Mixing
Heat equation in cylindrical coordinates at origin
$begingroup$
I'm trying to solve a heat equation in cylindrical coordinates
$$dfrac{partial u}{partial t} = a left(dfrac{partial^2 u}{partial r^2} + dfrac{1}{r} dfrac{partial u}{partial r} + dfrac{1}{r^2} dfrac{partial^2 u}{partial phi^2} + dfrac{partial^2 u}{partial z^2} right),~~~~~~~~~~~~~~ (r, phi, z) in Omega.$$
with Neumann boundary conditions
$$ left.dfrac{partial u}{partial n} right|_{partial Omega} = f(u, r, phi, z),$$
and initial conditions
$$u(0, r, phi, z) = g(r, phi, z).$$
The region $Omega$ is just a cylinder
$$Omega ={ (r, phi, z) : 0leq r leq R, ~~ 0 leq phi < 2pi, ~~ 0 leq z leq Z}.$$
Functions $f$ and $g$ are known and $f$ is a non-linear function with respect to $u$. Due to its non-linearity I cannot solve the problem analytically and trying to use a finite-difference method:
$$begin{array}{lcl}
t_tau = Delta t cdot tau, & tau=0,cdots, T, &\
r_i = Delta r cdot i, & i=0,cdots, I, & (r_I = R),\
phi_j = Delta phi cdot j, & i=0,cdots, J, & (phi_J = 2pi - Delta phi),\
z_k = Delta z cdot k, & k=0,cdots, K, & (z_K = Z),
end{array}$$
And
$$u^tau_{i,j,k} = u(t_tau, r_i, phi_j, z_k).$$
So
$$dfrac{u^{tau+1}_{i,j,k} - u^{tau}_{i,j,k}}{Delta t} = aleft( dfrac{u^tau_{i+1,j,k} - 2 u^tau_{i,j,k} + u^tau_{i-1,j,k}}{Delta r^2} + dfrac{1}{Delta r cdot i} dfrac{u^tau_{i+1,j,k} - u^tau_{i-1,j,k}}{2 Delta r} + dfrac{1}{(Delta r cdot i)^2} dfrac{u^tau_{i,j+1,k} - 2 u^tau_{i,j,k} + u^tau_{i,j-1,k}}{Delta phi^2} + dfrac{u^tau_{i,j,k+1} - 2 u^tau_{i,j,k} + u^tau_{i,j,k-1}}{Delta z^2}right).$$
Next, I use boundary conditions and periodicity of the solution to find $u^tau_{I+1, j, k}$, $~~u^tau_{-1, j, k}, ~~$ etc. For example:
$$ dfrac{u^tau_{I+1,j,k} - u^tau_{I-1,j,k}}{2 Delta r} = f(u^tau_{I,j,k}, Delta r cdot I, Delta phi cdot j, Delta z cdot k).$$
However, I have a hard time with Laplacian at $r=0$, i.e., writing the difference equations for $u^tau_{0, j, k}$.
I know that if the problem was symmetric with respect to $phi$, then
$$dfrac{partial^2 u}{partial phi^2} equiv 0.$$
Moreover,
$$left. dfrac{partial u}{partial r}right|_{r=0} = 0,$$
and, therefore,
$$left. dfrac{1}{r} dfrac{partial u}{partial r}right|_{r=0} = dfrac{partial^2 u}{partial r^2}.$$
So, the heat equation at $r=0$ would transform into
$$dfrac{partial u}{partial t} = a left(2dfrac{partial^2 u}{partial r^2} + dfrac{partial^2 u}{partial z^2} right),$$
which would allow me to just use a separate equations for $u^tau_{0, j, k}$.
What should be the heat equation at $r=0$ in my more general case?
