The Heat Equation and Separation of Variables, differential equations lesson

A metal rod starts with an uneven temperature profile while both of its ends are held at a fixed temperature. Which features of the profile fade first, and what is left after a long time? Here $u(x, t)$ is the temperature at position $x$ and time $t$ , measured from a chosen reference so that negative values are allowed, and subscripts denote partial derivatives, so $u_t = \partial u/\partial t$ and $u_{xx} = \partial^2 u/\partial x^2$ . This lesson writes the solution as a sum of spatial modes that each decay at their own rate, and it shows how insulated ends and nonzero end temperatures change the modes and the final state.

Key terms

Heat equation

Definition The one-dimensional heat equation for a temperature $u(x, t)$ on an interval $0 \le x \le L$ is

u_t = \kappa\,u_{xx}, \qquad \kappa > 0,

where the constant $\kappa$ is the thermal diffusivity, with units of length squared per time. For a uniform rod, Fourier’s law gives the heat flux $\phi = -k\,u_x$ with the thermal conductivity $k$ , and the heat balance of a short piece of the rod gives $\rho c_p\,u_t = -\phi_x$ , where $\rho$ is the density and $c_p$ the specific heat. Together these give the heat equation with

\kappa = \frac{k}{\rho\,c_p}.

A well-posed problem on the interval adds the initial temperature $u(x, 0) = f(x)$ and one boundary condition at each end, and it is solved forward in time, for $t \ge 0$ .

In plain words The temperature rises where the profile curves upward and falls where it curves downward. Peaks cool, valleys warm, and sharp features smooth out. A larger diffusivity means faster smoothing.

Example Take $\kappa = 2$ on $0 \le x \le \pi$ with both ends held at $0$ , and the initial profile $\sin x$ , which has its peak at $\pi/2$ . Its second derivative is $-\sin x < 0$ at every interior point, so the whole interior starts to cool. The function $u = e^{-2t}\sin x$ shows this exactly, since $u_t = -2e^{-2t}\sin x$ and $2u_{xx} = -2e^{-2t}\sin x$ . The opposite profile $-\sin x$ is a valley, and it warms toward $0$ at the same rate.

Separation of variables for PDEs

Definition Separation of variables looks for solutions of a linear homogeneous partial differential equation with linear homogeneous boundary conditions in the product form $u(x, t) = X(x)\,T(t)$ , a function of position alone times a function of time alone. Write $T' = dT/dt$ and $X'' = d^2X/dx^2$ . For the heat equation, substitution gives $X\,T' = \kappa\,X''\,T$ , and dividing by $\kappa XT$ where it is nonzero gives

\frac{T'(t)}{\kappa\,T(t)} = \frac{X''(x)}{X(x)} = -\lambda.

The left side depends only on $t$ and the middle only on $x$ , so both equal one constant, written $-\lambda$ . The spatial factor solves the eigenvalue problem $X'' + \lambda X = 0$ with the boundary conditions, and the time factor solves $T' = -\kappa\lambda T$ , so $T = e^{-\kappa\lambda t}$ up to a constant factor. Since the equation and the boundary conditions are linear and homogeneous, finite sums of these product solutions are again solutions. The initial condition is met by an infinite sum whose coefficients are the coefficients of $f$ in the orthogonal eigenfunctions. For a piecewise continuous $f$ those coefficients are bounded, so for every $t > 0$ the factors $e^{-\kappa\lambda_n t}$ make the series converge together with all its termwise derivatives, and the sum solves the equation and the boundary conditions for $t > 0$ .

In plain words Each building block keeps a fixed spatial shape and only changes its size in time. The boundary conditions decide which shapes are allowed, each allowed shape decays at its own rate, and the initial profile decides how much of each shape to use.

Example For $\kappa = 2$ on $0 \le x \le \pi$ with $u(0, t) = u(\pi, t) = 0$ , the problem $X'' + \lambda X = 0$ with $X(0) = X(\pi) = 0$ has the eigenvalues $\lambda_n = n^2$ and the eigenfunctions $\sin nx$ for $n = 1, 2, 3, \dots$ . The initial profile $f(x) = \sin x + 2\sin 2x$ is already a sum of two of them, so orthogonality gives the coefficient $1$ for $n = 1$ , the coefficient $2$ for $n = 2$ , and $0$ for every other $n$ . Hence

\begin{aligned} u &= e^{-2t}\sin x \\[0.4em] &\quad + 2e^{-8t}\sin 2x. \end{aligned}

To check, $u_t = -2e^{-2t}\sin x - 16e^{-8t}\sin 2x$ and $u_{xx} = -e^{-2t}\sin x - 8e^{-8t}\sin 2x$ , so $u_t = 2u_{xx}$ . At $t = 0$ the profile is $f$ , and both sines vanish at $0$ and at $\pi$ , so the end conditions hold too.

