Is the steady state solution of the Heat Equation with Dirichlet boundary conditions always 0?
$begingroup$
A heat equation problem with Dirichlet boundary conditions on the domain $[x_1,x_2]$
$$frac{delta u}{delta t} = k frac{delta^2 u}{delta x^2}$$
$$u(x_1,t) = u(x_2,t) = 0$$
Would have eigenfunctions corresponding to sines. Whereas Neumann boundary conditions $$u'(x_1,t) = u'(x_2,t) = 0$$
Results in cosine eigenfunctions.
When evaluating steady-state solutions, with Neumann conditions we can solve for the first coefficients of the Fourier Cosine Series expansion. In the case of Dirichlet conditions, the sine series do not have a $A_0$ coefficient describing steady-state.
Intuitively the steady-state solution for Dirichlet conditions should always decay to zero, as we are allowing heat exchange on the borders and the solution for u has a time-dependent exponential decay. This is constrained with the Neumann conditions (no heat exchange at borders = isolated body), having a non-trivial steady-state.
Is this intuition correct? Is the steady-state solution for the heat equation with Dirichlet B.C always zero? Or is there something I am missing?
heat-equation
asked Dec 3 '18 at 23:01
hirschmehirschme
285
$endgroup$
add a comment |
$begingroup$
A heat equation problem with Dirichlet boundary conditions on the domain $[x_1,x_2]$
$$frac{delta u}{delta t} = k frac{delta^2 u}{delta x^2}$$
$$u(x_1,t) = u(x_2,t) = 0$$
Would have eigenfunctions corresponding to sines. Whereas Neumann boundary conditions $$u'(x_1,t) = u'(x_2,t) = 0$$
Results in cosine eigenfunctions.
When evaluating steady-state solutions, with Neumann conditions we can solve for the first coefficients of the Fourier Cosine Series expansion. In the case of Dirichlet conditions, the sine series do not have a $A_0$ coefficient describing steady-state.
Intuitively the steady-state solution for Dirichlet conditions should always decay to zero, as we are allowing heat exchange on the borders and the solution for u has a time-dependent exponential decay. This is constrained with the Neumann conditions (no heat exchange at borders = isolated body), having a non-trivial steady-state.
Is this intuition correct? Is the steady-state solution for the heat equation with Dirichlet B.C always zero? Or is there something I am missing?
heat-equation
asked Dec 3 '18 at 23:01
hirschmehirschme
285
$endgroup$
add a comment |
$begingroup$
A heat equation problem with Dirichlet boundary conditions on the domain $[x_1,x_2]$
$$frac{delta u}{delta t} = k frac{delta^2 u}{delta x^2}$$
$$u(x_1,t) = u(x_2,t) = 0$$
Would have eigenfunctions corresponding to sines. Whereas Neumann boundary conditions $$u'(x_1,t) = u'(x_2,t) = 0$$
Results in cosine eigenfunctions.
When evaluating steady-state solutions, with Neumann conditions we can solve for the first coefficients of the Fourier Cosine Series expansion. In the case of Dirichlet conditions, the sine series do not have a $A_0$ coefficient describing steady-state.
Intuitively the steady-state solution for Dirichlet conditions should always decay to zero, as we are allowing heat exchange on the borders and the solution for u has a time-dependent exponential decay. This is constrained with the Neumann conditions (no heat exchange at borders = isolated body), having a non-trivial steady-state.
Is this intuition correct? Is the steady-state solution for the heat equation with Dirichlet B.C always zero? Or is there something I am missing?
heat-equation
asked Dec 3 '18 at 23:01
hirschmehirschme
285
$endgroup$
A heat equation problem with Dirichlet boundary conditions on the domain $[x_1,x_2]$
$$frac{delta u}{delta t} = k frac{delta^2 u}{delta x^2}$$
$$u(x_1,t) = u(x_2,t) = 0$$
Would have eigenfunctions corresponding to sines. Whereas Neumann boundary conditions $$u'(x_1,t) = u'(x_2,t) = 0$$
Results in cosine eigenfunctions.