heat-equation laplacian cylindrical-coordinates
asked Jan 8 at 0:22
JohnJohn
113
$endgroup$
add a comment |
$begingroup$
I'm trying to solve a heat equation in cylindrical coordinates
$$dfrac{partial u}{partial t} = a left(dfrac{partial^2 u}{partial r^2} + dfrac{1}{r} dfrac{partial u}{partial r} + dfrac{1}{r^2} dfrac{partial^2 u}{partial phi^2} + dfrac{partial^2 u}{partial z^2} right),~~~~~~~~~~~~~~ (r, phi, z) in Omega.$$
with Neumann boundary conditions
$$ left.dfrac{partial u}{partial n} right|_{partial Omega} = f(u, r, phi, z),$$
and initial conditions
$$u(0, r, phi, z) = g(r, phi, z).$$
The region $Omega$ is just a cylinder
$$Omega ={ (r, phi, z) : 0leq r leq R, ~~ 0 leq phi < 2pi, ~~ 0 leq z leq Z}.$$
Functions $f$ and $g$ are known and $f$ is a non-linear function with respect to $u$. Due to its non-linearity I cannot solve the problem analytically and trying to use a finite-difference method:
$$begin{array}{lcl}
t_tau = Delta t cdot tau, & tau=0,cdots, T, &\
r_i = Delta r cdot i, & i=0,cdots, I, & (r_I = R),\
phi_j = Delta phi cdot j, & i=0,cdots, J, & (phi_J = 2pi - Delta phi),\
z_k = Delta z cdot k, & k=0,cdots, K, & (z_K = Z),
end{array}$$
And
$$u^tau_{i,j,k} = u(t_tau, r_i, phi_j, z_k).$$
So
$$dfrac{u^{tau+1}_{i,j,k} - u^{tau}_{i,j,k}}{Delta t} = aleft( dfrac{u^tau_{i+1,j,k} - 2 u^tau_{i,j,k} + u^tau_{i-1,j,k}}{Delta r^2} + dfrac{1}{Delta r cdot i} dfrac{u^tau_{i+1,j,k} - u^tau_{i-1,j,k}}{2 Delta r} + dfrac{1}{(Delta r cdot i)^2} dfrac{u^tau_{i,j+1,k} - 2 u^tau_{i,j,k} + u^tau_{i,j-1,k}}{Delta phi^2} + dfrac{u^tau_{i,j,k+1} - 2 u^tau_{i,j,k} + u^tau_{i,j,k-1}}{Delta z^2}right).$$
Next, I use boundary conditions and periodicity of the solution to find $u^tau_{I+1, j, k}$, $~~u^tau_{-1, j, k}, ~~$ etc. For example:
$$ dfrac{u^tau_{I+1,j,k} - u^tau_{I-1,j,k}}{2 Delta r} = f(u^tau_{I,j,k}, Delta r cdot I, Delta phi cdot j, Delta z cdot k).$$
However, I have a hard time with Laplacian at $r=0$, i.e., writing the difference equations for $u^tau_{0, j, k}$.
I know that if the problem was symmetric with respect to $phi$, then
$$dfrac{partial^2 u}{partial phi^2} equiv 0.$$
Moreover,
$$left. dfrac{partial u}{partial r}right|_{r=0} = 0,$$
and, therefore,
$$left. dfrac{1}{r} dfrac{partial u}{partial r}right|_{r=0} = dfrac{partial^2 u}{partial r^2}.$$
So, the heat equation at $r=0$ would transform into
$$dfrac{partial u}{partial t} = a left(2dfrac{partial^2 u}{partial r^2} + dfrac{partial^2 u}{partial z^2} right),$$
which would allow me to just use a separate equations for $u^tau_{0, j, k}$.
What should be the heat equation at $r=0$ in my more general case?
heat-equation laplacian cylindrical-coordinates
asked Jan 8 at 0:22
JohnJohn
113
$endgroup$
$begingroup$
How viable is expanding $u = sum_{-infty}^infty u_n(r,z) exp(inphi)$ and then cutting off the sum at $n = pm J/2$?
