# Steady State Heat Equation with Source

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1. Nov 9, 2015

I am trying to solve the steady state heat equation with a heat source. I am starting out in 1 dimension (my book gives the solution in 2, but I'm just trying to get a feel right now) and I have a heat source Q, located at 0. It radiates heat through an infinite medium. So what would the steady state solution be?

$\frac{\partial U}{\partial t} = k_0\frac{\partial^2U}{\partial^2 x} + Q$

So since this is steady state, I figure I can set the left hand side to zero to get the equation:

$0 = k_0\frac{\partial^2U}{\partial^2 x} + Q$

In my book it gives the solution that looks something like (2d solution, radial coordinates): $U(x) = \frac{Q}{2\pi}ln(x)$ I know that that equation isn't exactly right, but the natural log term is in there. I am not sure how you can solve the heat equation to get the natural log term, I can only guess that it is by integrating 1/x at some point, but I'm not sure as to the process.

Also, the book is on geo-exchange systems and they are referring to a pipe underground radiating energy. They refer to this as a "line source" and they are modeling it in 2d at first.

2. Nov 9, 2015

### Geofleur

If you want the heat source to be located exactly at the origin and nowhere else, you will need $Q \delta(x)$, where $\delta(x)$ is the Dirac delta function. The way you have it, there is a heat source of strength $Q$ everywhere along the $x$-axis. The units for $Q$ would have to be changed too, but let's not worry about that unless you really want to work with the delta function source.

To find out what source $U(x)=\frac{Q}{2\pi}\ln x$ corresponds to, use it to calculate $\frac{d^2 U}{dx^2}$ and see what differential equation that implies.

3. Nov 9, 2015

### Geofleur

Actually, for your problem we probably should deal with the delta function source, but lets sidestep the unit issue for the moment. The steady state heat equation in cylindrical coordinates, with a source right at $r = 0$, reads $\frac{d}{dr}\left( r \frac{dU}{dr} \right) = \delta(r) Q$, where $Q$ is constant. If we integrate both sides of this from $0$ to $r$, we get

$\int_0^{r} \frac{d}{dr'}\left( r' \frac{dU}{dr'} \right) dr' = \int_0^{r} \delta(r') Q dr' = Q + C_1$.

The integration constant $C_1$ shows up because one of the integration limits, $r$, is really indefinite. Since $Q + C_1$ is just a constant, we might as well absorb the $Q$ into the $C_1$. Then, carrying out the integration on the left hand side of the above equation, we have

$r \frac{dU}{dr}=C_1$.

Now move $r$ over to the other side of the equation and integrate again to get

$U(r) = C_1 \ln r + C_2$,

where $C_2$ is another integration constant. So a delta function source at $r = 0$ really does lead to a natural logarithm. But here is the interesting thing. Suppose we had included no source at all, and had instead simply written

$\frac{d}{dr}\left(r\frac{dU}{dr}\right) = 0$.

We would have gotten the same result! Often, when solving differential equations with no source, a solution corresponding to a point source sneaks in!

4. Nov 10, 2015

I have the book in hand now and the exact answer is:

$T(r) = \frac{q}{2\pi \lambda}ln(r)$ where lambda is the thermal conductivity of the material. So I see why you use the dirac delta function, and from your work, I'm inferring that if we integrate it from 0 to infinity in polar coordinates we get 1. But we don't necessarily need it in this case.

So I guess one way we can get Q at the origin is by solving for the boundary conditions in $r\frac{dU}{dr} = c_1$? I feel more intuitive than rigorous doing this, but it seems as though the change in heat with respect to r should be proportional to $\frac{Q}{2\pi \lambda r}$. Since in 2-space the increase in perimeter is proportional to 2pi*r. At least the units work out in this case.

Thank you for your timely and thorough response. Hopefully I'll be able to work through this whole book.

5. Nov 10, 2015

### Geofleur

Here is a way to incorporate the line source as a boundary condition. Consider a section of the line source of height $H$ and surround it with an imaginary cylinder of the same height and with a radius $r$. Let $\mathbf{f}$ be the heat flux and $Q$ be the heat released from the line source per unit height and per unit time. Then we can write the heat flow through the cylinder as

$\int_S \mathbf{f} \cdot d\mathbf{S} = QH$.

where $d\mathbf{S}$ is an element of cylinder area. Assume that the heat flux is independent of $\phi$ and $z$ (we are using cylindrical coordinates). Then $\mathbf{f} = f(r)\mathbf{\hat{r}}$, and the integral becomes

$\int_0^H \int_0^{2\pi} f(r) r d\phi dz = 2\pi H f(r) r = QH$.

Solving for $f(r)$ we find that the heat flux we want is $\mathbf{f} = \frac{Q}{2\pi r}\mathbf{\hat{r}}$. Notice that the heat flux goes as $\frac{1}{r}$. If we were dealing with a point source instead of a line source and had surrounded it with a sphere, we would have gotten heat flux $\propto \frac{1}{r^2}$.

