0
$\begingroup$

We have the following system that describes the heat conduction in a rectangular region: $$\begin{cases} u_{xx}+u_{yy}+S=u_t \\ u(a,y,t)=0 \\ u_x(x,b,t)=0 \\ u_y(0,y,t)=0 \\ u(x,0,t) = 0 \\ u(x,y,0) = f(x,y) \end{cases} $$ $S$ is source term (independent of time and space ) placed at the origin. enter image description here

How can I solve this system analytically using separation of variables? or is there another method to solve it(analytically)?
EDIT: I am looking for the solution when $S$ is present because that's what makes the challenge .

$\endgroup$
  • 2
    $\begingroup$ Boundary condition for u(x, 0) is missing. In addition, that the two Neumann boundary conditions are given in direction tangential to the boundary is strange to me -- it's just another way of saying fixed value. Please double check. $\endgroup$ – Taozi Dec 22 '16 at 2:11
  • $\begingroup$ @Taozi: Do you think it's an ill-posed problem? (To be honest: I think so) $\endgroup$ – Han de Bruijn Dec 26 '16 at 18:07
  • $\begingroup$ What are the condition imposed on $f$? $\endgroup$ – Hetebrij Dec 26 '16 at 18:25
  • $\begingroup$ Sorry. I don't see how time $t$ can play any role. Therefore it seems to me that you already have the solution, namely $\,u(x,y)=f(x,y)$ . Anyway, I don't understand anything of your problem. $\endgroup$ – Han de Bruijn Dec 26 '16 at 18:50
  • $\begingroup$ @HandeBruijn Sorry I forgot $u_t$ in RHS $\endgroup$ – Freshman42 Dec 26 '16 at 19:06
-2
$\begingroup$

EDIT This is as complete an answer as can be given with the ill-specified problem. Please see the bottom for a new discussion of the forcing term.

Interpreting your $S$ term as a source located at the origin $(S=S_0\delta(x)\delta(y))$, we have: $$ u_{xx} + u_{yy} = u_{t} + S_0\delta(x)\delta(y) $$ So, for all points not precisely the origin, we have the homogeneous equation $$ u_{xx} + u_{yy} = u_{t} $$ We don't yet get to what to do exactly at the origin. For now, let us try the separation of variables approach, guessing a form: $$ u(x,y,t) = X(x)Y(y)T(t) $$

