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?
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 !