I'm asked to find the equilibrium solution of the heat equation $u_{t} = \nabla^{2} u$ in the ring $0< a Now, I know how to find equilibrium solutions of normal looking heat equations of the form $u_{t} = \alpha u_{rr}$, but $u_{t} = \nabla^{2} u = u_{tt} + u_{rr}$ is very strange. I imagine that I need to make a change of variables. The question is, what change of variables would work so that I could then proceed with solving this problem? Also, am I interpreting $\nabla^{2}u$ wrong? Since this is a ring, should it be in polar coordinates? That part was not clear to me from reading the question
Need help with change of variables in very strange heat equation problem
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pde
heat-equation
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0Are you sure that's the right formula? It suggests that the time-derivative of the heat has no dependence on the angle coordinate in polar coordinates! – 2017-01-11
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0@JohnHughes this is exactly the formula as given to me. But on reading it, I wasn't sure if it was in rectangular coordinates, and they were just calling $x$ $r$ or if it was actually in polar coordinates. – 2017-01-11
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0Addressing your revision: you're interpreting it wrong. $\nabla^{2}u$ means $u_{xx} + u_{yy}$, and you need to convert this to polar coordinates if you hope to express your answer in polar coordinates (which you should!). – 2017-01-11
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0@JohnHughes question for you: what are my independent and dependent variables here? – 2017-01-11
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0Your independent variables are $t$ and $x$. – 2017-01-11
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0I thought $r$ was my independent variable and $t$ was my dependent one. – 2017-01-11
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0No. The original equation is casted in Cartesian coordinate $(x,y)$, so before you do any transformation, your independent variable is $x,y,t$. (Apologise for forgettting $y$ in the previous comment). After you make the transformation from Cartesian to polar coordinate $(r,\theta)$, your independent variables now become $r,\theta,t$. – 2017-01-11
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0My apologies, I did not notice that your boundary conditions are given in terms of polar coordinates. But what I said still holds true, $r,\theta,t$ is your independent variable initially. Once you look for equilibrium solution, $u$ does not depend on $t$ anymore, other $u_t$ wouldn't be zero everywhere in your domain. – 2017-01-11
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0@CheeHan so $u_{t}$ does equal zero everywhere in the domain? I think maybe you meant to say "otherwise $u_{t}$ wouldn't be zero everywhere in your domain". Just making sure that's what you meant. – 2017-01-12
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0Yes you are right. I had a long day........ – 2017-01-12
2 Answers
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To find equilibrium solution, you need to solve $\nabla^2 u = 0$ with the given domain and boundary condition. Now, in Cartesian coordinate $(x,y)$, $\nabla^2=\partial_{xx} + \partial_{yy}$. In polar coordinate $(r,\theta)$, it is given by $$ \nabla^2 = \frac{1}{r}\frac{\partial}{\partial r}\left(r\frac{\partial}{\partial r}\right) + \frac{1}{r^2}\frac{\partial^2}{\partial\theta^2}. \label{eq:1}\tag{1}$$ Now, use the method of separation of variables. More precisely, we guess an ansatz of the form $$ u(r,\theta) = R(r)\Theta(\theta), $$ but since the boundary condition is independent of $\theta$, we guess $$ u(r,\theta) = R(r). $$ Substituting this into $\nabla^2 u=0$ and using \eqref{eq:1} yields $$ \frac{1}{r}(rR')' = 0 \implies rR' = C\implies R = C\ln(r) + D. $$ To find the constants $C,D$, we apply the boundary conditions (BCs) $R'(a) = 1$ and $R'(b) + R(b) = 2$. The first BC yields $$ 1 = R'(a) = \frac{C}{a}\implies C=a,$$ while the second BC yields $$ 2 = \frac{C}{b} + C\ln(b) + D = \frac{a}{b} + a\ln(b) + D\implies D = 2 - \frac{a}{b} - a\ln(b). $$ Hence, the equilibrium solution has the form $$ u(r,\theta) = u(r) = \frac{a}{r} + 2- \frac{a}{b} - a\ln(b). $$
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0I could add more details about how to carry out the separation of variables method, but I am assuming you know how to do that (: – 2017-01-11
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0so I can't assume then that my problem is $\theta$ independent like JohnHughes suggested? – 2017-01-11
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0so if what you're saying is true, what happens to my boundary conditions, since they have $t$ in them? – 2017-01-11
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0I just added the boundary condition @JessyCat. The point is when you are looking for equilibrium solution, $u$ does not depend on $t$ anymore. – 2017-01-11
