# Talk:Heat equation

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## What is u?

Nowhere does it say what u physically means... 76.254.50.99 (talk) 15:38, 2 September 2009 (UTC)

The first line covers it really... "The heat equation is an important partial differential equation which describes the distribution of heat (or variation in temperature)" It is the temperature at a point in space and time —Preceding unsigned comment added by 132.181.247.158 (talk) 16:06, 7 September 2009 (UTC)

The first line is nonsense. Anyone would understand "equation which describes the distribution of heat" as saying that u is heat. Which is of course false. 76.254.50.99 (talk) 20:29, 7 September 2009 (UTC)

Why is u first the temperature, then it becomes 'excess temperature' when introducing the radiative term? This treatment is misleading. — Preceding unsigned comment added by 79.119.49.89 (talk) 05:37, 3 August 2013 (UTC)

## New applications

I added a sentence mentioning the equations use in image analysis. I have also once seen it mentioned in a textbook on population genetics. Can anyone corroborate that this is a common usage? In that case that should be mentioned too on this page. 130.235.35.201

## Notation

To me it is confusing that temperature is denoted u while internal energy is denoted x. Isn't the convention that temperature is called T and internal energy U?

Do you think:

${\displaystyle {\partial u \over \partial t}=k\left({\partial ^{2}u \over \partial x^{2}}+{\partial ^{2}u \over \partial y^{2}}+{\partial ^{2}u \over \partial z^{2}}\right)}$

would be better? Dmn 20:40, 28 Mar 2004 (UTC)

yes I think it would be better if the article mentioned the equation at least once in this form (and I wouldn't mind if it was in this form throughout the whole article)ThorinMuglindir 17:32, 30 October 2005 (UTC)
Done, on 02:21, 23 February 2006 (UTC), by anon IP 150.140.211.76. (comment by Oleg Alexandrov (talk) 02:21, 23 February 2006 (UTC))

you are missing convection, and the equation is not in its genrelised form (you have written it in cartesian) you should probably write it in terms of the Material derivative and Laplacian. also i believe it is convention to write the derivitives on the right hand side and the source terms on the left giving:
${\displaystyle {\frac {Du}{Dt}}-k\nabla ^{2}u=q}$
RichardMathie (talk) 20:45, 10 August 2008 (UTC)

## Section title "Solving the heat equation using Fourier series" seems misnamed

The section titled "Solving the heat equation using Fourier series" nowhere mentions Fourier series. It does, however, mention that Fourier solved the problem in the manner described in the section, using separation of variables. Perhaps the section should be titled "Solving the heat equation using separation of variables"? 192.114.23.210 (talk) 10:16, 3 March 2011 (UTC)

## solving heat equations with fourier

is there a solved example where we solve/prove the heat equation using the fourier series

Is the question the solution of ${\displaystyle {\partial u \over \partial t}=D{\partial ^{2}u \over \partial x^{2}}}$ which is ${\displaystyle {1 \over {\sqrt {4\pi Dt}}}e^{-{x^{2} \over {4Dt}}}}$ ?

If it is, there it is.

Shouldn't the method of using Fourier transformations (mentioned below this line!) and convolutions be included in the Methods of solution section?

We start with the partial differential equation (pde), one dimensional diffusion equation

${\displaystyle {\partial u \over \partial t}=D{\partial ^{2}u \over \partial x^{2}}}$

Take Fourier transform the pde with respect to x using ${\displaystyle u(k,t)=\int _{-\infty }^{\infty }e^{-ikx}u(x,t)dx}$ . And so

${\displaystyle \int _{-\infty }^{\infty }e^{-ikx}{\partial ^{2} \over \partial x^{2}}u(x,t)=\int _{-\infty }^{\infty }(-k^{2})e^{-ikx}u(x,t)dx=-k^{2}u(k,t)}$ by separation by parts.

The pde becomes ${\displaystyle {\partial \over \partial t}u(k,t)=-k^{2}Du(k,t)}$. This is an 1st order differential equation. The solution is simply ${\displaystyle u(k,t)=C(k)e^{-Dk^{2}t}}$

First we need to find what is ${\displaystyle C(k)}$. For ${\displaystyle t=0}$, we see that ${\displaystyle C(k)=u(k,0)}$ and

${\displaystyle u(k,0)=\int _{-\infty }^{\infty }e^{-ikx}u(x,0)dx}$

The initial condition is ${\displaystyle u(x,0)=\delta (x)}$. Since it is a delta function the answer will be the Green function.