We can use the result that $f = \frac{Q}{2\pi r}$ to evaluate the constant $C_1$ in our solution $U(r) = C_1 \ln r + C_2$. We have

$-\lambda \frac{dU}{dr} = f = \frac{Q}{2 \pi r} \rightarrow \frac{dU}{dr} = -\frac{Q}{2 \pi \lambda r}$.

Because $\frac{dU}{dr} = \frac{C_1}{r}$, we get $C_1 = -\frac{Q}{2\pi \lambda}$. Trying to incorporate the boundary condition directly by evaluating $\frac{dU}{dr}$ at the origin would not work, because $\frac{dU}{dr}$ blows up at $r = 0$. This is a way to get around that problem.

6. Nov 14, 2015

I was going over your math on post #4 and it looks like you missed the heat equation a bit. In cylindrical coordinates, the heat equation gives,

$\frac{1}{r}\frac{d}{dr}(r\frac{dT}{dr}) = 0$ I set it as a homogeneous differential equation because I'm really not skilled enough to solve the non-homogeneous one... This simplifies to->
$\frac{d^2T}{dr} + \frac{1}{r}\frac{dT}{dr} = 0$

Which has the solution just as you put it: $c_1 ln(r) + c_2$. I'm not comfortable solving the non-homogeneous case, but as you pointed out above, it's not a big deal because the boundary conditions are found later.

OK, that makes sense... I guess this whole thing is a little frustrating because intuitively it doesn't make a lot of sense. I understand how after an infinite amount of time a heat generating point will have infinite temperature, but as the radius gets larger and larger, the temperature approaches -infinity, there is no reference to the starting temperature of the medium. And at steady state, wouldn't the whole medium just be super hot anyway with nowhere for the energy to go but out?

7. Nov 16, 2015

### Geofleur

For the non-homogeneous case, I started from $\frac{d}{dr}\left( r \frac{dU}{dr} \right) = Q\delta(r)$ without showing how I got to that equation in the first place - sorry about that! Here are the missing steps:

Let's begin with $\frac{1}{r}\frac{d}{dr}\left( r \frac{dU}{dr} \right) = \mathcal{Q} \delta(r \ \hat{\mathbf{r}})$. Here, the curly $\mathcal{Q}$ is the heat per unit time emitted from the line source along the $z$-axis, and I have written the delta function with a vector argument, where $\hat{\mathbf{r}}$ is the unit vector that points radially outward, perpendicular to the axis of the cylinder. If we convert this delta function into one written in terms of the coordinate $r$ only (that is, not in terms of a vector quantity), then we get the equation with which I started in the earlier post. The key is that $\delta(r \ \hat{\mathbf{r}})$ is zero everywhere except along the $z$-axis and satisfies

$\int_0^H \int_0^R \int_0^{2\pi} \delta(r \ \hat{\mathbf{r}}) r d\phi dr dz = 1$,

when integrated over the volume of a cylinder of arbitrary radius $R$ and height $H$. The above integral will be unity only if we set

$\delta( r \ \hat{\mathbf{r}}) = \frac{\delta(r)}{2\pi H r}$.

Putting this result into $\frac{1}{r}\frac{d}{dr}\left( r \frac{dU}{dr} \right) = \mathcal{Q} \delta(r \ \hat{\mathbf{r}})$ yields

$\frac{d}{dr}\left( r \frac{dU}{dr} \right) = \mathcal{Q} \frac{\delta(r)}{2\pi H}$,

and if we define $Q = \mathcal{Q}/(2\pi H)$ we get the starting equation from post #3.

Indeed, we have assumed that there is a steady state for this physical situation, but there isn't necessarily! In an infinite medium, heat might just propagate outward from the line source forever. The temperature at any point, no matter how far away, might eventually increase. If we instead say that, at some maximum, finite radius, the cylinder has a boundary that is held at some temperature, then we can choose the boundary conditions to satisfy that constraint, and we don't care what happens past that radius anymore (or if we do, we have to solve the heat equation separately for that domain, and that equation may well be time dependent).

Also, there are no true line sources of heat, there will really be some finite inner radius as well. We keep the line source as long as we are willing to tolerate that the solution may not be physical too nearby!

I hope that this helps to clarify things.

8. Nov 18, 2015

My book sets up opposite "mirror" line sources of heat to give a boundary condition at the ground surface (I'm reading about horizontal piping systems). I can see how you would be able to use the steady state equation to make something like that work as long as the boundary conditions are accounted for. As far as the log term tending towards infinity as r approaches zero, I guess is sort of similar to Newton's Law of Gravity. $F = G\frac{m_1m_2}{r^2}$ Which works to describe two masses at reasonable distance, but point masses would experience an infinite force as r approaches zero. So it's not a perfect model, but it works well enough. The log term is maybe the same way.