Dirichlet and Neumann modes

Definition On $0 \le x \le L$ , the Dirichlet conditions $u(0, t) = u(L, t) = 0$ hold both end temperatures at zero. The separated problem then has the eigenvalues $\lambda_n = (n\pi/L)^2$ and the Dirichlet modes $X_n = \sin(n\pi x/L)$ for $n = 1, 2, 3, \dots$ . The Neumann conditions $u_x(0, t) = u_x(L, t) = 0$ describe insulated ends, through which no heat flows. They give the same eigenvalues $\lambda_n = (n\pi/L)^2$ with the Neumann modes $X_n = \cos(n\pi x/L)$ for $n = 0, 1, 2, \dots$ , including the zero mode $\lambda_0 = 0$ with the constant $X_0 = 1$ . In both cases

u = \sum_n c_n\,e^{-\kappa\lambda_n t}X_n(x),

with the half-range coefficients of $f$ ,

\begin{gathered} b_n = \tfrac{2}{L}\int_0^L f(x)\sin\frac{n\pi x}{L}\,dx \\[0.4em] a_n = \tfrac{2}{L}\int_0^L f(x)\cos\frac{n\pi x}{L}\,dx \end{gathered}

Dirichlet problems use $c_n = b_n$ . Neumann problems use $c_n = a_n$ for $n \ge 1$ and $c_0 = a_0/2$ , which is the mean value of $f$ .

In plain words Zero end temperatures call for sines, which vanish at the ends. Insulated ends call for cosines, which are flat at the ends, and the cosines include the constant, which never decays because heat cannot leave.

Example Take $\kappa = 2$ on $0 \le x \le \pi$ with insulated ends and $f(x) = 3 + \cos x$ . The cosine coefficients are $a_0 = 6$ , so $c_0 = 3$ , and $a_1 = 1$ , with all others zero, so

u = 3 + e^{-2t}\cos x.

Its slope $u_x = -e^{-2t}\sin x$ vanishes at $0$ and at $\pi$ , and $u_t = -2e^{-2t}\cos x = 2u_{xx}$ . The cosine fades and the constant $3$ remains. Leaving out the zero mode would predict decay to $0$ , which is wrong for this problem.

Conserved mean temperature

Definition Let $u$ solve $u_t = \kappa\,u_{xx}$ on $0 \le x \le L$ for $t > 0$ , with $u$ , $u_x$ , $u_t$ and $u_{xx}$ continuous up to the ends. Integrating the equation over the interval gives

\frac{d}{dt}\int_0^L u\,dx = \kappa\,\Big[\,u_x\,\Big]_{x=0}^{x=L}.

With insulated ends both values of $u_x$ are zero, so the total $\int_0^L u\,dx$ is constant, and so is the mean temperature

\bar{u} = \frac{1}{L}\int_0^L u\,dx = \frac{a_0}{2}.

For a piecewise continuous $f$ , every Neumann mode with $n \ge 1$ decays, and $u$ approaches the constant $\bar{u}$ uniformly as $t \to \infty$ . With Dirichlet ends the boundary fluxes are in general nonzero, and the total is not conserved.

In plain words Insulation means that no heat crosses the ends, so the heat already in the rod has nowhere to go. It spreads out evenly, and the final temperature is the average of the initial one.

Example For $u = 3 + e^{-2t}\cos x$ on $0 \le x \le \pi$ , the cosine integrates to zero, so $\int_0^\pi u\,dx = 3\pi$ at every time, and the mean is $3$ , which is also the limit. For the Dirichlet solution $e^{-2t}\sin x + 2e^{-8t}\sin 2x$ the total is

\int_0^\pi u\,dx = 2e^{-2t},

which decays because heat leaves through the ends. Its derivative $-4e^{-2t}$ matches the flux formula, since $2\big[u_x\big]_0^\pi = -4e^{-2t}$ .

Boundary lifting

Definition For constant end temperatures $u(0, t) = A$ and $u(L, t) = B$ , the lift

\ell(x) = A + (B - A)\frac{x}{L}

is the steady solution with these end values, since $\ell'' = 0$ . The difference $v = u - \ell$ satisfies $v_t = \kappa\,v_{xx}$ with zero end values and $v(x, 0) = f(x) - \ell(x)$ . That is a Dirichlet problem, solved by the sine series, and $u = \ell + v$ tends to $\ell$ as $t \to \infty$ . If the end temperatures $A(t)$ and $B(t)$ vary in time, the same formula gives a lift $\ell(x, t)$ , and

v_t = \kappa\,v_{xx} + \kappa\,\ell_{xx} - \ell_t,

where $\ell_{xx} = 0$ for this linear lift, so the changing ends act as a source $-\ell_t$ in the equation for $v$ . For $u$ to be continuous at the corners $(0, 0)$ and $(L, 0)$ , the data must agree there, $f(0) = A(0)$ and $f(L) = B(0)$ .