So from the above equation simply ${\displaystyle u(k,0)=1=C(k)}$. And so

${\displaystyle u(k,t)=e^{-Dk^{2}t}}$. This is the solution of the pde but we need to convert it back to x space.

The inverse Fourier will be ${\displaystyle u(x,t)={1 \over 2\pi }\int _{-\infty }^{\infty }e^{ikx}u(k,t)dk}$

So,${\displaystyle u(x,t)={1 \over 2\pi }\int _{-\infty }^{\infty }e^{ikx}e^{-Dk^{2}t}dk={1 \over 2\pi }\int _{-\infty }^{\infty }e^{-Dk^{2}t+ikx}dk}$

Let's play a little with the exponential part.

${\displaystyle -Dk^{2}t+ikx=-Dt(k^{2}-{ikx \over Dt})=-Dt\left[(k-{ix \over 2Dt})^{2}+{x^{2} \over 4D^{2}t^{2}}\right]=-Dt\left[(k-{ix \over 2Dt})^{2}\right]-{x^{2} \over 4Dt}}$

So, ${\displaystyle u(x,t)={1 \over 2\pi }e^{-{x^{2} \over 4Dt}}\int _{-\infty }^{\infty }e^{-Dt(k-{ix \over 2Dt})^{2}}dk}$

Set ${\displaystyle k-{ix \over 2Dt}=s}$ so ${\displaystyle dk=ds}$. The integral is ${\displaystyle \int _{-\infty }^{\infty }e^{-Dts^{2}}ds={\sqrt {\pi \over Dt}}}$ That gives the solution.

${\displaystyle u(x,t)={1 \over 2\pi }e^{-{x^{2} \over 4Dt}}{\sqrt {\pi \over Dt}}={1 \over {\sqrt {4\pi Dt}}}e^{-{x^{2} \over 4Dt}}}$ --71.96.115.55 15:51, 30 October 2005 (UTC)

## the matrix A governing heat transfer

Do the eigenvectors of the matrix have physical meaning? ie. do they yield the direction of highest heat flow?

--24.84.203.193 28 June 2005 05:21 (UTC)

## Laplace operator

Let ${\displaystyle u=u(x_{1},\ldots ,x_{n},t)}$. Would be better to say:

"when we write ${\displaystyle u_{t}=k\Delta u}$, we consider
${\displaystyle \Delta u=u_{x_{1}x_{1}}+\cdots +u_{x_{n}x_{n}}}$
${\displaystyle \Delta u=u_{x_{1}x_{1}}+\cdots +u_{x_{n}x_{n}}+u_{tt}}$"? --nosig
Would my recent addition make it more clear? Oleg Alexandrov 22:24, 20 July 2005 (UTC)

## diffusion

I'll change the following sentence:

"The heat equation also describes other physical processes, such as diffusion."

Because the heat equation governs the diffusion of heat, which is already a diffusive process. --anon

I don't think that's a good idea. People usually don't think of heat as diffusion. Saying that the heat equation governs both heat and diffusion would be best of understanding I think, even if a bit reduntant. Oleg Alexandrov (talk) 09:34, 25 October 2005 (UTC)

## Heat and Schrodinger equations

This article has now become very confusing with its discussion of the Schrodinger eqn. Mentioning imaginary time later on in the article might have been OK.--CSTAR 12:24, 25 October 2005 (UTC)

Agree with C*. Also, there is some good new material in this article, but it needs better integration and more work. Oleg Alexandrov (talk) 12:56, 25 October 2005 (UTC)
I agree that the reference to the Schrödinger equation should be removed. As I understand it the theory for the Schrödinger equation is very different from the theory for the geat equation.