When evaluating steady-state solutions, with Neumann conditions we can solve for the first coefficients of the Fourier Cosine Series expansion. In the case of Dirichlet conditions, the sine series do not have a $A_0$ coefficient describing steady-state.
Intuitively the steady-state solution for Dirichlet conditions should always decay to zero, as we are allowing heat exchange on the borders and the solution for u has a time-dependent exponential decay. This is constrained with the Neumann conditions (no heat exchange at borders = isolated body), having a non-trivial steady-state.
Is this intuition correct? Is the steady-state solution for the heat equation with Dirichlet B.C always zero? Or is there something I am missing?
heat-equation
heat-equation
asked Dec 3 '18 at 23:01
hirschmehirschme
285
asked Dec 3 '18 at 23:01
hirschmehirschme
285
asked Dec 3 '18 at 23:01
hirschmehirschme
285
asked Dec 3 '18 at 23:01
hirschmehirschme
285
asked Dec 3 '18 at 23:01
hirschmehirschme
285
285
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Your reasoning is correct. The steady state solutions can be obtained by setting $partial u/partial t=0$, leading to
$$
u = c_1 x+c_2.
$$
For the homogeneous Dirichlet B.C., the only solution is the trivial one (i.e., $u=0$. For the Neumann B.C., a uniform solution $u=c_2$ exists. Also, the steady state solution in this case is the mean temperature in the initial condition. Integrating the heat equation along $x_1<x<x_2$,
$$
frac{partial}{partial t} int_{x_1}^{x_2} u dx =k left.frac{partial u}{partial x} right|_{x_1}^{x_2}=0,
$$
i.e., the total energy $E=int_D u dx$ inside the domain is constant. For the steady state solution,
$$
int_{x_1}^{x_2} u(tto infty,x) dx=int_{x_1}^{x_2} u(t=0,x) dx
$$
$$
c_2=frac{1}{x_2-x_1} int_{x_1}^{x_2} u_0(x) dx,
$$
in which $u_0(x)$ is the initial condition for $u$.
answered Dec 3 '18 at 23:50
rafa11111rafa11111
1,1291417
$endgroup$
1
$begingroup$
This is very helpful. How are you getting $u = c_1 x + c_2$ after setting $delta u / delta t = 0$? The solution I obtain for $u$, for general $lambda > or < 0$ is $u(x,t) = sum_n A_n sin(lambda_n x) e^{-lambda_n t}$. I could get your solution only by setting $lambda = 0$ and getting eigenfunctions for that: $f(x) = c_1 x + c_2$
$endgroup$
– hirschme
Dec 4 '18 at 16:08
1
$begingroup$
If $partial u/partial t=0$, the 'remaining' equation is $partial^2 u/partial x^2=0$. Furthermore, since there is no dependence on $t$ anymore, we can change the partial derivative to total derivative, i.e., $$frac{d^2 u}{dx^2}=0.$$ Integrating twice, we have $u=c_1 x+c_2$. See that this solution does not correspond to the solution for the original equation for any time $t$, but is the solution of the steady state equation. One can even say that, if $u$ does satisfy $$frac{partial u}{partial t} = k frac{partial ^2 u}{partial x^2},$$ then $$lim_{tto infty} u(x,t) = c_1x+c_2.$$
$endgroup$
– rafa11111
Dec 4 '18 at 16:19
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%2f3024832%2fis-the-steady-state-solution-of-the-heat-equation-with-dirichlet-boundary-condit%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
$begingroup$
Your reasoning is correct. The steady state solutions can be obtained by setting $partial u/partial t=0$, leading to
$$
u = c_1 x+c_2.
