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Okay so I am studying for my PDE course and I am convering Fourier transforms. In fact I am using fourier transforms to find a solution to the heat equation on an infinite length rod.

After going through the derivation of the fourier transform and applying it to the heat equation we get the following solution: $$u(x,t) = a(x) \ast K(x,t)$$ where K(x,t) is the heat kernel so $$K(x,t) = \frac{1}{\sqrt{4\pi kt}}e^{\frac{-x^2}{4kt}}$$

Okay so given an initial condition $u(x,0) = F(x)$ we can not simply plug it in since when $t=0$ the denominator is undefined. It is at this point where I am getting confused

Our professor told us that to solve the heat equation using the IC (initial condition) we have to study what the convolution is doing. So lets see what happens when we take the limit as $t \rightarrow 0$. $$\lim_{t\rightarrow 0}\ \ a \ast K(x,t) = \lim_{t\rightarrow 0} \int_{-\infty}^{\infty} a(x')K(x-x', t) dt $$

okay so what is $x'$? Where did it come from? and why does our professor do this? I do know for a fact that the integration over the infinite bounds means the kernal has area 1 due to the fact that it is a Guassian function.

After a little bit of notes he ends up with the following $$\lim_{t\rightarrow 0}\int_{-\infty}^\infty a(x')K(x-x',t) dx' = a(x)$$ and then he says that we can now replace the $a(x)$ with $F(x)$ (our initial condition). Okay well the last line makes sense but the limit of the integral dosn't make sense.

But how does this relate to solving the heat equation and ultimately how does this end up as the Dirac-Delta function ?

So if anyone can explain with a little intuition what it happening. Maybe a graph (i have a graph in my notes but I am confused about it). Thanks

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Note that the convolution have to be done w.r.t. the space variable, hence $x^\prime$ is the variable in the convolution integral. In other words, the convolution integral has to be written as $\int_{-\infty}^\infty a(x^\prime) K(x-x^\prime ,t)\ \text{d} x^\prime$ (I'm assuming that the space variable $x$ lies in $\mathbb{R}$) and the limit you want to evaluate actually is: $$\lim_{t\to 0^+} \int_{-\infty}^\infty a(x^\prime)\ K(x-x^\prime ,t)\ \text{d} x^\prime\; .$$ –  Pacciu Dec 9 '11 at 1:45
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2 Answers

As $t \to 0$ the kernel $K(x,t)$ tends to the Dirac delta function $\delta(x)$ (which is not actually a function, but a so-called distribution, or generalized function). You can recognize this at least informally because the area under $K(x,t)$ stays constant, equal to $1$, but as $t \to 0$ it becomes more and more concentrated at the point $x =0$, with a higher and higher peak there.

Formally, what this convergence statement means is that the equation that you wrote down holds, namely that $$a(x) = \lim_{t \to 0} \int_{-\infty}^{\infty} a(x') K(x-x',t)\, dx'.$$

(Note also that in your first convoluton integral, there is a typo, as pointed out by Pacciu; the $dt$ there should be $dx'$.)

Now think about $K(x,t)$ as $t$ tends from $0$ to $\infty$: it flows from being a peak concentrated at $x = 0$ to a more-and-more shallow graph that spreads out over the whole $x$-axis. You should think of this as heat flowing from a point source at the origin and slowly spreading out over the whole axis. (Imagine blasting a point on a long steel beam with a blow torch for an instant, and then think about how the heat will diffuse along the beam.)

Now when you have initial conditions $a(x)$, this describes heat being applied at $t = 0$ not just at the point $x = 0$, but along the whole $x$-axis, according to the density $a(x)$. The convolution $a(x)*K(x,t)$ then describes how this heat has diffused through the line at time $t$; as sos440 writes in their answer, it is the superposition of the diffusion of the heat from each point $x$ that was present at time $t = 0$.

(If we took $a(x)$ to be $\delta(x)$, then we would be back at the situation of all the heat being initially concentrated at the single point $x = 0$; mathematically this corresponds to the formula $\delta(x)*K(x,t) = K(x,t)$ --- i.e. the $\delta$ function is the identity for convolution.)

