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I am asked to solve the following problem:

A long, slim, waterpipe with radius 1cm runs water with a constant temperature 60 degrees celsius. The pipe is surrounded by 3cm insulation. Outside of the insulation, the room temperature is 20 degrees celsius. What is the temperature in the insulation after a long period of time? Neglect resistance between materials.

I set up the problem as $$\frac{\partial u}{\partial t} - a\frac{\partial^2u}{\partial x^2} = 0, \quad 1<x<4, \ t>0,$$ $$u(1,t)=60,\quad u(4,t)=20, \quad t>0.$$ and realizing the stationary solution is asked, i.e. $\frac{\partial u}{\partial t} = 0,$ the problem becomes $$\frac{\partial^2u}{\partial x^2} = 0, \quad 1<x<4$$ $$u(1)=60,\quad u(4)=20$$

If I solve this with the method I am supposed to use, I get the correct answer $$60-\frac{20}{\ln(2)}\ln(x).$$

But the function $u(x) = -\frac{40}{3}x+\frac{220}{3}$, arrived at by simply assuming the form $y=kx+m$, also solves the equation. Why can this not be a solution aswell?

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    $\begingroup$ Are you sure $60 - 20 \frac{20}{\ln{}2} \ln{}x$ verifies $\partial_{xx} u=0$ ? $\endgroup$
    – Velobos
    Commented Aug 10, 2020 at 16:36
  • $\begingroup$ @Velobos I haven't checked it, but this is the answer given in my textbook. Is it wrong? $\endgroup$
    – user810259
    Commented Aug 10, 2020 at 16:46
  • $\begingroup$ I'm not sure that you started with the correct differential equation. Your variable $x$ is a polar coordinate, and the Laplacian operator that occurs in the heat equation has a different expression in polar coordinates than cartesian coordinates. Specifically, I'd expect not only $\partial^2u/\partial x^2$ but also $\frac1x\partial u/\partial x$. $\endgroup$
    – Andreas Blass
    Commented Aug 10, 2020 at 17:00

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The set up of your problem is wrong : we can't see the cylindrical aspect of the problem. Your heat equation should be :

$$\frac{\partial u}{\partial t}(r,\theta,t) - a \Delta u(r,\theta,t) = 0, \quad 1<r<4, \theta \in [0,2\pi] , \ t>0, $$ with $u(1,\theta,t)=60$ and $u(4,\theta,t)=20$, for all $\theta \in [0,2\pi]$, for $t>0$.

You can now use the cylindrical laplacian : $$\Delta u = \frac{1}{r} \frac{\partial}{\partial r} \left( r \frac{\partial u}{\partial r}\right) + \frac{1}{r^2} \frac{\partial u}{\partial \theta}$$

Since $u$ doesn't depend on $\theta$ because of the geometry, you have to find $u(r,\theta)=u(r)$ such as :

$$\frac{1}{r} \frac{\partial}{\partial r} \left( r \frac{\partial u}{\partial r}\right)(r) = 0, \quad 1 \leq r \leq 4, \quad \forall \theta$$ with $u(1)=20$ and $u(4)=60$.

Solving this problem should give you your textbook answer !

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    $\begingroup$ Minor technical correction: the reason for the lack of dependence on $\theta$ is because the boundary conditions don't have a dependence on $\theta$, not solely because of the symmetry of the geometry. $\endgroup$
    – Ian
    Commented Aug 10, 2020 at 17:02
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    $\begingroup$ Note also that a solution of the form $u(r,\theta) = m r + b$, as proposed by the OP, is not a solution of Laplace's equation in cylindrical coordinates (unless $m = 0$.) $\endgroup$
    – Michael Seifert
    Commented Aug 10, 2020 at 17:14

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