$endgroup$
– eyeballfrog
Jan 8 at 0:45
add a comment |
$begingroup$
I'm trying to solve a heat equation in cylindrical coordinates
$$dfrac{partial u}{partial t} = a left(dfrac{partial^2 u}{partial r^2} + dfrac{1}{r} dfrac{partial u}{partial r} + dfrac{1}{r^2} dfrac{partial^2 u}{partial phi^2} + dfrac{partial^2 u}{partial z^2} right),~~~~~~~~~~~~~~ (r, phi, z) in Omega.$$
with Neumann boundary conditions
$$ left.dfrac{partial u}{partial n} right|_{partial Omega} = f(u, r, phi, z),$$
and initial conditions
$$u(0, r, phi, z) = g(r, phi, z).$$
The region $Omega$ is just a cylinder
$$Omega ={ (r, phi, z) : 0leq r leq R, ~~ 0 leq phi < 2pi, ~~ 0 leq z leq Z}.$$
Functions $f$ and $g$ are known and $f$ is a non-linear function with respect to $u$. Due to its non-linearity I cannot solve the problem analytically and trying to use a finite-difference method:
$$begin{array}{lcl}
t_tau = Delta t cdot tau, & tau=0,cdots, T, &\
r_i = Delta r cdot i, & i=0,cdots, I, & (r_I = R),\
phi_j = Delta phi cdot j, & i=0,cdots, J, & (phi_J = 2pi - Delta phi),\
z_k = Delta z cdot k, & k=0,cdots, K, & (z_K = Z),
end{array}$$
And
$$u^tau_{i,j,k} = u(t_tau, r_i, phi_j, z_k).$$
So
$$dfrac{u^{tau+1}_{i,j,k} - u^{tau}_{i,j,k}}{Delta t} = aleft( dfrac{u^tau_{i+1,j,k} - 2 u^tau_{i,j,k} + u^tau_{i-1,j,k}}{Delta r^2} + dfrac{1}{Delta r cdot i} dfrac{u^tau_{i+1,j,k} - u^tau_{i-1,j,k}}{2 Delta r} + dfrac{1}{(Delta r cdot i)^2} dfrac{u^tau_{i,j+1,k} - 2 u^tau_{i,j,k} + u^tau_{i,j-1,k}}{Delta phi^2} + dfrac{u^tau_{i,j,k+1} - 2 u^tau_{i,j,k} + u^tau_{i,j,k-1}}{Delta z^2}right).$$
Next, I use boundary conditions and periodicity of the solution to find $u^tau_{I+1, j, k}$, $~~u^tau_{-1, j, k}, ~~$ etc. For example:
$$ dfrac{u^tau_{I+1,j,k} - u^tau_{I-1,j,k}}{2 Delta r} = f(u^tau_{I,j,k}, Delta r cdot I, Delta phi cdot j, Delta z cdot k).$$
However, I have a hard time with Laplacian at $r=0$, i.e., writing the difference equations for $u^tau_{0, j, k}$.
I know that if the problem was symmetric with respect to $phi$, then
$$dfrac{partial^2 u}{partial phi^2} equiv 0.$$
Moreover,
$$left. dfrac{partial u}{partial r}right|_{r=0} = 0,$$
and, therefore,
$$left. dfrac{1}{r} dfrac{partial u}{partial r}right|_{r=0} = dfrac{partial^2 u}{partial r^2}.$$
So, the heat equation at $r=0$ would transform into
$$dfrac{partial u}{partial t} = a left(2dfrac{partial^2 u}{partial r^2} + dfrac{partial^2 u}{partial z^2} right),$$
which would allow me to just use a separate equations for $u^tau_{0, j, k}$.
What should be the heat equation at $r=0$ in my more general case?
heat-equation laplacian cylindrical-coordinates
asked Jan 8 at 0:22
JohnJohn
113
$endgroup$
I'm trying to solve a heat equation in cylindrical coordinates
$$dfrac{partial u}{partial t} = a left(dfrac{partial^2 u}{partial r^2} + dfrac{1}{r} dfrac{partial u}{partial r} + dfrac{1}{r^2} dfrac{partial^2 u}{partial phi^2} + dfrac{partial^2 u}{partial z^2} right),~~~~~~~~~~~~~~ (r, phi, z) in Omega.$$
with Neumann boundary conditions
$$ left.dfrac{partial u}{partial n} right|_{partial Omega} = f(u, r, phi, z),$$
and initial conditions
$$u(0, r, phi, z) = g(r, phi, z).$$
The region $Omega$ is just a cylinder
$$Omega ={ (r, phi, z) : 0leq r leq R, ~~ 0 leq phi < 2pi, ~~ 0 leq z leq Z}.$$
Functions $f$ and $g$ are known and $f$ is a non-linear function with respect to $u$. Due to its non-linearity I cannot solve the problem analytically and trying to use a finite-difference method:
$$begin{array}{lcl}
t_tau = Delta t cdot tau, & tau=0,cdots, T, &\
r_i = Delta r cdot i, & i=0,cdots, I, & (r_I = R),\
phi_j = Delta phi cdot j, & i=0,cdots, J, & (phi_J = 2pi - Delta phi),\
z_k = Delta z cdot k, & k=0,cdots, K, & (z_K = Z),
end{array}$$
And
$$u^tau_{i,j,k} = u(t_tau, r_i, phi_j, z_k).$$
So
$$dfrac{u^{tau+1}_{i,j,k} - u^{tau}_{i,j,k}}{Delta t} = aleft( dfrac{u^tau_{i+1,j,k} - 2 u^tau_{i,j,k} + u^tau_{i-1,j,k}}{Delta r^2} + dfrac{1}{Delta r cdot i} dfrac{u^tau_{i+1,j,k} - u^tau_{i-1,j,k}}{2 Delta r} + dfrac{1}{(Delta r cdot i)^2} dfrac{u^tau_{i,j+1,k} - 2 u^tau_{i,j,k} + u^tau_{i,j-1,k}}{Delta phi^2} + dfrac{u^tau_{i,j,k+1} - 2 u^tau_{i,j,k} + u^tau_{i,j,k-1}}{Delta z^2}right).$$
Next, I use boundary conditions and periodicity of the solution to find $u^tau_{I+1, j, k}$, $~~u^tau_{-1, j, k}, ~~$ etc. For example:
$$ dfrac{u^tau_{I+1,j,k} - u^tau_{I-1,j,k}}{2 Delta r} = f(u^tau_{I,j,k}, Delta r cdot I, Delta phi cdot j, Delta z cdot k).$$
However, I have a hard time with Laplacian at $r=0$, i.e., writing the difference equations for $u^tau_{0, j, k}$.