Added in response to a question in the comments below:

Imagine for a moment that we had a certain amount of heat $A_i$ initially applied at the points $x_i$, for $i = 1, \ldots, n$. When one unit of heat is placed at $x =0$, it diffuses according to $K(x,t)$. So the amount $A_i$ of heat at $x_i$ diffuses according to $A_i K(x-x_i,t)$. (I am just changing the variable in $K(x,t)$ to shifts its centre from $x = 0$ to $x = x_i$, and scaling it by the amount $A_i$.)

So the total heat at a point $x$ and time $t$ will be $\sum_{i = 1}^n A_i K(x-x_i,t)$. (I am just adding up the heat which has arrived at the point $x$ at time $t$ from each of the points $x_1, \ldots, x_n$.)

Now imagine that instead of just having heat concentrated at $n$ point sources at time $t$, we instead have heat distributed throughout the line with density $a(x)$, so that the amount of heat in the (infinitesimally) small interval $[x',x' + dx']$ is $a(x') dx'$. Then the above sum becomes the integral $\int_{-\infty}^{\infty} a(x') K(x-x',t) dx'$, i.e. $a(x) * K(x,t)$.

Hence the amount of heat at a point $x$ at time $t$ is exactly given by $a(x) * K(x,t)$, as your professor explained.

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okay, thankyou. What is $x'$ though in the limit integral? and what is $a(x)$. You say its the density but density of what? Also the last line, the convolution of $\delta(x) \ast K(x,t) = K(x,t)$ is true for what conditions? Or is it always true? Also my initial condition is $F(x)$ not $a(x)$. not that I know what either of them are. –  Tyler Hilton Dec 9 '11 at 3:51
@Tyler: Dear Tyler, The variable $x'$ in the limit integral is a dummy variable; see my edit for a little more explanation. As for $a(x)$, it is the linear density of heat. Regarding the convolution formula for $\delta(x)$, it is essentially a tautology. You shouldn't worry too much about precise conditions unless you are actually studying distributions as a rigorous theory --- in which case the precise conditions will be described in the course of your studies. Regards, –  Matt E Dec 9 '11 at 3:55
Okay, to be honest I am still confused. Maybe I am looking for a more rigourous theory but that confuses me even more. I was wondering if you have any resources that explains that in MORE laymen terms, I'd appreciate that. It would be very helpful. –  Tyler Hilton Dec 9 '11 at 4:00
so to verify some points. The area under the kernel is ALWAYS 1? Why is that? Is that because its a Gaussian function? What connects them together? A second major question. You say my initial condition is $a(x)$. But its not. my initial condition is $F(x)$. But the $a(x)$ is also there in the convolution. How does $a(x) = F(x)$ happen at the end? –  Tyler Hilton Dec 9 '11 at 4:10
@Tyler: Dear Tyler, You wrote yourself that the area under the kernel is always one. (This is conservation of energy: if you have one unit of heat concentrated at time $t = 0$, then as it diffuses over the line, the total heat energy has to remain at one unit.) Of course, you can verify it just by doing the integral. Regarding initial conditions, as you pointed out in your question, you have to take $a(x) = F(x)$ --- when you take the limit as $t \to 0$ of the convolution $a(x) * K(x,t)$, you get $a(x)$. –  Matt E Dec 9 '11 at 4:36
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Heat equation describes the dispersion of heat by inspecting the way how the temperature evolves in time.

By either physical intuition or mathematical analysis on the behavior of the heat equation, we find that law of superposition holds. Then, as we think of point mass or point charge in physics, we can think of something like a point temperature $T$ concentrated at a point (in physical sense, at least) and see how it evolves in time first. For any other cases, now we think of given temperature distribution as an aggregation of infinitesimal point temperatures $dT$ and then just add up all the solutions $du(x, t)$ corresponding to each $dT$. The law of superposition will tell us that the resulting sum $u = \int \; du$ is the solution.

Your concern about the limiting integral is just natural, but you do not have worry about it. If you have right intuition, then your calculation will be justified if you learn suitable theories. For example, you may refer to the theory concerning the Dirac delta measure and the corresponding approximation to the identity. In fact, the theory of elliptic and parabolic PDE is one of the most active and vastest area in the mathematics, so you may find many enlightning explanations to the heat equation.

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