$$
For the homogeneous Dirichlet B.C., the only solution is the trivial one (i.e., $u=0$. For the Neumann B.C., a uniform solution $u=c_2$ exists. Also, the steady state solution in this case is the mean temperature in the initial condition. Integrating the heat equation along $x_1<x<x_2$,
$$
frac{partial}{partial t} int_{x_1}^{x_2} u dx =k left.frac{partial u}{partial x} right|_{x_1}^{x_2}=0,
$$
i.e., the total energy $E=int_D u dx$ inside the domain is constant. For the steady state solution,
$$
int_{x_1}^{x_2} u(tto infty,x) dx=int_{x_1}^{x_2} u(t=0,x) dx
$$
$$
c_2=frac{1}{x_2-x_1} int_{x_1}^{x_2} u_0(x) dx,
$$
in which $u_0(x)$ is the initial condition for $u$.
answered Dec 3 '18 at 23:50
rafa11111rafa11111
1,1291417
$endgroup$
1
$begingroup$
This is very helpful. How are you getting $u = c_1 x + c_2$ after setting $delta u / delta t = 0$? The solution I obtain for $u$, for general $lambda > or < 0$ is $u(x,t) = sum_n A_n sin(lambda_n x) e^{-lambda_n t}$. I could get your solution only by setting $lambda = 0$ and getting eigenfunctions for that: $f(x) = c_1 x + c_2$
$endgroup$
– hirschme
Dec 4 '18 at 16:08
1
$begingroup$
If $partial u/partial t=0$, the 'remaining' equation is $partial^2 u/partial x^2=0$. Furthermore, since there is no dependence on $t$ anymore, we can change the partial derivative to total derivative, i.e., $$frac{d^2 u}{dx^2}=0.$$ Integrating twice, we have $u=c_1 x+c_2$. See that this solution does not correspond to the solution for the original equation for any time $t$, but is the solution of the steady state equation. One can even say that, if $u$ does satisfy $$frac{partial u}{partial t} = k frac{partial ^2 u}{partial x^2},$$ then $$lim_{tto infty} u(x,t) = c_1x+c_2.$$
$endgroup$
– rafa11111
Dec 4 '18 at 16:19
add a comment |
$begingroup$
Your reasoning is correct. The steady state solutions can be obtained by setting $partial u/partial t=0$, leading to
$$
u = c_1 x+c_2.
$$
For the homogeneous Dirichlet B.C., the only solution is the trivial one (i.e., $u=0$. For the Neumann B.C., a uniform solution $u=c_2$ exists. Also, the steady state solution in this case is the mean temperature in the initial condition. Integrating the heat equation along $x_1<x<x_2$,
$$
frac{partial}{partial t} int_{x_1}^{x_2} u dx =k left.frac{partial u}{partial x} right|_{x_1}^{x_2}=0,
$$
i.e., the total energy $E=int_D u dx$ inside the domain is constant. For the steady state solution,
$$
int_{x_1}^{x_2} u(tto infty,x) dx=int_{x_1}^{x_2} u(t=0,x) dx
$$
$$
c_2=frac{1}{x_2-x_1} int_{x_1}^{x_2} u_0(x) dx,
$$
in which $u_0(x)$ is the initial condition for $u$.
answered Dec 3 '18 at 23:50
rafa11111rafa11111
1,1291417
$endgroup$
1
$begingroup$
This is very helpful. How are you getting $u = c_1 x + c_2$ after setting $delta u / delta t = 0$? The solution I obtain for $u$, for general $lambda > or < 0$ is $u(x,t) = sum_n A_n sin(lambda_n x) e^{-lambda_n t}$. I could get your solution only by setting $lambda = 0$ and getting eigenfunctions for that: $f(x) = c_1 x + c_2$
$endgroup$
– hirschme
Dec 4 '18 at 16:08
1
$begingroup$
If $partial u/partial t=0$, the 'remaining' equation is $partial^2 u/partial x^2=0$. Furthermore, since there is no dependence on $t$ anymore, we can change the partial derivative to total derivative, i.e., $$frac{d^2 u}{dx^2}=0.$$ Integrating twice, we have $u=c_1 x+c_2$. See that this solution does not correspond to the solution for the original equation for any time $t$, but is the solution of the steady state equation. One can even say that, if $u$ does satisfy $$frac{partial u}{partial t} = k frac{partial ^2 u}{partial x^2},$$ then $$lim_{tto infty} u(x,t) = c_1x+c_2.$$
$endgroup$
– rafa11111
Dec 4 '18 at 16:19
add a comment |
$begingroup$
Your reasoning is correct. The steady state solutions can be obtained by setting $partial u/partial t=0$, leading to
$$
u = c_1 x+c_2.