I know that if the problem was symmetric with respect to $phi$, then
$$dfrac{partial^2 u}{partial phi^2} equiv 0.$$
Moreover,
$$left. dfrac{partial u}{partial r}right|_{r=0} = 0,$$
and, therefore,
$$left. dfrac{1}{r} dfrac{partial u}{partial r}right|_{r=0} = dfrac{partial^2 u}{partial r^2}.$$
So, the heat equation at $r=0$ would transform into
$$dfrac{partial u}{partial t} = a left(2dfrac{partial^2 u}{partial r^2} + dfrac{partial^2 u}{partial z^2} right),$$
which would allow me to just use a separate equations for $u^tau_{0, j, k}$.
What should be the heat equation at $r=0$ in my more general case?
heat-equation laplacian cylindrical-coordinates
heat-equation laplacian cylindrical-coordinates
asked Jan 8 at 0:22
JohnJohn
113
asked Jan 8 at 0:22
JohnJohn
113
edited Jan 8 at 17:09
John
asked Jan 8 at 0:22
JohnJohn
113
asked Jan 8 at 0:22
JohnJohn
113
asked Jan 8 at 0:22
JohnJohn
113
113
$begingroup$
How viable is expanding $u = sum_{-infty}^infty u_n(r,z) exp(inphi)$ and then cutting off the sum at $n = pm J/2$?
$endgroup$
– eyeballfrog
Jan 8 at 0:45
add a comment |
$begingroup$
How viable is expanding $u = sum_{-infty}^infty u_n(r,z) exp(inphi)$ and then cutting off the sum at $n = pm J/2$?
$endgroup$
– eyeballfrog
Jan 8 at 0:45
$begingroup$
How viable is expanding $u = sum_{-infty}^infty u_n(r,z) exp(inphi)$ and then cutting off the sum at $n = pm J/2$?
$endgroup$
– eyeballfrog
Jan 8 at 0:45
$begingroup$
How viable is expanding $u = sum_{-infty}^infty u_n(r,z) exp(inphi)$ and then cutting off the sum at $n = pm J/2$?
$endgroup$
– eyeballfrog
Jan 8 at 0:45
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
I think I came up with the answer myself. I'll leave it here if anyone meets the similar problem.
Suppose $f(x, y, z)$ is a sufficiently smooth function defined in Cartesian coordinates and $g(r, phi, z) = f(r cosphi, rsinphi, z)$ is the same function in cylindrical coordinates. We have
$$left. nabla^2 f right|_{begin{array}{l}x=rcosphi\y=rsinphi\z=zend{array}} = nabla^2 g.$$
Using second-order central difference approximation for the second derivatives:
$$left. nabla^2 f right|_{begin{array}{l}x=0\y=0end{array}} approx dfrac{f(Delta x, 0, z) - 2f(0, 0, z) + f(-Delta x, 0, z)}{(Delta x)^2} + dfrac{f(0, Delta y, z) - 2f(0, 0, z) + f(0, -Delta y, z)}{(Delta y)^2} + dfrac{f(0, 0, z+Delta z) - 2f(0, 0, z) + f(0, 0, z-Delta z)}{(Delta z)^2} = dfrac{g(Delta r, 0, z) - 2g(0,phi,z)+g(Delta r,pi,z)}{(Delta r)^2} + dfrac{g(Delta r, frac{pi}{2}, z) - 2g(0,phi,z)+g(Delta r,frac{3pi}{2},z)}{(Delta r)^2} + dfrac{g(0, 0, z+Delta z) - 2g(0, phi, z) + g(0, 0, z-Delta z)}{(Delta z)^2},$$
where $phi$ can be anything, since $g(0,phi_1,z)=g(0,phi_2,z)$ for any $phi_1, phi_2$ as they correspond to the same point.