## Fundamental solution for the heat operator

I think the formula for the solution to the heat operator on Rn should come after the Fourier series solution. And moreover, its relation to the Fourier method should be clarified. --CSTAR 02:12, 26 October 2005 (UTC)

Agree with that. This because the Fourier series thing is simpler, so it should come before. Oleg Alexandrov (talk) 03:21, 26 October 2005 (UTC)

## heat equation and special relativity

Is there an invariant form of the heat equation? --MarSch 10:10, 26 October 2005 (UTC)

Why is the title Heat Equation not Diffusion Equation? Heat is just one example of diffusion.

I couldn't agree more on this. If we're speaking physiscs, heat transfers are just one among many diffusive phenomena, which all involve this equation of a generalization of this equation.ThorinMuglindir 17:40, 30 October 2005 (UTC)

## Restructure the article

1. Heat equation in a finite 1-dimensional medium. Derivation from Fourier's law.
2. Solution by Fourier series
3. Extension to Heat equation in three dimensional regions
4. Fundamental solution (as for example discussed in the German WP page)[1]
5. General Heat equation in inhomogeneous anisotropic media. Derivation as in article now.
6. Friedrichs extension of Laplacian and solution by Borel functional calculus
7. Heat eqn on manifolds.

--CSTAR 16:03, 30 October 2005 (UTC)

As a physicist I (of course) disagree with this overly mathematical treatment, that leaves aside other diffusion phenomena. This equation is relevant to all diffusion calculations, of which the heat equation is nothing but a particular case.ThorinMuglindir 17:27, 30 October 2005 (UTC)
Well the solutions are obviously the same. What is it that you object to? Certainly heat flow in anistropic media is physically important. If you want to say that diffusion is a similar mathematical problem, well that's OK, but maybe you should have a different article with a discussion of the physics of diffusion. This is also related to the Wiener process. Also I think the treatment of Schrodinger eqn here is misplaced. There already is a separate article on the Feynman-Kac formula.--CSTAR 18:17, 30 October 2005 (UTC)
The Feynman-Kac formula is apparently about any partial differential equation and stochastic processes, not Schrodinger equations and stochastic processes. Have you seen how vague it is? As it reads all we can say is that this might be related to Green functions of the Scrodinger equation, or not. It doesn't mention the analogy between Schrodinger and diffusion equations either. If you feel you can demonstrate to me that this is related to the Green functions, I might well learn something, but then I'll just note that this formalism needs not be deployed to introduce green functions of the diffusion and Scrodinger equations.ThorinMuglindir 19:30, 30 October 2005 (UTC)
Now my objections to the proposed organisation of the rewrite: If this article is to be completely rewritten, first I object to leaving heat equation as its title. The title should be diffusion equation, because in physics heat transfers are just a subset of diffusion. Second, although the term fundamental solutions might be mentioned, the sections about Green functions should mention Green functions instead, because that's the term physicists and engineers use. Third, still about Green functions, I object to introducing them in the framework of heat transfer: they are better introduced in the framework of particle diffusion, because the intial condition corresponding a Dirac delta function is more easily understood in this framework, as it corresponds to the initial condition of a brownian particle in a known position. This initial condition is at the base of the decomposition of the solution on Green funtions. Can you explain simply what a temperature field corresponding to a Dirac delta function means physically? Can you realize that initial condition in an experiment?ThorinMuglindir 19:30, 30 October 2005 (UTC)
About your proposal of moving contents to a new article: this is too early to speak of this. We all agree that the diffusion equation can be applied to particle diffusion, heat conduction, and many other things. Let's try to find a common ground. My position remains that the article is not that bad as is, though some minor things could be reorganised, and renaming it to diffusion equation would be benefitial, though it would require some careful work.ThorinMuglindir 19:30, 30 October 2005 (UTC)
A word about my proposed title change: imagine a guy who doesn't know too much about physics. He types heat equation in the search engine, and is directly redirected to an article called diffusion equation. He's just learned something, because heat exchange is diffusion, while not all diffusion is heat exchange. In the way it stands now, there's nothing to learn, and someone who types diffusion equation on the search engine may be led to thinking that he's obtained the particular when he was searching for the general.ThorinMuglindir 19:30, 30 October 2005 (UTC)
The Feynman Kac formula is a path integral representation of the fundamental solution of the Schrodinger eqn extended to imaginary time; N.B. this isn't what currently is in the Feynman-Kac formula article, but that article should in my view also be rewritten. Green's functions and fundamental solutions are basically the same thing. Your section on the Schrodinger eqn is is the extension of the fundamental solution for the heat operator to imaginary time.--CSTAR 20:58, 30 October 2005 (UTC)
OK. As I wrote to Oleg on my talk page, I also intended to mention in this page that Green functions are the a simplification of path integrals for the case where there is no source term or other extra term in the equation, such as the equation that the heat equation article is about. You'll note path integrals are a much more complex mathematical object. And, someone who needs info about Green functions won't always be interested in path integrals. There are numeric methods which are based on the Green functions alone.ThorinMuglindir 21:19, 30 October 2005 (UTC)

