabla^2V=frac{1}{r}frac{partial}{partial r}left(rfrac{partial V}{partial r}right)+frac{1}{r^2}frac{partial^2 V}{partial phi^2}+frac{partial^2 V}{partial z^2}
$$
in cylindrical coordinates. $frac{partial^2 V}{partial z^2}=0$ and $V(r,phi) = R(r)Phi(phi)$. Plugging in $V(r,phi)$ and using separation of variables:
$$
frac{r}{R}frac{d}{dr}left(rfrac{dR}{dr}right)+frac{1}{Phi}frac{d^2Phi}{dphi^2}=0
$$
Now I have two ordinary differential equations. Using a separation constant $k^2$, I can solve for $Phi(phi)$ easily:
$$
frac{d^2Phi}{dphi^2}=-Phi k^2 quad Rightarrow quad Phi(phi)=Asin(kphi)+Bcos(kphi)
$$
Here's where I'm stuck, solving for $R(r)$
$$
frac{r}{R}frac{d}{dr}left(rfrac{dR}{dr}right)=k^2
$$
I tried applying the product rule and writing it out like this:
$$
rR''+R'-frac{k^2}{r}R=0
$$
but I don't see how to get past this. I'm eventually supposed to end up with something in this form:
$$
V(r, phi)=A_0+B_0ln(r)+sum^infty_{k=1}[r^k(A_kcos(kphi)+B_ksin(kphi))+r^{-k}(C_kcos(kphi)+D_ksin(kphi)]
$$
Some of the constants might be different but this is what I'm going for. A Fourier thing happened somewhere.

ordinary-differential-equations

share|cite|improve this question

edited 1 hour ago

griffin175

asked 6 hours ago

griffin175griffin175

84

$endgroup$

add a comment | 

$begingroup$

I know this looks like a physics thing, but it's purely the math that I'm stuck on. I'm attempting to write Laplace's equation, $nabla^2V=0$, in cylindrical coordinates for a potential, $V(r,phi,z)$, independent of $z$.
By definition (in cylindrical),
$$
nabla^2V=frac{1}{r}frac{partial}{partial r}left(rfrac{partial V}{partial r}right)+frac{1}{r^2}frac{partial^2 V}{partial phi^2}+frac{partial^2 V}{partial z^2}
$$
in cylindrical coordinates. $frac{partial^2 V}{partial z^2}=0$ and $V(r,phi) = R(r)Phi(phi)$. Plugging in $V(r,phi)$ and using separation of variables:
$$
frac{r}{R}frac{d}{dr}left(rfrac{dR}{dr}right)+frac{1}{Phi}frac{d^2Phi}{dphi^2}=0
$$
Now I have two ordinary differential equations. Using a separation constant $k^2$, I can solve for $Phi(phi)$ easily:
$$
frac{d^2Phi}{dphi^2}=-Phi k^2 quad Rightarrow quad Phi(phi)=Asin(kphi)+Bcos(kphi)
$$
Here's where I'm stuck, solving for $R(r)$
$$
frac{r}{R}frac{d}{dr}left(rfrac{dR}{dr}right)=k^2
$$
I tried applying the product rule and writing it out like this:
$$
rR''+R'-frac{k^2}{r}R=0
$$
but I don't see how to get past this. I'm eventually supposed to end up with something in this form:
$$
V(r, phi)=A_0+B_0ln(r)+sum^infty_{k=1}[r^k(A_kcos(kphi)+B_ksin(kphi))+r^{-k}(C_kcos(kphi)+D_ksin(kphi)]
$$
Some of the constants might be different but this is what I'm going for. A Fourier thing happened somewhere.

ordinary-differential-equations

share|cite|improve this question

edited 1 hour ago

griffin175

asked 6 hours ago

griffin175griffin175

84

$endgroup$

share|cite|improve this question

edited 1 hour ago

griffin175

asked 6 hours ago

griffin175griffin175

84

share|cite|improve this question

edited 1 hour ago

griffin175

asked 6 hours ago

griffin175griffin175

84

asked 6 hours ago

griffin175griffin175

84

asked 6 hours ago

griffin175griffin175

84

1 Answer
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active

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0

$begingroup$

The radial equation is

$$ r^2R'' + rR' - k^2R = 0 $$

If you didn't know, this is called a Cauchy-Euler equation, with well-known solutions of the form $x^n$. Plugging in this form results in the characteristic equation

$$ m(m-1) + m - k^2 = 0 implies m^2-k^2 = 0 implies m = pm k $$

If $kne0$, the general solution is

$$ R_k(r) = A_kr^k + B_kr^{-k} $$

If $k=0$, we have a double root and the solution has the form

$$ R_0(r) = A_0 + B_0ln r $$

You may verify this by directly solving

$$ r^2R'' + rR' = 0 implies frac{R''}{R'} = -frac{1}{r} implies ln (R') = -ln r + c implies R' = frac{B}{r} implies dots $$

share|cite|improve this answer

answered 5 hours ago

DylanDylan

13.6k31027

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  That explains a lot, half the battle is just knowing what it's called. How does $R_k$ get written in terms of sin and cos when combined with $Phi$ in the infinite sum?
  $endgroup$
  – griffin175
  4 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  If $k=0$, then $Phi$ is constant, otherwise it's just direct multiplication $$ V(r,phi) = A_0 + B_0ln r + sum_{k=1}^infty (A_k r^k + B_kR^{-k})(C_kcos(kphi) + D_ksin(kphi)) $$
  $endgroup$
  – Dylan
  4 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I was missing the $r^{-k}$ in my infinite sum that prevented me from seeing this connection. Thanks!
  $endgroup$
  – griffin175
  1 hour ago
  