Thus, the finite difference equation that corresponds to the heat equation in the cylindrical coordinates at $r=0$ is the following (using notations from the question):
$$ dfrac{u^{tau+1}_{0,j,k} - u^{tau}_{0,j,k}}{Delta t} = a left( dfrac{u^tau_{1,0,k} - 2u^tau_{0,j,k} + u^tau_{1,beta,k}}{(Delta r)^2} + dfrac{u^tau_{1,alpha,k} - 2u^tau_{0,j,k} + u^tau_{1,gamma,k}}{(Delta r)^2} + dfrac{u^tau_{0,j,k+1} - 2u^tau_{0,j,k} + u^tau_{0,j,k-1}}{(Delta z)^2}right),$$
where $alpha cdot Deltaphi = pi/2$, $~~beta cdot Deltaphi = pi,~~$ and $~~gamma cdot Deltaphi = 3pi/2~~$ (so the grid should be appropriately chosen). And once again, $u^tau_{0,j,k}$ doesn't really depend on $j$.
answered Jan 8 at 5:23
JohnJohn
113
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3065670%2fheat-equation-in-cylindrical-coordinates-at-origin%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
I think I came up with the answer myself. I'll leave it here if anyone meets the similar problem.
Suppose $f(x, y, z)$ is a sufficiently smooth function defined in Cartesian coordinates and $g(r, phi, z) = f(r cosphi, rsinphi, z)$ is the same function in cylindrical coordinates. We have
$$left. nabla^2 f right|_{begin{array}{l}x=rcosphi\y=rsinphi\z=zend{array}} = nabla^2 g.$$
Using second-order central difference approximation for the second derivatives:
$$left. nabla^2 f right|_{begin{array}{l}x=0\y=0end{array}} approx dfrac{f(Delta x, 0, z) - 2f(0, 0, z) + f(-Delta x, 0, z)}{(Delta x)^2} + dfrac{f(0, Delta y, z) - 2f(0, 0, z) + f(0, -Delta y, z)}{(Delta y)^2} + dfrac{f(0, 0, z+Delta z) - 2f(0, 0, z) + f(0, 0, z-Delta z)}{(Delta z)^2} = dfrac{g(Delta r, 0, z) - 2g(0,phi,z)+g(Delta r,pi,z)}{(Delta r)^2} + dfrac{g(Delta r, frac{pi}{2}, z) - 2g(0,phi,z)+g(Delta r,frac{3pi}{2},z)}{(Delta r)^2} + dfrac{g(0, 0, z+Delta z) - 2g(0, phi, z) + g(0, 0, z-Delta z)}{(Delta z)^2},$$
where $phi$ can be anything, since $g(0,phi_1,z)=g(0,phi_2,z)$ for any $phi_1, phi_2$ as they correspond to the same point.
Thus, the finite difference equation that corresponds to the heat equation in the cylindrical coordinates at $r=0$ is the following (using notations from the question):
$$ dfrac{u^{tau+1}_{0,j,k} - u^{tau}_{0,j,k}}{Delta t} = a left( dfrac{u^tau_{1,0,k} - 2u^tau_{0,j,k} + u^tau_{1,beta,k}}{(Delta r)^2} + dfrac{u^tau_{1,alpha,k} - 2u^tau_{0,j,k} + u^tau_{1,gamma,k}}{(Delta r)^2} + dfrac{u^tau_{0,j,k+1} - 2u^tau_{0,j,k} + u^tau_{0,j,k-1}}{(Delta z)^2}right),$$
where $alpha cdot Deltaphi = pi/2$, $~~beta cdot Deltaphi = pi,~~$ and $~~gamma cdot Deltaphi = 3pi/2~~$ (so the grid should be appropriately chosen). And once again, $u^tau_{0,j,k}$ doesn't really depend on $j$.
answered Jan 8 at 5:23
JohnJohn
113
$endgroup$
add a comment |
$begingroup$
I think I came up with the answer myself. I'll leave it here if anyone meets the similar problem.