## Mistake?

It seems that there is a mistake in section 1 and in what follows this section. Why there is k under the square root? There is no k in Eq.6.

You're absolutely right. It should be fixed now. Thanks very much for mentioning this. Cheers, Jitse Niesen (talk) 19:10, 10 February 2006 (UTC)

## k should be the thermal diffusivity

In my opinion k should be the thermal diffusivity not the thermal conductivity

${\displaystyle k={\frac {K}{\rho c_{p}}}}$

where

• ${\displaystyle K}$ is the thermal conductivity
• ${\displaystyle \rho }$ is the material density
• ${\displaystyle c_{p}}$ is the material heat capacity

The preceding unsigned comment was added by Dapanara (talk • contribs) . Daniele

There needs to be some definition of k. Currently the article states "k is a material-specific quantity depending on the thermal conductivity , the density and the heat capacity." Why not give the actual definition rather than this vague and unenlightening statement? —The preceding unsigned comment was added by 69.107.101.47 (talk) 18:12, August 23, 2007 (UTC)

There is a mix of thermal conductivity (in Europe often defined by λ (unit = W/(m.ºK), and the thermal diffusivity, most often noted by α (unit m2/sec. In a lot of the equations the symbol k is used initially defined as the thermal conductivity, where α was menat. It would be a good idea to list the symbols used, and give their eaxct definition.

As an example. At the end of the one dimensional paragraph this statement appears:

The coefficient k/(cpρ) is called thermal diffusivity and is often denoted α.

Then in the fundamental one dimensional solution this equation appears:

In one variable, the Green's function is a solution of the initial value problem

${\displaystyle {\begin{cases}u_{t}(x,t)-ku_{xx}(x,t)=0&-\infty

where δ is the Dirac delta function. The solution to this problem is the fundamental solution

${\displaystyle \Phi (x,t)={\frac {1}{\sqrt {4\pi kt}}}\exp \left(-{\frac {x^{2}}{4kt}}\right).}$

This should be:

${\displaystyle {\begin{cases}u_{t}(x,t)-\alpha u_{xx}(x,t)=0&-\infty

where δ is the Dirac delta function. The solution to this problem is the fundamental solution

${\displaystyle \Phi (x,t)={\frac {1}{\sqrt {4\pi \alpha t}}}\exp \left(-{\frac {x^{2}}{4\alpha t}}\right).}$

Gerben49 (talk) 11:21, 20 December 2011 (UTC)

Yes, note however this issue is about the physical model, and should not enter the more mathematical sections where the PDE is treated. While the coefficient in front of ${\displaystyle \Delta }$ has an important meaning in each physical or probabilistic model, from the mathematical point of view, of course, it is just a positive real number, that can be denoted by whatever letter, and that may be assumed to be 1 with no loss of generality, as explained in the introduction. (Also, the notation ${\displaystyle u(x,t=0)}$ is never used in mathematics, and should be avoided. The meaning of the Dirac ${\displaystyle \delta }$ is a distribution, and needs no special notation). --pma 14:41, 20 December 2011 (UTC)

## Clarification

I think it might be a good idea for someone to explain what situations the heat equation works in. For example, it may just be me, but I didn't understand whether what was being talked about with "propagation" was whether the heat came from a point source, or a source of finite volume is. The equation doesn't make sense to me because it seems like you could have to rooms full of air that were "isotropic" and "homogeneous" and they still could be different temperatures and have different levels rates of change of temperatures. Right?

and also, wouldn't k need units of some sort?