$$
For the homogeneous Dirichlet B.C., the only solution is the trivial one (i.e., $u=0$. For the Neumann B.C., a uniform solution $u=c_2$ exists. Also, the steady state solution in this case is the mean temperature in the initial condition. Integrating the heat equation along $x_1<x<x_2$,
$$
frac{partial}{partial t} int_{x_1}^{x_2} u dx =k left.frac{partial u}{partial x} right|_{x_1}^{x_2}=0,
$$
i.e., the total energy $E=int_D u dx$ inside the domain is constant. For the steady state solution,
$$
int_{x_1}^{x_2} u(tto infty,x) dx=int_{x_1}^{x_2} u(t=0,x) dx
$$
$$
c_2=frac{1}{x_2-x_1} int_{x_1}^{x_2} u_0(x) dx,
$$
in which $u_0(x)$ is the initial condition for $u$.
answered Dec 3 '18 at 23:50
rafa11111rafa11111
1,1291417
$endgroup$
Your reasoning is correct. The steady state solutions can be obtained by setting $partial u/partial t=0$, leading to
$$
u = c_1 x+c_2.
$$
For the homogeneous Dirichlet B.C., the only solution is the trivial one (i.e., $u=0$. For the Neumann B.C., a uniform solution $u=c_2$ exists. Also, the steady state solution in this case is the mean temperature in the initial condition. Integrating the heat equation along $x_1<x<x_2$,
$$
frac{partial}{partial t} int_{x_1}^{x_2} u dx =k left.frac{partial u}{partial x} right|_{x_1}^{x_2}=0,
$$
i.e., the total energy $E=int_D u dx$ inside the domain is constant. For the steady state solution,
$$
int_{x_1}^{x_2} u(tto infty,x) dx=int_{x_1}^{x_2} u(t=0,x) dx
$$
$$
c_2=frac{1}{x_2-x_1} int_{x_1}^{x_2} u_0(x) dx,
$$
in which $u_0(x)$ is the initial condition for $u$.
answered Dec 3 '18 at 23:50
rafa11111rafa11111
1,1291417
edited Dec 4 '18 at 15:09
answered Dec 3 '18 at 23:50
rafa11111rafa11111
1,1291417
answered Dec 3 '18 at 23:50
rafa11111rafa11111
1,1291417
answered Dec 3 '18 at 23:50
rafa11111rafa11111
1,1291417
1,1291417
1
$begingroup$
This is very helpful. How are you getting $u = c_1 x + c_2$ after setting $delta u / delta t = 0$? The solution I obtain for $u$, for general $lambda > or < 0$ is $u(x,t) = sum_n A_n sin(lambda_n x) e^{-lambda_n t}$. I could get your solution only by setting $lambda = 0$ and getting eigenfunctions for that: $f(x) = c_1 x + c_2$
$endgroup$
– hirschme
Dec 4 '18 at 16:08
1
$begingroup$
If $partial u/partial t=0$, the 'remaining' equation is $partial^2 u/partial x^2=0$. Furthermore, since there is no dependence on $t$ anymore, we can change the partial derivative to total derivative, i.e., $$frac{d^2 u}{dx^2}=0.$$ Integrating twice, we have $u=c_1 x+c_2$. See that this solution does not correspond to the solution for the original equation for any time $t$, but is the solution of the steady state equation. One can even say that, if $u$ does satisfy $$frac{partial u}{partial t} = k frac{partial ^2 u}{partial x^2},$$ then $$lim_{tto infty} u(x,t) = c_1x+c_2.$$
$endgroup$
– rafa11111
Dec 4 '18 at 16:19
add a comment |
1
$begingroup$
This is very helpful. How are you getting $u = c_1 x + c_2$ after setting $delta u / delta t = 0$? The solution I obtain for $u$, for general $lambda > or < 0$ is $u(x,t) = sum_n A_n sin(lambda_n x) e^{-lambda_n t}$. I could get your solution only by setting $lambda = 0$ and getting eigenfunctions for that: $f(x) = c_1 x + c_2$
