The problem with no walls
On a finite rod, separation of variables worked because the two ends quantized the allowed shapes into a discrete list of sine modes. Now take those walls away. We want the temperature on an infinitely long rod: the Cauchy problem for the heat equation, u_t = k u_xx for all x and all t > 0, with a single piece of data — the initial profile u(x,0) = f(x). There are no boundary conditions to impose, only the mild request that u stays bounded and decays far away. With no walls to pick out special frequencies, every frequency is now allowed, and the discrete sum over modes must become a continuous integral.
That is precisely the job of the tool from the previous guide. The Fourier transform rebuilds any reasonable f(x) as a continuous superposition of pure waves e^(i xi x), one for every wavenumber xi, weighted by the transform F(xi) = (integral of f(x) e^(-i xi x) dx). Where a finite rod gave a Fourier series, the infinite line gives a Fourier integral — same idea, continuum of modes. The strategy writes itself: transform the whole problem in x, solve whatever is left, and transform back.
Transform in x and the PDE becomes an ODE
The magic move is the differentiation rule: under the Fourier transform in x, taking a space derivative becomes multiplying by i xi. So u_x turns into i xi times the transform, and u_xx — two derivatives — turns into (i xi)^2 = -xi^2 times it. Let U(xi,t) denote the transform of u(x,t) taken in x at each fixed time t. The time derivative just rides along, since transforming in x and differentiating in t commute. Apply the transform to both sides of u_t = k u_xx and the second derivative in space dissolves into a plain multiplication.
u_t = k u_xx (PDE in x, t)
transform in x: u_xx -> -xi^2 U
U_t = k (-xi^2) U = -k xi^2 U (one ODE in t, per xi)
solve the ODE (xi is just a constant here):
U(xi, t) = U(xi, 0) e^(-k xi^2 t)
= F(xi) e^(-k xi^2 t)Look at what we are left with: U_t = -k xi^2 U. For each fixed xi this is the simplest ODE in the world — a single first-order linear equation, exactly the decay law you met two rungs down. Its solution is a pure exponential, U(xi,t) = F(xi) e^(-k xi^2 t), where F(xi) = U(xi,0) is the transform of the initial data f. The infinite rod, far from being harder than the finite one, is in some ways cleaner: there is no eigenvalue hunt, just one explicit formula valid for every frequency at once.
Transforming back gives a Gaussian
We have the answer in frequency space; now we carry it home with the inverse Fourier transform. The product U(xi,t) = F(xi) · e^(-k xi^2 t) is a transform multiplied by another transform. Multiplication on the frequency side is the signature of convolution on the physical side — that is the convolution theorem, the second of the transform's two great rules. So u(x,t) is the convolution of f with whatever function has e^(-k xi^2 t) as its Fourier transform. We just need to recognize that function.
Here a beautiful fact rescues us: the Fourier transform of a Gaussian is again a Gaussian. The bell curve e^(-k xi^2 t) in frequency is the transform of a bell curve in x — a Gaussian whose width grows with time. Doing the bookkeeping carefully, the partner function is G(x,t) = (1 / sqrt(4 pi k t)) · e^(-x^2 / (4 k t)), a spreading bell of unit area. This G is the heat kernel, the single most important object in the theory of diffusion, and the entire solution is its convolution with the initial data.
The resulting convolution formula, u(x,t) = (integral over y of G(x - y, t) f(y) dy), has a wonderfully physical reading. Think of the initial profile f as a crowd of tiny heat spikes, one at each point y, each of strength f(y). The kernel G(x - y, t) is how a single spike at y has spread out by time t: a Gaussian bump centred at y, getting wider and flatter as t grows. The full temperature at (x,t) is just the sum of all those spreading bumps, each weighted by how hot its source was. Diffusion is nothing but every point smearing its initial heat outward into an ever-wider bell, and the bells overlapping.
The kernel is a response to a point
What is G by itself? Set the initial data to a single concentrated spike — all the heat piled at the origin, nothing anywhere else. That idealized spike is the Dirac delta, and feeding f = delta into the convolution simply returns u(x,t) = G(x,t). So the heat kernel is the temperature you get from a unit of heat dumped at one point: it is the response to a point source, the heat equation's fundamental solution. Be honest about what kind of object this is — the delta is not a function in the ordinary sense but a distribution, an idealized point of infinite density and unit total, and the kernel is the smooth response that this idealization triggers.
This is why convolution against G solves everything: by linearity, an arbitrary f is a superposition of point spikes, so the response to f is the same superposition of point responses G. The kernel also has a clean shape of its own. At each time it is an exact Gaussian whose width is proportional to sqrt(k t) — so its spread grows like the square root of time, the unmistakable signature of self-similar diffusion. To go four times as far, a diffusing front needs sixteen times as long; this sqrt(t) scaling, not a steady speed, is the honest law of how heat creeps outward.
What the Gaussian confesses about diffusion
Now read three deep facts straight off the formula. First, infinite speed. The Gaussian G(x,t) is strictly positive for every x the instant t > 0, however small. So a spike at the origin warms a point a mile away immediately — at t = 0.0001 the temperature there is fantastically tiny but not zero. The heat equation has infinite propagation speed: information leaks everywhere at once. That is physically a fiction (real heat respects light speed), and it is exactly the price diffusion pays for being a smoothing parabolic equation rather than a finite-speed hyperbolic one like the wave equation.
Second, smoothing. The instant t > 0, u(x,t) is a convolution against an infinitely smooth Gaussian, and convolving with something smooth makes the result smooth — even if f was a jagged step or a spike. So a ragged initial temperature becomes perfectly smooth immediately; the smoothing effect we glimpsed in the e^(-k xi^2 t) factor is now visible as the bell curve doing the blurring. Third, and as a direct consequence, irreversibility. Run the formula backward to negative t and 4kt goes negative, so the Gaussian's exponent flips sign and the integral blows up; the backward heat equation is ill-posed. You cannot un-blur a photograph or un-mix milk into coffee, and the kernel is the reason written in mathematics.
The recipe, and where it reaches its edge
- Transform in x. Apply the Fourier transform to u_t = k u_xx; the differentiation rule sends u_xx to -xi^2 U.
- Solve the ODE. With xi frozen, U_t = -k xi^2 U gives U(xi,t) = F(xi) e^(-k xi^2 t), where F is the transform of f.
- Recognize the partner. The factor e^(-k xi^2 t) is the transform of the Gaussian heat kernel G(x,t) = 1/sqrt(4 pi k t) e^(-x^2/(4 k t)).
- Invert by convolution. A product of transforms is a convolution, so u(x,t) = integral of G(x - y, t) f(y) dy — the answer in closed form.
Be honest about the fine print. The Fourier transform method needs the whole line (or a periodic domain) — a genuine boundary, like the edge of a finite rod, breaks the clean rule because e^(i xi x) does not respect that wall. For a half-line you switch to the sine or cosine transform, the very subject of the next guide. The method also leans on the equation having constant coefficients in x, so the derivative-to-multiplication rule applies cleanly; coefficients that vary in space do not transform to a simple algebraic factor.
Even so, what we have built is no small thing. Two transform rules — derivatives become multiplication, products become convolutions — turned a partial differential equation into a one-line ODE and handed back a closed-form solution for any initial temperature on an infinite rod. The heat kernel that fell out is the same Gaussian you will meet again as a fundamental solution, as the probability density of Brownian motion, and as a Green's function a couple of rungs up. To add a heat source along the way rather than just an initial condition, you stack these point responses through time with Duhamel's principle — but that is the same kernel doing the work, integrated once more.