This article starts out overly technical from the beginning. Anyone without a degree in physics or math will get virtually nothing out of this article as they are derailed from the get-go, is that what is desired? One could easily write a simple conceptual paragraph or two for the lay person, another section on the 1D heat equation for say undergrads, and then get into all the gory math detail you wanted later. Mentioning parabolic partial differential equations, causality, the Riemann conjecture, and Ricci flow in the "General-audience description" is pretty hilarious.

## Animated gif

I think the animated gif is cool. However can you make it stop after the first cycle. It tends to be as distracting as some animated gif ads when you are trying to read the text.

I think you're probably right. I'll take it off for now, and see about adding a non-animated thumbnail that links to the animated version. --Wtt 22:38, 1 May 2007 (UTC)
Well you can always press Esc to stop animated gif --Novwik (talk) 19:03, 22 December 2007 (UTC)
The animation illustrates what heat flow is about. It can be stopped, as mentioned above, and besides, it is next to the table of contents. If you actually scroll down to the article text, you won't see it. Oleg Alexandrov (talk) 04:43, 23 December 2007 (UTC)

## Heat capacity

In the derivation of the 1-dimensional equation, the heat capacity at constant pressure ${\displaystyle c_{p}}$ is used. I think that the stated condition of "no work" usually refers to constant volume, so ${\displaystyle c_{v}}$ should be used. Is a different definition of zero work being used? Or should this be changed?

Of course, these quantities are very similar for liquids and solids, while heat flow in gases is generally dominated by convection. But if ${\displaystyle c_{p}}$ and ${\displaystyle c_{v}}$ are to be used interchangeably because the difference is irrelevant, why not just use ${\displaystyle c}$? Woodford (talk) 15:07, 29 August 2008 (UTC)

I agree, the text is misleading as formulated. ${\displaystyle c_{v}}$ should be used for internal heat, i.e. the ${\displaystyle dT/dt}$ term, and ${\displaystyle c_{p}}$ should be used for heat due to material flowing across the surface of the control volume (such as advection in CFD). Since we are not treating advection here, we should only be using ${\displaystyle c_{v}}$. In general, you should NEVER use ${\displaystyle c_{p}}$ unless you need to account for ${\displaystyle PV}$ work due to material crossing the control volume surface. But it could be generalized to just a ${\displaystyle c=c_{p}=c_{v}}$ for the special case of an incompressible substance.

## information propagation at infinite speed

in the "General-audience description" "and they are stuck together end to end, then instantaneously the temperature at the point of connection is 50 and the graph of the temperature is smoothly running from 0 to 100. This is not physically possible, since there would then be information propagation at infinite speed,"

I believe this section could do with being reworded, as "at the point of connection" the distance between them is 0, so mathematically information is not propagating at infinite speed, true, the limit of the rate of change goes to infinity as the distance goes to 0 for a finite temperature difference, but isn't that simply a statement that 2 parts of the same object, 0 distance apart, can't be different temperatures?

the reason we don't observe infinite heat flows is because materials are not continuous. They do however come quite close to it - if you touch a hot object and a cold object the interface comes to equal temperature quite quickly (although the rest of the body of the materials may not for quite some time).

Andy t roo (talk) 04:45, 22 September 2008 (UTC)

I took out that silly diversion. Dicklyon (talk) 05:10, 22 September 2008 (UTC)

## Animated GIF caption

I was checking the matlab code for the animation and it sets the box boundary conditions as T=0. Why? The caption says "hot body is placed in a box of cold water". I would expect that a hot body, let's say initial T=100 placed in a box with initial T=0 would equalize to a T between 0 and 100. But with boundary T=0 this will not happen. With no source and boundary T=0, the steady state solution is T=0 all over the box. Maybe the boundary condition should be something like dT/dx = 0? Italo Tasso (talk) 00:29, 25 November 2008 (UTC)

## Notation with Integral

In the section of "Heat conduction in non-homogeneous anisotropic media" appears this equation: ${\displaystyle q_{t}(V)=\int _{V}Q(t,x)\,dx\quad }$

But this notation is a little bit misleading, it seems like a one-variable integral. But it is a volume integral, isn't it?