Suppose $f(x, y, z)$ is a sufficiently smooth function defined in Cartesian coordinates and $g(r, phi, z) = f(r cosphi, rsinphi, z)$ is the same function in cylindrical coordinates. We have
$$left. nabla^2 f right|_{begin{array}{l}x=rcosphi\y=rsinphi\z=zend{array}} = nabla^2 g.$$
Using second-order central difference approximation for the second derivatives:
$$left. nabla^2 f right|_{begin{array}{l}x=0\y=0end{array}} approx dfrac{f(Delta x, 0, z) - 2f(0, 0, z) + f(-Delta x, 0, z)}{(Delta x)^2} + dfrac{f(0, Delta y, z) - 2f(0, 0, z) + f(0, -Delta y, z)}{(Delta y)^2} + dfrac{f(0, 0, z+Delta z) - 2f(0, 0, z) + f(0, 0, z-Delta z)}{(Delta z)^2} = dfrac{g(Delta r, 0, z) - 2g(0,phi,z)+g(Delta r,pi,z)}{(Delta r)^2} + dfrac{g(Delta r, frac{pi}{2}, z) - 2g(0,phi,z)+g(Delta r,frac{3pi}{2},z)}{(Delta r)^2} + dfrac{g(0, 0, z+Delta z) - 2g(0, phi, z) + g(0, 0, z-Delta z)}{(Delta z)^2},$$
where $phi$ can be anything, since $g(0,phi_1,z)=g(0,phi_2,z)$ for any $phi_1, phi_2$ as they correspond to the same point.
Thus, the finite difference equation that corresponds to the heat equation in the cylindrical coordinates at $r=0$ is the following (using notations from the question):
$$ dfrac{u^{tau+1}_{0,j,k} - u^{tau}_{0,j,k}}{Delta t} = a left( dfrac{u^tau_{1,0,k} - 2u^tau_{0,j,k} + u^tau_{1,beta,k}}{(Delta r)^2} + dfrac{u^tau_{1,alpha,k} - 2u^tau_{0,j,k} + u^tau_{1,gamma,k}}{(Delta r)^2} + dfrac{u^tau_{0,j,k+1} - 2u^tau_{0,j,k} + u^tau_{0,j,k-1}}{(Delta z)^2}right),$$
where $alpha cdot Deltaphi = pi/2$, $~~beta cdot Deltaphi = pi,~~$ and $~~gamma cdot Deltaphi = 3pi/2~~$ (so the grid should be appropriately chosen). And once again, $u^tau_{0,j,k}$ doesn't really depend on $j$.
answered Jan 8 at 5:23
JohnJohn
113
$endgroup$
add a comment |
$begingroup$
I think I came up with the answer myself. I'll leave it here if anyone meets the similar problem.
Suppose $f(x, y, z)$ is a sufficiently smooth function defined in Cartesian coordinates and $g(r, phi, z) = f(r cosphi, rsinphi, z)$ is the same function in cylindrical coordinates. We have
$$left. nabla^2 f right|_{begin{array}{l}x=rcosphi\y=rsinphi\z=zend{array}} = nabla^2 g.$$
Using second-order central difference approximation for the second derivatives:
$$left. nabla^2 f right|_{begin{array}{l}x=0\y=0end{array}} approx dfrac{f(Delta x, 0, z) - 2f(0, 0, z) + f(-Delta x, 0, z)}{(Delta x)^2} + dfrac{f(0, Delta y, z) - 2f(0, 0, z) + f(0, -Delta y, z)}{(Delta y)^2} + dfrac{f(0, 0, z+Delta z) - 2f(0, 0, z) + f(0, 0, z-Delta z)}{(Delta z)^2} = dfrac{g(Delta r, 0, z) - 2g(0,phi,z)+g(Delta r,pi,z)}{(Delta r)^2} + dfrac{g(Delta r, frac{pi}{2}, z) - 2g(0,phi,z)+g(Delta r,frac{3pi}{2},z)}{(Delta r)^2} + dfrac{g(0, 0, z+Delta z) - 2g(0, phi, z) + g(0, 0, z-Delta z)}{(Delta z)^2},$$
where $phi$ can be anything, since $g(0,phi_1,z)=g(0,phi_2,z)$ for any $phi_1, phi_2$ as they correspond to the same point.
