Heat Equation

Author

John Robin Inston

Published

September 25, 2026

1 Heat Equation

The (homogeneous) heat equation describes the diffusion of heat in a general \(n\)-dimensional continuous medium. The same equation governs any quantity that spreads out from high to low concentration, such as dye in a liquid or a pollutant in air. It is the prototypical parabolic [[partial-differential-equations|PDE]].

Denoting the heat of material at location \(\boldsymbol{x}\) at time \(t\) by \(u(t,\boldsymbol{x})\) the heat equation (diffusion equation) gives the heat diffusion dynamics as \[ u_{t}-k\Delta u=0,\qquad k>0, \] where \(\Delta\) denotes the Laplace Operator and \(k\) is the diffusivity. The non-homogeneous heat equation is \(u_{t}-k\Delta u=f\), where \(f\) represents a heat source.

In one dimension this is \(u_t=ku_{xx}\), where \(u(t,x)\) is the temperature at position \(x\) and time \(t\) in a long, thin, perfectly insulated bar. Scaling time by \(k\) allows us to take \(k=1\) without loss of generality.

1.1 Intuition

The Laplacian measures how much \(u\) at a point differs from its average over a small neighbourhood. In one dimension \[ u_{xx}(x)=\lim_{h\to0}\frac{u(x+h)+u(x-h)-2u(x)}{h^2} =\lim_{h\to0}\frac{2}{h^2}\left(\frac{u(x+h)+u(x-h)}{2}-u(x)\right), \] so \(u_t=ku_{xx}\) says that each point moves towards the average of its neighbours, at a rate proportional to how far it is from that average.

Where \(u\) is a local peak (\(u_{xx}<0\)) the temperature falls.
Where \(u\) is a local trough (\(u_{xx}>0\)) the temperature rises.
Where \(u\) is linear (\(u_{xx}=0\)) nothing changes, as heat flowing in from the hot side exactly matches the heat flowing out to the cold side.

The result is that bumps flatten out and sharp features are smoothed immediately, which is why the equation is irreversible. Probabilistically, \(u\) is the density of a cloud of particles each performing independent random walks (Brownian Motion), and the equation describes how the cloud spreads.

2 Derivation

To derive the heat (diffusion) equation consider the diffusion of dye through a liquid in a thin pipe, where the dye moves from high concentration to low concentration. Let \(u(t,x)\) be the concentration (mass per unit length) of the dye at position \(x\) in the pipe at time \(t\). The total mass of dye in the pipe from \(x_{0}\) to \(x_{1}\) at time \(t\) is given by \[ M(t)=\int_{x_{0}}^{x_{1}}{u(x,t)}~d{x} \implies \frac{\partial M}{\partial t}=\int_{x_{0}}^{x_{1}}{u_{t}(x,t)}~d{x}. \]

Conservation of mass. Dye is neither created nor destroyed, so the mass changes only through the flux \(J(t,x)\) (mass per unit time, positive to the right) through the two ends: \[ \frac{\partial M}{\partial t}=J(t,x_0)-J(t,x_1)=-\int_{x_0}^{x_1}J_x(t,x)~dx. \] Comparing the two expressions for \(\partial M/\partial t\), which hold for all \(x_0<x_1\), gives the continuity equation \[ u_t+J_x=0. \]

Fick’s (Fourier’s) law. Material flows from high to low concentration at a rate proportional to the steepness of the gradient, \[ J=-ku_x,\qquad k>0. \] For heat this is Fourier’s law, with \(u\) the temperature, \(J\) the heat flux, and \(k=\kappa/(\rho c)\) built from the conductivity \(\kappa\), the density \(\rho\) and the specific heat \(c\).

Substituting gives \(u_t=ku_{xx}\). In \(n\) dimensions the same argument on a region \(\Omega\) uses the divergence theorem: \[ \frac{d}{dt}\int_\Omega u~dx=-\int_{\partial\Omega}\vec J\cdot\vec n~dS =-\int_\Omega\nabla\cdot\vec J~dx, \] so \(u_t+\nabla\cdot\vec J=0\), and with \(\vec J=-k\nabla u\) (see Gradient) this becomes \(u_t=k\nabla\cdot\nabla u=k\Delta u\). If the medium is also moving with velocity \(\vec V\), then \(\vec J=u\vec V-k\nabla u\) and one gets the transport-diffusion equation \(u_t+\vec V\cdot\nabla u=k\Delta u\) (see Transport Equation).

3 Initial and Boundary Conditions

The heat equation is first order in time, so one initial condition \(u(0,x)=u_0(x)\) is required. On a bar \(0<x<L\) we also need one condition at each end. The standard ones are:

Dirichlet: the temperature is prescribed, \(u(t,0)=a\). (End held in a bath.)
Neumann: the flux is prescribed, \(u_x(t,0)=g\). With \(g=0\) this is a perfectly insulated end.
Robin: \(u_x+\alpha u=g\), where the end loses heat to the surroundings at a rate proportional to the temperature difference.

A laterally insulated rod of length \(L\) has initial temperature \(50\). At \(t=0\) the end \(x=0\) is placed in icy water (\(0\)) and the end \(x=L\) in boiling water (\(100\)). The problem is \[ \begin{cases} u_t=ku_{xx}, & 0<x<L,\ t>0,\\ u(0,x)=50,\\ u(t,0)=0,\quad u(t,L)=100. \end{cases} \]

4 Properties

Let \(u\) solve \(u_t=ku_{xx}\) on \([0,L]\). Then:

Linearity: if \(u_{1}\) and \(u_{2}\) are solutions, then so is \(c_{1}u_{1}+c_{2}u_{2}\) for any constants \(c_{1}\) and \(c_{2}\).
Maximum principle: the maximum and minimum of \(u\) over \([0,T]\times[0,L]\) are attained at \(t=0\) or at \(x=0\) or \(x=L\).
Uniqueness: for given initial and Dirichlet data there is at most one solution.
Smoothing and infinite speed: for \(t>0\) the solution is smooth (indeed analytic in \(x\)), and a disturbance at one point is felt instantaneously everywhere.
Conservation (insulated ends): if \(u_x(t,0)=u_x(t,L)=0\) then \(\int_0^Lu~dx\) is constant in time.