Plugging into the original equation (and dropping the variable dependencies for notational simplicity) we have: $$ X''YT + XY''T = XYT' $$ Dividing by $XYT$ everywhere gives $$ \frac{X''}{X} + \frac{Y''}{Y} = \frac{T'}{T} $$ By the usual argument, the left hand side is a function of only the spatial coordinates and the right hand side is only a function of only the time coordinate, yet they are equal to each other. Therefore, each side must be equal to some constant, which we will denote as $-\lambda$. Let us consider the time equation first: $$ \frac{T'}{T} = -\lambda \;\;\Rightarrow\;\; T' + \lambda T = 0 $$ The general solution to this equation (up to a multiplicative constant we'll account for later) is: $$ T(t) = e^{-\lambda t} $$ Now consider the spatial equation $$ \frac{X''}{X} + \frac{Y''}{Y} = -\lambda $$ Moving terms gives $$ \frac{X''}{X} = -\left(\lambda+\frac{Y''}{Y}\right) $$ By the same argument as above, each side must also be a constant, which I will denote as $-k_x^2$. So for the $x$ equation we have: $$ \frac{X''}{X} = -k_x^2 \;\;\Rightarrow\;\; X'' + k_x^2 X = 0 $$ The general solution here is a sinusoid with spatial frequency $k_x$: $$ X(x) = A\cos(k_x x) + B\sin(k_x x) $$ Now we apply the boundary conditions, which, as written in the original question, are under-specified. I'm going to assume you meant two insulating walls (where the temperature gradient normal to the wall vanishes) and two zero temperature walls (where the temperature itself vanishes). For the $x$ dependence, these are: $$ u_x(0,y,t) = 0 \\ u(a,y,t) = 0 $$ Plugging in the separated forms reveals these correspond to $$ X'(0) = 0 \;\;\Rightarrow\;\; -Ak_x\sin(0) + Bk_x\cos(0) = 0 \;\;\Rightarrow\;\; B=0 $$ and $$ X(a) = 0 \;\;\Rightarrow\;\; A\cos(k_x a) = 0 \;\;\Rightarrow\;\; k_x=\frac{(n-1/2)\pi}{a}\;\forall n\in\mathbb{Z} $$ Now we turn to the $Y$ dependence. We have $$ -\left(\lambda+\frac{Y''}{Y}\right) = -k_x^2 \;\Rightarrow\; Y'' + (\lambda-k_x^2)Y = 0 $$ Now define the constant $k_y^2 = (\lambda-k_x^2)$ so that the solution to the above equation is $$ Y(y) = C\cos(k_y y) + D\sin(k_y y) $$ Now applying the assumed boundary conditions $$ u(x,0,t) = 0 \\ u_y(x,b,t) = 0 $$ to the separated form gives $$ Y(0)=0\;\Rightarrow\; C\cos(0) + D\sin(0) = 0 \;\Rightarrow\; C=0 $$ and $$ Y'(b) = 0 \;\Rightarrow\; Dk_y\cos(k_y b) = 0 \;\Rightarrow k_y=\frac{(m-1/2)\pi}{b}\;\forall m\in\mathbb{Z} $$ Now note that $k_x$ and $k_y$ can only take on discrete values as given above, and that $\lambda = k_x^2+k_y^2$ by the definition above as well. So for any two indices $(m,n)$, the following is a solution of the equation $$ u_{m,n} = C_{m,n} e^{-(k_x^2+k_y^2) t}\cos(k_x x)\sin(k_y y) $$ And now by linearity, a superposition of those is also a solution $$ u(x,y,t) = \sum_n^\infty\sum_m^\infty C_{m,n} e^{-(k_x^2+k_y^2) t}\cos(k_x x)\sin(k_y y) $$ There's two final things to consider, the delta function forcing and the initial condition. First, the $t=0$ initial condition $$ u(x,y,0) = \sum_n^\infty\sum_m^\infty C_{m,n} \cos(k_x x)\sin(k_y y) = f(x,y) $$ This tells me the $C_{n,m}$ must be chosen to be like the fourier coefficients of the $f(x,y)$ function, something like $$ C_{n,m} = \int_0^a\int_0^b f(x,y) \cos(k_x x)\sin(k_y y) dx dy $$ So now we have a full solution to the homogeneous problem with $S=0$, or equivalently, one that is valid precisely everywhere except $x=y=0$. Now we need to know how to handle that singularity at the origin. The way we normally handle this is to force a derivative discontinuity into the solution by flipping signs of the solution at a point in such a way that taking laplacian results in an impulse. However, I claim that the boundary conditions given are incompatible with a forcing at the origin. To see why, consider that $u_x(0,y,t) = 0$ says that there is no heat flux through the left wall. However, the source is touching that wall, meaning that the source is touching a perfectly insulating wall; how can heat from a source flow through an insulating wall? It cannot! The source has no effect in this problem because of the boundary conditions given. Therefore, the homogeneous solution given is the solution to the problem.

If you can clarify the boundary conditions and source specification and post in another question, you might get a better answer that helps you know how to solve these problems. For what it's worth, I can see a path forward with a point source if the boundary conditions were such that the modes had the form $\sin(k_x x)\sin(k_y y)$. As it is, we had to guess at what boundary conditions you meant because the ones you gave don't lead to a well-posed problem.

$\endgroup$
  • $\begingroup$ Given the poor specifications of the problem, this is a very worthwhile (+1) attempt. $\endgroup$ – Han de Bruijn Dec 28 '16 at 22:13
  • $\begingroup$ @HandeBruijn it is easy to solve $u_{xx}+u_{yy}=u_t$ the probem is when there is a source term! $\endgroup$ – Freshman42 Dec 28 '16 at 23:20
  • $\begingroup$ @freshman42, If it was easy to get this far and you had a question about the particular solution under a delta function source, perhaps you could have worked out the above in your question then clarified that you didn't know how to work with the source term. Doing so would have clarified the questions others had about the boundary conditions. $\endgroup$ – rajb245 Dec 29 '16 at 0:04
  • $\begingroup$ Would the downvoters care to give a decent answer of their own, please? Thanks. $\endgroup$ – Han de Bruijn Dec 30 '16 at 9:21
  • $\begingroup$ @HandeBruijn thanks for coming to the defense of this answer. If the point of the site is to archive valuable answers to interesting questions, then I believe the downvotes cannot be justified. The tooltip when you go to downvote an answer says "The answer is not useful." That is of course an opinion, one that you and I do not share with the downvoters. $\endgroup$ – rajb245 Dec 30 '16 at 16:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.