Thus, the finite difference equation that corresponds to the heat equation in the cylindrical coordinates at $r=0$ is the following (using notations from the question):
$$ dfrac{u^{tau+1}_{0,j,k} - u^{tau}_{0,j,k}}{Delta t} = a left( dfrac{u^tau_{1,0,k} - 2u^tau_{0,j,k} + u^tau_{1,beta,k}}{(Delta r)^2} + dfrac{u^tau_{1,alpha,k} - 2u^tau_{0,j,k} + u^tau_{1,gamma,k}}{(Delta r)^2} + dfrac{u^tau_{0,j,k+1} - 2u^tau_{0,j,k} + u^tau_{0,j,k-1}}{(Delta z)^2}right),$$
where $alpha cdot Deltaphi = pi/2$, $~~beta cdot Deltaphi = pi,~~$ and $~~gamma cdot Deltaphi = 3pi/2~~$ (so the grid should be appropriately chosen). And once again, $u^tau_{0,j,k}$ doesn't really depend on $j$.
answered Jan 8 at 5:23
JohnJohn
113
$endgroup$
I think I came up with the answer myself. I'll leave it here if anyone meets the similar problem.
Suppose $f(x, y, z)$ is a sufficiently smooth function defined in Cartesian coordinates and $g(r, phi, z) = f(r cosphi, rsinphi, z)$ is the same function in cylindrical coordinates. We have
$$left. nabla^2 f right|_{begin{array}{l}x=rcosphi\y=rsinphi\z=zend{array}} = nabla^2 g.$$
Using second-order central difference approximation for the second derivatives:
$$left. nabla^2 f right|_{begin{array}{l}x=0\y=0end{array}} approx dfrac{f(Delta x, 0, z) - 2f(0, 0, z) + f(-Delta x, 0, z)}{(Delta x)^2} + dfrac{f(0, Delta y, z) - 2f(0, 0, z) + f(0, -Delta y, z)}{(Delta y)^2} + dfrac{f(0, 0, z+Delta z) - 2f(0, 0, z) + f(0, 0, z-Delta z)}{(Delta z)^2} = dfrac{g(Delta r, 0, z) - 2g(0,phi,z)+g(Delta r,pi,z)}{(Delta r)^2} + dfrac{g(Delta r, frac{pi}{2}, z) - 2g(0,phi,z)+g(Delta r,frac{3pi}{2},z)}{(Delta r)^2} + dfrac{g(0, 0, z+Delta z) - 2g(0, phi, z) + g(0, 0, z-Delta z)}{(Delta z)^2},$$
where $phi$ can be anything, since $g(0,phi_1,z)=g(0,phi_2,z)$ for any $phi_1, phi_2$ as they correspond to the same point.
Thus, the finite difference equation that corresponds to the heat equation in the cylindrical coordinates at $r=0$ is the following (using notations from the question):
$$ dfrac{u^{tau+1}_{0,j,k} - u^{tau}_{0,j,k}}{Delta t} = a left( dfrac{u^tau_{1,0,k} - 2u^tau_{0,j,k} + u^tau_{1,beta,k}}{(Delta r)^2} + dfrac{u^tau_{1,alpha,k} - 2u^tau_{0,j,k} + u^tau_{1,gamma,k}}{(Delta r)^2} + dfrac{u^tau_{0,j,k+1} - 2u^tau_{0,j,k} + u^tau_{0,j,k-1}}{(Delta z)^2}right),$$
where $alpha cdot Deltaphi = pi/2$, $~~beta cdot Deltaphi = pi,~~$ and $~~gamma cdot Deltaphi = 3pi/2~~$ (so the grid should be appropriately chosen). And once again, $u^tau_{0,j,k}$ doesn't really depend on $j$.
answered Jan 8 at 5:23
JohnJohn
113
answered Jan 8 at 5:23
JohnJohn
113
answered Jan 8 at 5:23
JohnJohn
113
answered Jan 8 at 5:23
JohnJohn
113
113
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3065670%2fheat-equation-in-cylindrical-coordinates-at-origin%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
How viable is expanding $u = sum_{-infty}^infty u_n(r,z) exp(inphi)$ and then cutting off the sum at $n = pm J/2$?
$endgroup$
– eyeballfrog
Jan 8 at 0:45