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0and yes, you may assume that your solution is independent of $\theta$ too, since the boundary condition does not depend on $\theta$. Put in this way, one of the boundary condition says $u_r(a,t) = 1$, this means that at the ring $r=a$, $u_r$ must equal to 1 for all $\theta\in[0,2\pi)$. – 2017-01-11
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0also right now I"m looking at the case when $\mu = 0$. I solved the equation in $r$ and got $R(r) = c_{1}\ln(r) + c_{2}$, where $c_{1}$ and $c_{2}$ are constants of integration. Then, I aplied the B.C. $R^{\prime}(a) = \frac{c_{1}}{a} = 1$, which implies that $c_{1} = a$. Then, when $R^{\prime}(b) + R(b) = 2$, this is $\frac{c_{1}}{b}+c_{1}\ln(b) + c_{2}$, but since $c_{1} = a$, $c_{2} = 2 - \frac{a}{b}-a\ln(b)$, is that correct? – 2017-01-12
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0maybe you should say something about separation of variables. I don't think I'm handling the case where $\mu = 0$ correctly. – 2017-01-12
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0just tell me is this an eigenvalue problem or not? – 2017-01-12
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0I really wish you could work out some more details about this. I just had a really bad panic attack about this problem. – 2017-01-12
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0I just put some details into the answer. I believe my answer coincides with yours @JessyCat (: – 2017-01-12
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I suspect that you were given
$$
u_{t} = \nabla^{2} u,
0< a Now what exactly $u(a, t)$ means isn't clear to me, but I suspect it's this: The problem has circular symmetry, so although $u$ would normally be expressed (in polar coordinates) as $u(r, \theta, t)$, the author's omitted $\theta$ entirely. The Laplace operator in polar coordinates (see this wiki article) looks like
$$
\nabla^2 u = \frac{1}{r} \frac{\partial}{\partial r}(r u_r) + \frac{1}{r^2} u_{\theta\theta}
$$
By independence of $\theta$, the last term is trivial, and your equation becomes
$$
u_t = \frac{1}{r} \frac{\partial}{\partial r}(r u_r)
$$
where $u = u(r, t)$ satisfies $u_r(a, t) = 1$ for all $t$, and $u_r(b, t) + u(b, t) = 2$ for all $t$. My guess is that this is amenable to some sort of separation of variables, but it's been 30 years since i've solved a differential equation for real. :)
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0..and it looks as if @Chee Han agrees with me about the separation of vars. :) – 2017-01-11
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0Didn't know you were writing up a solution! – 2017-01-11
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0And I didn't know that YOU were. Fortunately, we seem to agree. All the best! – 2017-01-11
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0@JohnHughes check out page 8 of this bad boy: http://www.math.ttu.edu/~gilliam/ttu/s10/m3351_s10/c14_2d_disk_heat.pdf – 2017-01-11
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0FORTUNATELY for you, there's no $\theta$ dependence, so this reduces to "big messy fraction = 0" , which gets much simpler to solve. – 2017-01-11
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0@JohnHughes the only thing is then, should I have $u_{t} = \frac{1}{r}u_{r}(r,t) + u_{rr}(r,t)+ \frac{1}{r^{2}}u_{tt}(r,t)$ or $u_{t} = \frac{1}{r}u_{r} + u_{rr}+ \frac{1}{r^{2}}u_{\theta\theta}$? – 2017-01-11
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0You want to find equilibrium solution, so $u_t=0$. Equilibrium solution also means stationary solution. – 2017-01-11
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0You should certainly not have $u_{tt}$. You have $u_{\theta\theta}$, which, because the solution is independent of $\theta$, must be 0. – 2017-01-11
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0And chee's point is well-taken: the left hand side can be set to 0 because you're asked for the equilibrium solution (which I missed when I read the problem). – 2017-01-11
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0@JohnHughes and so basically you're solving the Laplace Equation and not the Heat Equation – 2017-01-11
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0Yes, that's right. – 2017-01-11
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0@JohnHughes what happens to my boundary conditions then? – 2017-01-11
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0Your boundary conditions are already expressed in polar coordinates: at $r = a$ you have one condition; at $r = b$ you have another. And since $u$ is independent of time (steady state solution!) and angle (symmetry!), that's all you need. It's a 1D problem at this point. (And this is the last I'm going to say about it: you'll learn something by working the rest out yourself.) – 2017-01-11