I think the followin notation would be better:

${\displaystyle q_{t}(V)=\int _{V}Q(t,x,y,z)\,dV\quad }$

or ${\displaystyle q_{t}(V)=\int _{V}Q(t,x,y,z)\,dxdydz\quad }$.

Not at all misleading: the use of "dx" for integrals wrto n-dimensional Lebesgue measure is customary between all mathematicians. That the variable "x" belongs to the domain of the integrand is in any case clear. The first solution you wrote is a bit strange, and should be avoided, for it mixes the notation for the domain of integration V and the notation for the measure and the variables. The last you wrote, with dxdydz, should be better used together with a triple iterated integral, which is correct but would make the notation unnecessarily complicated and unnecessarily coordinate-dependent (the use of the 3 dimensional Lebesgue measure also allows the simpler notation you wrote, which is also the more correct. --pma 17:16, 21 April 2010 (UTC)
That may be true with respect to mathematicians, but because this article is neither pure math, nor pure physics, but applied math/theoretical physics in its relevance, it would be useful to include a more direct terminology. there are many alternatives to "dx" with the context being the guide... I think the most obvious one is: "dnx" which is used all the time in physics and mathematical physics texts. so that you would have "d3x" for R3. There are other alternatives such as dσ, dτ, dζ, and dξ which have various traditional meanings (not all of them relavent to this specific discussion of dimensionality)68.6.76.31 (talk) 03:15, 21 June 2011 (UTC)68.6.76.31 (talk) 03:17, 21 June 2011 (UTC)

## Proposal to split solution and some tables

The article is too long and technical in my opinion. I suggest to split an article on solution to the heat equation which would include methods using Fourier series, Fourier transform, fundamental solutions. (Igny (talk) 23:25, 28 December 2008 (UTC))

Sounds like a good idea. Go for it. siℓℓy rabbit (talk) 23:31, 28 December 2008 (UTC)
How about the following names Solution to heat equation (Fourier methods) and Fundamental solutions to heat equation?(Igny (talk) 19:33, 29 December 2008 (UTC))
Both suggestions seem eminently reasonable. siℓℓy rabbit (talk) 20:11, 29 December 2008 (UTC)

## discrete finite-element analysis/simulation

Seems like this article should at least mention/linkto discrete finite-element analysis/simulation. -96.237.10.106 (talk) 14:36, 25 May 2009 (UTC)

## Water Capacity Calculation

Hello All,

here i met one question about the water capability calculation. For exzample, one heat exchanger, the water flow rate is 100GPM. Water entering temperature is 50F and the leaving temperature is 40F. Water entering pressure is 50 PSIA. The differential pressure though the heat exchanger is 10PSID.

How to calculate the water capability?

Which method we should use? Q=CpM(DeltaT) Cp-Water specific heat M-Water mass flow DeltaT-Entering T -Leaving T Note: Do we need to calculate the water facor led to temperature rising?

Q=M(Hi-Ho) M-Water mass flow Hi-Entering water enthalpy Ho-Leaving water enthalpy

Do not know which method we should use. Please help. thanks, Yours Lkvyang (talk) 07:48, 15 July 2009 (UTC)

## Possible error in: Problem on (0,∞) with homogeneous initial conditions and non-homogeneous Dirichlet boundary conditions

I'm not sure the solution given can be correct. If I substitute x=0, I get u = 0, not h(t). For the case h(t) = 1, I find u(x,t) = erfc(x/sqrt(4t)) but do not get this answer by using the general form on the page. Does someone have a reference for this? Thank you-- 134.34.19.174 (talk) 18:39, 22 November 2009 (UTC)