Proof. (3) If \(u_1,u_2\) are two solutions then \(w=u_1-u_2\) solves the problem with zero data. Let \(E(t)=\int_0^Lw^2~dx\ge0\). Then, integrating by parts and using \(w=0\) at \(x=0,L\), \[ E'(t)=2\int_0^Lww_t~dx=2k\int_0^Lww_{xx}~dx=-2k\int_0^Lw_x^2~dx\le0. \] Since \(E(0)=0\) and \(E\ge0\) is non-increasing, \(E\equiv0\), so \(w\equiv0\).

(5) \(\frac{d}{dt}\int_0^Lu~dx=k\int_0^Lu_{xx}~dx=k[u_x]_0^L=0\).

Note: The maximum principle matches intuition: heat cannot pile up in the interior of the bar, because every interior point is moving towards the average of its neighbours.

5 Solutions

5.1 Steady State

A solution with \(u_t=0\) satisfies \(u_{xx}=0\), so is linear in \(x\). For the rod problem above the steady state is \(u_\infty(x)=100x/L\), which is where the temperature settles as \(t\to\infty\). Writing \(u=u_\infty+v\) then reduces to a problem for \(v\) with zero boundary data.

5.2 Separation of Variables

For \(u_t=ku_{xx}\) on \((0,L)\) with \(u(t,0)=u(t,L)=0\) look for solutions of the form \(u=T(t)X(x)\). Substitution gives \[ \frac{T'}{kT}=\frac{X''}{X}=-\lambda, \] since the left side depends only on \(t\) and the right only on \(x\), so both are constant. The boundary conditions give \(X''+\lambda X=0\), \(X(0)=X(L)=0\) which has non-trivial solutions only for \[ \lambda_n=\left(\frac{n\pi}{L}\right)^2,\qquad X_n(x)=\sin\frac{n\pi x}{L},\qquad n=1,2,\dots \] and \(T_n(t)=e^{-k\lambda_nt}\). By linearity \[ u(t,x)=\sum_{n=1}^\infty b_n\,e^{-k(n\pi/L)^2t}\sin\frac{n\pi x}{L},\qquad b_n=\frac{2}{L}\int_0^Lu_0(x)\sin\frac{n\pi x}{L}~dx, \] where the \(b_n\) are the Fourier sine coefficients of the initial data.

If \(u_0(x)=\sin(\pi x/L)\) then only \(b_1=1\) is non-zero and \[ u(t,x)=e^{-k\pi^2t/L^2}\sin\frac{\pi x}{L}. \] The profile keeps its shape and simply decays, with time constant \(L^2/(k\pi^2)\). Higher modes \(\sin(n\pi x/L)\) decay \(n^2\) times faster, which explains the smoothing: rough features are made of high modes which vanish first.

5.3 The Whole Line

The heat kernel (fundamental solution) on \(\mathbb{R}\) is \[ \Phi(t,x)=\frac{1}{\sqrt{4\pi kt}}\,e^{-x^2/4kt},\qquad t>0. \]

For bounded continuous \(u_0\) the problem \(u_t=ku_{xx}\), \(u(0,x)=u_0(x)\) on \(\mathbb{R}\) has the solution \[ u(t,x)=\int_{-\infty}^\infty\Phi(t,x-y)\,u_0(y)~dy=(\Phi(t,\cdot)*u_0)(x). \]

Proof. \(\Phi\) solves the equation. Differentiating, \[ \Phi_t=\Phi\left(-\frac{1}{2t}+\frac{x^2}{4kt^2}\right),\qquad k\Phi_{xx}=k\Phi\left(-\frac{1}{2kt}+\frac{x^2}{4k^2t^2}\right) =\Phi\left(-\frac{1}{2t}+\frac{x^2}{4kt^2}\right), \] so \(\Phi_t=k\Phi_{xx}\), and the same holds for the translates \(\Phi(t,x-y)\) and hence, differentiating under the integral, for \(u\).

\(\Phi\) has unit mass. By the Gaussian Integral, \(\int_{-\infty}^\infty e^{-(x-y)^2/4kt}~dy=\sqrt{4\pi kt}\), so \(\int\Phi~dy=1\). The kernel concentrates at \(x\) as \(t\to0\) (its width is \(\sim\sqrt{kt}\) and its height grows) so it approximates a delta function, and \(u(t,x)\to u_0(x)\).

Note: \(\Phi(t,\cdot)\) is the density of the Normal Distribution \(\mathcal{N}(0,2kt)\). The solution is the average of the initial data against a Gaussian whose width \(\sqrt{2kt}\) grows like the square root of time. This is the diffusive scaling of Brownian Motion. The tails are positive for every \(t>0\), which is the infinite propagation speed.

Let \(u_0=\delta_0\) be an initial unit of heat concentrated at the origin. Then \(u(t,x)=\Phi(t,x)\): a Gaussian whose peak \(1/\sqrt{4\pi kt}\) falls like \(t^{-1/2}\) while its width grows like \(t^{1/2}\), so the total heat stays equal to \(1\).