Consider an iron cylinder. The heat doesn't go through the cylinder surface, but at the circular cross-section in the endpoints. The temperature of the surroundings is $15\ \mathrm{^\circ C}$. The initial temperature of the iron stick is $100\ \mathrm{^\circ C}$. How to set up the differential equation? $T(0,t)$ and $T(l,t)$ doesn't fit anymore.

  • $\begingroup$ Are you asking how the cylinder cools down, when its surface is insulated and heat is lost only through the cross sections at the ends of the cylinder? $\endgroup$ – Gert Mar 22 '20 at 15:48
  • $\begingroup$ @Gert Yes, I cannot find an equation to describe the amount of heat into the cross-section. $\endgroup$ – yuanming luo Mar 22 '20 at 16:09
  • $\begingroup$ OK, Will try and answer. $\endgroup$ – Gert Mar 22 '20 at 16:11
  • $\begingroup$ The confusion is the endpoint of the iron cylinder loss heat and gets heat at the same point, I only know the rate of losing heat at a certain temperature but I cannot determine the rate of gaining heat. $\endgroup$ – yuanming luo Mar 22 '20 at 16:23
  • $\begingroup$ Are you familiar with the transient heat conduction equation in 1D? $\endgroup$ – Chet Miller Mar 22 '20 at 17:40

So we have a cylinder of height $H$ and radius $R$:


Surrounding temperature is $T_e=15\text{ C}$ and initial, uniform temperature of the cylinder is $T_0=100\text{ C}$.

The cylinder is perfectly insulated, except for both circular ends, which lose heat through convection (only).

The go to equation for this kind of problem is Fourier's heat equation:

$$T_t=\alpha \nabla^2 T$$

Because of the geometry of the problem this equation begs for cylindrical coordinates:

$$\frac{\partial T}{\partial t}=\alpha \Big[\frac{1}{r}\frac{\partial}{\partial r}\Big(r\frac{\partial T}{\partial r}\Big)+\frac{1}{r^2}\frac{\partial^2 T}{\partial \phi^2}+\frac{\partial^2 T}{\partial x^2}\Big]$$

Due to symmetry considerations the $\phi$ term isn't needed and developing the partial derivatives, we get (using shorthand partials):


Furthermore, because of the insulation there's no radial temperature gradient, so:




So we're looking for a function:


which describes the spatial distribution of $T$, as well as its evolution in time $t$. Such a function would also allow to calculate the heat flux through the ends of the cylinder (as a function of time).

All real-world differential equations require boundary conditions and this case is no different.

Initial condition:


Convection at ends:

$$T_t(r,+\frac{H}{2},t)=h(T-T_e)$$ and: $$T_t(r,-\frac{H}{2},t)=h(T-T_e)$$

where $h$ is the convection heat transfer coefficient.

It's advisable to make a small substitution:


So we have:




Once $u(x,t)$ is found, we can convert it back to $T(x,t)$.


We assume (Ansatz):

$$u(x,t)=X(x)\Theta(t)$$ Inserting into the PDE we get:

$$X\Theta'=\alpha \Theta X''$$ Divide by $u(x,t)=X(x)\Theta(t)$:

$$\frac{\Theta''}{\alpha \Theta}=\frac{X''}{X}=-k^2$$

where $k$ is a Real number. So we have two ODEs:

$$\frac{\Theta''}{\alpha \Theta}=-k^2\tag{1}$$


$(1)$ solves to:

$$\Theta(t)=C \exp{(-k^2 \alpha t)}$$

And $(2)$ solves to:

$$X(x)=A \sin(kx)+B \cos(kx)\tag{3}$$

'All that is left to do' is determine the integration constants $A$, $B$ and $C$.

Unfortunately the BCs are of the Neumann type and thus non-homogeneous (non-zero). This generally creates an unsightly mess with no easy way from which to extricate $A$ and $B$.

Instead I'll try the much simpler case where the ends are kept at constant temperature $u=0$, so: $$u(-\frac{H}{2},t)=u(\frac{H}{2},t)=0$$ This also means: $$X(-\frac{H}{2})\Theta(t)=X(\frac{H}{2})\Theta(t)=0$$ Assume $\Theta(t) \neq 0$, thus: $$X(-\frac{H}{2})=X(\frac{H}{2})=0$$ Insert into $(3)$:

$$A \sin(k\frac{H}{2})+B \cos(k\frac{H}{2})\tag{4}=0$$

$$A \sin(-k\frac{H}{2})+B \cos(-k\frac{H}{2})=0$$ From the last equation: $$A \sin(k\frac{H}{2})-B \cos(k\frac{H}{2})=0\tag{5}$$ Now add $(4)$ to $(5)$:


Assume $A \neq 0$, then:


This happens for:

$$k\frac{H}{2}=n \pi$$ Or: $$k=\frac{2n\pi}{H}$$ For $n=1,2,3,4,...$

With $\sin(k\frac{H}{2})=0$, then $B=0$. So we get:

$$X_n(x)=A_n\sin\Big(\frac{2n\pi x}{H}\Big)$$


$$u_n(x,t)=D_n \exp{\Big[-\Big(\frac{2n\pi}{H}\Big)^2 \alpha t\Big]}\sin\Big(\frac{2n\pi x}{H}\Big)$$

Applying the superposition principle:

$$u(x,t)=\sum_{n=1}^{+\infty}D_n \exp{\Big[-\Big(\frac{2n\pi}{H}\Big)^2 \alpha t\Big]}\sin\Big(\frac{2n\pi x}{H}\Big)$$

The coefficients $D_n$ are obtained with the initial condition and the Fourier series.

For $t=0$, $u(x,0)=85$ and:

$$85=\sum_{n=1}^{+\infty}D_n \sin\Big(\frac{2n\pi x}{H}\Big)$$


$$D_n=\frac{2}{H}\int_{-H/2}^{+H/2}85 \sin\Big(\frac{2n\pi x}{H}\Big)\text{d}x$$

  • $\begingroup$ Edit made: eliminated the confusing dual use o the symbol $T$. $\endgroup$ – Gert Mar 22 '20 at 18:12
  • $\begingroup$ Now justify why the Ansatz yields a complete solution :p $\endgroup$ – user224659 Mar 22 '20 at 18:52
  • $\begingroup$ I know that there is no easy justification for this in the general case.. Always left me kind of unsatisfied when first encountering this kind of problem in classical electrodynamics. $\endgroup$ – user224659 Mar 22 '20 at 18:55
  • $\begingroup$ Trust me: the dreaded non-homogeneous BCs are a bigger problem! $\endgroup$ – Gert Mar 22 '20 at 19:48
  • $\begingroup$ This is a 1D problem, and the temperature does not vary with r. It is only a function of x and t. $\endgroup$ – Chet Miller Mar 22 '20 at 21:45

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