  
  
  
  

add a comment | 

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0

$begingroup$

The radial equation is

$$ r^2R'' + rR' - k^2R = 0 $$

If you didn't know, this is called a Cauchy-Euler equation, with well-known solutions of the form $x^n$. Plugging in this form results in the characteristic equation

$$ m(m-1) + m - k^2 = 0 implies m^2-k^2 = 0 implies m = pm k $$

If $kne0$, the general solution is

$$ R_k(r) = A_kr^k + B_kr^{-k} $$

If $k=0$, we have a double root and the solution has the form

$$ R_0(r) = A_0 + B_0ln r $$

You may verify this by directly solving

$$ r^2R'' + rR' = 0 implies frac{R''}{R'} = -frac{1}{r} implies ln (R') = -ln r + c implies R' = frac{B}{r} implies dots $$

share|cite|improve this answer

answered 5 hours ago

DylanDylan

13.6k31027

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  That explains a lot, half the battle is just knowing what it's called. How does $R_k$ get written in terms of sin and cos when combined with $Phi$ in the infinite sum?
  $endgroup$
  – griffin175
  4 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  If $k=0$, then $Phi$ is constant, otherwise it's just direct multiplication $$ V(r,phi) = A_0 + B_0ln r + sum_{k=1}^infty (A_k r^k + B_kR^{-k})(C_kcos(kphi) + D_ksin(kphi)) $$
  $endgroup$
  – Dylan
  4 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I was missing the $r^{-k}$ in my infinite sum that prevented me from seeing this connection. Thanks!
  $endgroup$
  – griffin175
  1 hour ago
  

  
  
  
  

add a comment | 

0

$begingroup$

The radial equation is

$$ r^2R'' + rR' - k^2R = 0 $$

If you didn't know, this is called a Cauchy-Euler equation, with well-known solutions of the form $x^n$. Plugging in this form results in the characteristic equation

$$ m(m-1) + m - k^2 = 0 implies m^2-k^2 = 0 implies m = pm k $$

If $kne0$, the general solution is

$$ R_k(r) = A_kr^k + B_kr^{-k} $$

If $k=0$, we have a double root and the solution has the form

$$ R_0(r) = A_0 + B_0ln r $$

You may verify this by directly solving

$$ r^2R'' + rR' = 0 implies frac{R''}{R'} = -frac{1}{r} implies ln (R') = -ln r + c implies R' = frac{B}{r} implies dots $$

share|cite|improve this answer

answered 5 hours ago

DylanDylan

13.6k31027

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  That explains a lot, half the battle is just knowing what it's called. How does $R_k$ get written in terms of sin and cos when combined with $Phi$ in the infinite sum?
  $endgroup$
  – griffin175
  4 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  If $k=0$, then $Phi$ is constant, otherwise it's just direct multiplication $$ V(r,phi) = A_0 + B_0ln r + sum_{k=1}^infty (A_k r^k + B_kR^{-k})(C_kcos(kphi) + D_ksin(kphi)) $$
  $endgroup$
  – Dylan
  4 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I was missing the $r^{-k}$ in my infinite sum that prevented me from seeing this connection. Thanks!
  $endgroup$
  – griffin175
  1 hour ago
  

  
  
  
  

add a comment | 

$begingroup$

The radial equation is

$$ r^2R'' + rR' - k^2R = 0 $$

If you didn't know, this is called a Cauchy-Euler equation, with well-known solutions of the form $x^n$. Plugging in this form results in the characteristic equation

$$ m(m-1) + m - k^2 = 0 implies m^2-k^2 = 0 implies m = pm k $$

If $kne0$, the general solution is

$$ R_k(r) = A_kr^k + B_kr^{-k} $$

If $k=0$, we have a double root and the solution has the form

$$ R_0(r) = A_0 + B_0ln r $$

You may verify this by directly solving

$$ r^2R'' + rR' = 0 implies frac{R''}{R'} = -frac{1}{r} implies ln (R') = -ln r + c implies R' = frac{B}{r} implies dots $$

share|cite|improve this answer

answered 5 hours ago

DylanDylan

13.6k31027

$endgroup$

share|cite|improve this answer

answered 5 hours ago

DylanDylan

13.6k31027

answered 5 hours ago

DylanDylan

13.6k31027

answered 5 hours ago

DylanDylan

13.6k31027