Now I'm lazy to go and look for a reference, but I made the computation by myself. Let ${\displaystyle K}$ be the fundamental solution of ${\displaystyle \partial _{t}-k\partial _{xx}}$. I found as a solution with homogeneous IC and ${\displaystyle h(t)}$ as BC, a convolution of ${\displaystyle h(t)}$ with the distribution ${\displaystyle -2k\,\partial _{x}K}$ :
${\displaystyle u:=h*(-2k\,\partial _{x}K).}$
Now, the point is that for x>0 the exponential makes it a smooth convolution of functions; so we can write, for all x>0
${\displaystyle u(x,t)=\int _{0}^{t}{\frac {x}{\sqrt {4\pi k(t-s)^{\mathbf {3} }}}}\exp \left(-{\frac {x^{2}}{4k(t-s)}}\right)h(s)\,ds.}$
So by the way it seems that the formula in the article is indeed incorrect, because the exponent 3 is lacking. And this gives an explication to your question: we are not allowed to substitute x=0 in the integral to represent the convolution of h with the distribution ${\displaystyle -2k\,\partial _{x}K(t,0)}$ (notice that as a function, now it has a non-integrable singularity). Is true that the integral for u(t,x) converges to h(t) as x → 0, however. I'm pretty sure about my computation, but maybe one could make a double check in some classic text on the subject. --pma (talk) 22:57, 23 November 2009 (UTC)
Looking at the edit history, it seemed that this formula had quite a lively past. I have fixed the error in the article and added a few lines of explanation. --pma (talk) 20:25, 5 December 2009 (UTC)

## Need of standard notation

This article presents annoying incongruence in notations indeed. The Laplacian should be denoted Δu everywhere, avoiding double notations like ${\displaystyle \nabla ^{2}}$. Moreover, the awful and unclear notation "Δx" to denote an increment of the variable x should be banned once for all - it also fights with the notation for the Laplacian.

## Section "Physical problem and equation"

The section Physical problem and equation seems to have been edited by too many hands. This needs to be reorganized somehow. In particular, I think the basic discussion of the equation in the Three-dimensional problem section should be moved upwards in the article (minus irrelevant commentary on Black-Scholes and special relativity). The "General description" section could easily house this content. At present, it doesn't really discuss the equation itself, rather some "interesting properties" (like the maximum principle and smoothness properties). As an aside, shouldn't content specifically about the maximum principle be moved to the appropriate subsection? This doesn't help anyone understand the heat equation itself. Sławomir Biały (talk) 12:20, 1 October 2010 (UTC)

## Animated GIF is confusing

I find this most confusing. For a start we have no idea what each axis is. We have no idea what the shape of the hot body is. We have no idea what the colours mean. Finally if as I suspect the z-axis is the temperature, why does this not flatten out as thermal equilibrium is reached? Drkirkby (talk) 09:03, 10 February 2011 (UTC)

I agree; without labels, it is confusing. __ Just plain Bill (talk) 17:44, 10 February 2011 (UTC)

## λ → -λ

The section about solving the equation says: "Since the right hand side depends only on x and the left hand side only on t, both sides are equal to some constant value λ." Shouldn't it be -λ, as the rows below read "T'(t) = -λαT(t)" and "X''(x) = -λX(x)"? ✎ 16:33, 24 May 2011 (UTC)

## Schrodinger Equation Connection: Wick Rotation

whoever wrote the section did not take enough advanced physics to be told about the Wick Rotation. That is the connection between imaginary time and inverse temperature, which is what gives us a "sort of" relationship between classical diffusion and time evolution in quantum mechanics. See the Wick Rotation article.68.6.76.31 (talk) 03:06, 21 June 2011 (UTC)

## fundamental solution:need for 4*Pi*k*t vs. just t?

When the following script:

[snip]

Indicating that t works just as well as 4*Pi*k*t as the argument to the pow function in divisor of u. Is there any reason for using the more complicated expression?

Answering myself, the reason is that the differential operator is linear and derivative_of(c*f,x) = c*derivative_of(f,x) hence for any solution, f, of the linear homogeneous differential equation, c*f is also a solution of same equation.

Sorry for noise :( If I knew how to delete this post, I would.

Cppljevans (talk) 20:59, 21 September 2011 (UTC)

## Name

In my undergraduate (Engineering) studies, this equation was referred to as 'the Fourier equation'. Should we not mention this name? --Moemin05 (talk) 18:51, 28 May 2012 (UTC)

You are the first person I have ever heard that refers to it in that way. I think the Heat Equation is by far the most conventional name. Hamsterlopithecus (talk) 05:34, 29 December 2012 (UTC)

## Historical Background

I was looking for the historical background of the heat equation but this article doesn't seem to cover it. Is there a reason for that or should we make a History section? Hamsterlopithecus (talk) 05:41, 29 December 2012 (UTC)

## k wrongly used instead of α?

I think that starting from the paragraph 6 (Fundamental solutions) what is called k should be called α, according to the definitions given in the article until that point. — Preceding unsigned comment added by 141.89.115.240 (talk) 16:46, 24 February 2015 (UTC)

## Special case k=const

I think it's worth mentioning that the derivation assumes constant thermal conductivity, which is a special case. Hennui (talk) 14:24, 29 July 2015 (UTC)

## Vector/scalar inconsistency

In the paragraph where Fourier's law is introduced the 1D special case is presented.