$endgroup$
– hirschme
Dec 4 '18 at 16:08
1
$begingroup$
If $partial u/partial t=0$, the 'remaining' equation is $partial^2 u/partial x^2=0$. Furthermore, since there is no dependence on $t$ anymore, we can change the partial derivative to total derivative, i.e., $$frac{d^2 u}{dx^2}=0.$$ Integrating twice, we have $u=c_1 x+c_2$. See that this solution does not correspond to the solution for the original equation for any time $t$, but is the solution of the steady state equation. One can even say that, if $u$ does satisfy $$frac{partial u}{partial t} = k frac{partial ^2 u}{partial x^2},$$ then $$lim_{tto infty} u(x,t) = c_1x+c_2.$$
$endgroup$
– rafa11111
Dec 4 '18 at 16:19
1
1
$begingroup$
This is very helpful. How are you getting $u = c_1 x + c_2$ after setting $delta u / delta t = 0$? The solution I obtain for $u$, for general $lambda > or < 0$ is $u(x,t) = sum_n A_n sin(lambda_n x) e^{-lambda_n t}$. I could get your solution only by setting $lambda = 0$ and getting eigenfunctions for that: $f(x) = c_1 x + c_2$
$endgroup$
– hirschme
Dec 4 '18 at 16:08
$begingroup$
This is very helpful. How are you getting $u = c_1 x + c_2$ after setting $delta u / delta t = 0$? The solution I obtain for $u$, for general $lambda > or < 0$ is $u(x,t) = sum_n A_n sin(lambda_n x) e^{-lambda_n t}$. I could get your solution only by setting $lambda = 0$ and getting eigenfunctions for that: $f(x) = c_1 x + c_2$
$endgroup$
– hirschme
Dec 4 '18 at 16:08
1
1
$begingroup$
If $partial u/partial t=0$, the 'remaining' equation is $partial^2 u/partial x^2=0$. Furthermore, since there is no dependence on $t$ anymore, we can change the partial derivative to total derivative, i.e., $$frac{d^2 u}{dx^2}=0.$$ Integrating twice, we have $u=c_1 x+c_2$. See that this solution does not correspond to the solution for the original equation for any time $t$, but is the solution of the steady state equation. One can even say that, if $u$ does satisfy $$frac{partial u}{partial t} = k frac{partial ^2 u}{partial x^2},$$ then $$lim_{tto infty} u(x,t) = c_1x+c_2.$$
$endgroup$
– rafa11111
Dec 4 '18 at 16:19
$begingroup$
If $partial u/partial t=0$, the 'remaining' equation is $partial^2 u/partial x^2=0$. Furthermore, since there is no dependence on $t$ anymore, we can change the partial derivative to total derivative, i.e., $$frac{d^2 u}{dx^2}=0.$$ Integrating twice, we have $u=c_1 x+c_2$. See that this solution does not correspond to the solution for the original equation for any time $t$, but is the solution of the steady state equation. One can even say that, if $u$ does satisfy $$frac{partial u}{partial t} = k frac{partial ^2 u}{partial x^2},$$ then $$lim_{tto infty} u(x,t) = c_1x+c_2.$$
$endgroup$
– rafa11111
Dec 4 '18 at 16:19
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%2f3024832%2fis-the-steady-state-solution-of-the-heat-equation-with-dirichlet-boundary-condit%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