On the left hand side q is written in bold (vector) font. While it is true that a one-dimensional vector is a scalar, I think it will be clearer if either:

• A unit vector in the x-direction is added to the right hand side,

or

• q is written non-bold to clarify that it is scalar,

or

• The situation is otherwise clarified.

130.225.98.190 (talk) 13:59, 26 August 2015 (UTC)

## Agent-Based Model Interpretation

Hello all,

According to some agent-based modeling textbooks and classes (I'll see if I can find some references, until then I'll provide some math), one can roughly interpret the heat equation (for uniform materials) as follows. Let each atom, cell, or agent replace its own temperature with the average temperature around it each time interval. This is approximately correct. It follows from finite differences and euler integration.

In 2 dimensions using cartesian coordinates,
${\displaystyle \nabla ^{2}u=\left({\frac {\partial ^{2}u}{\partial x^{2}}}+{\frac {\partial ^{2}u}{\partial y^{2}}}\right)}$.
Using finite differences, the second partial derivatives are
${\displaystyle {\frac {\partial ^{2}u}{\partial x^{2}}}\approx u(x-h,y,t)+u(x+h,y,t)-2u(x,y,t)}$
${\displaystyle {\frac {\partial ^{2}u}{\partial y^{2}}}\approx u(x,y-h,t)+u(x,y+h,t)-2u(x,y,t)}$

Substituting in we have,
${\displaystyle \nabla ^{2}u=\left[u(x-h,y,t)+u(x+h,y,t)-2u(x,y,t)\right]+\left[u(x,y-h,t)+u(x,y+h,t)-2u(x,y,t)\right]}$
${\displaystyle \nabla ^{2}u=u(x-h,y,t)+u(x+h,y,t)+u(x,y-h,t)+u(x,y+h,t)-4u(x,y,t)}$

Integrating in time, using Euler's method we have:
${\displaystyle u(x,y,t+dt)=u(x,y,t)+dt\nabla ^{2}u}$
${\displaystyle u(x,y,t+dt)=u(x,y,t)+dt\left(u(x-h,y,t)+u(x+h,y,t)+u(x,y-h,t)+u(x,y+h,t)-4u(x,y,t)\right)}$.

If we choose dt appropriately the ${\displaystyle u(x,y,t)}$ term vanishes:
${\displaystyle u(x,y,t+1/4)={\frac {\left(u(x-h,y,t)+u(x+h,y,t)+u(x,y-h,t)+u(x,y+h,t)\right)}{4}}}$

In 3 dimensions, the appropriate ${\displaystyle dt}$ is 1/6.

This provides a very intuitive meaning for the heat equation. Given the appropriate time interval, the heat equation says each atom/cell/agent is just updating its temperature with the average surrounding temperature. I think this agent-based interpretation would be a useful addition to this article. This may even be suitable for the simple-English wiki.

If I or the community finds some references, so this isn't counted as original research, should we include this? Does this "intuitive" understanding as the averaging of nearby temperatures provide anything meaningful? If other users think it would be worthwhile including, where in the article should it go?Mouse7mouse9 06:36, 6 January 2016 (UTC)

## Assessment comment

The comment(s) below were originally left at Talk:Heat equation/Comments, and are posted here for posterity. Following several discussions in past years, these subpages are now deprecated. The comments may be irrelevant or outdated; if so, please feel free to remove this section.

 A fundamental PDE. Close to (or already is) B+. Silly rabbit 00:05, 12 June 2007 (UTC)

Last edited at 00:05, 12 June 2007 (UTC). Substituted at 17:21, 29 April 2016 (UTC)