In plain words Subtract a simple profile that already has the right end temperatures. What remains has zero end temperatures, which the sine modes handle, and adding the profile back restores the correct end values.

Example On $0 \le x \le \pi$ with $\kappa = 2$ , hold the ends at $u(0, t) = 1$ and $u(\pi, t) = 3$ , and start from $f(x) = 1 + 2x/\pi + \sin x$ . The lift is $\ell = 1 + 2x/\pi$ , so $v(x, 0) = \sin x$ and $v = e^{-2t}\sin x$ . Hence

u = 1 + \frac{2x}{\pi} + e^{-2t}\sin x,

which takes the values $1$ and $3$ at the ends, satisfies $u_t = -2e^{-2t}\sin x = 2u_{xx}$ , and tends to the straight-line profile from $1$ to $3$ . The data agree at the corners, since $f(0) = 1$ and $f(\pi) = 3$ .

Diffusive decay

Definition In the Dirichlet problem on $0 \le x \le L$ with a piecewise continuous $f$ , the mode $n$ decays like $e^{-\kappa(n\pi/L)^2 t}$ , at a rate proportional to $n^2$ . If $b_1 \ne 0$ , then

e^{\kappa\pi^2 t/L^2}\,u(x, t) \to b_1\sin\frac{\pi x}{L}

uniformly as $t \to \infty$ , so the solution comes to look like a decaying first mode with the time scale $L^2/(\kappa\pi^2)$ . For every $t > 0$ the profile is infinitely differentiable in $x$ , even when $f$ has jumps. For a solution with zero Dirichlet or zero Neumann conditions that is smooth up to the ends, the energy $H(t) = \tfrac{1}{2}\int_0^L u^2\,dx$ satisfies

H'(t) = -\kappa\int_0^L u_x^2\,dx \le 0,

because the boundary term $\kappa\big[u\,u_x\big]_0^L$ from integration by parts vanishes.

In plain words Fine wiggles die out much faster than broad ones. After a short time only the smoothest modes are visible, and after a long time only the slowest mode present in the data. A rod twice as long takes four times as long to smooth out.

Example In the solution $u = e^{-2t}\sin x + 2e^{-8t}\sin 2x$ , the second mode starts twice as large as the first but decays four times as fast. At $t = 0.5$ the amplitudes are $e^{-1} \approx 0.368$ and $2e^{-4} \approx 0.037$ , a ratio of $2e^{-3} \approx 0.10$ . The energy is

H(t) = \frac{\pi}{4}\left(e^{-4t} + 4e^{-16t}\right),

and its derivative $-\pi\left(e^{-4t} + 16e^{-16t}\right)$ equals $-2\int_0^\pi u_x^2\,dx$ , as the identity says.

Check yourself

Solve $u_t = u_{xx}$ on $0 \le x \le \pi$ with $u(0, t) = u(\pi, t) = 0$ and $u(x, 0) = \sin^3 x$ , using the identity $\sin^3 x = \tfrac{3}{4}\sin x - \tfrac{1}{4}\sin 3x$ .

Show the answer

The initial profile is already a sum of the Dirichlet modes $\sin x$ and $\sin 3x$ , with the eigenvalues $1$ and $9$ . Each coefficient picks up its own decay factor, so

\begin{aligned} u &= \tfrac{3}{4}e^{-t}\sin x \\[0.4em] &\quad - \tfrac{1}{4}e^{-9t}\sin 3x. \end{aligned}

Both terms satisfy $u_t = u_{xx}$ , both vanish at $0$ and at $\pi$ , and at $t = 0$ the sum is $\sin^3 x$ . By $t = 0.8$ the factor $e^{-9t}$ has fallen to $e^{-7.2} \approx 0.0007$ , while $e^{-t}$ is still about $0.45$ , so the profile already looks like a single sine arch.

Solve $u_t = u_{xx}$ on $0 \le x \le \pi$ with insulated ends and $u(x, 0) = x$ . A classmate expands $x$ in sines and predicts that the rod cools to $0$ . What went wrong, and what temperature does the rod approach?

Show the answer

Insulated ends call for the cosine modes, including the constant one. The sine series belongs to the problem with zero end temperatures, a different problem. The cosine coefficients are $a_0 = \tfrac{2}{\pi}\int_0^\pi x\,dx = \pi$ and, for $n \ge 1$ ,

\begin{aligned} a_n &= \frac{2}{\pi}\int_0^\pi x\cos nx\,dx \\[0.4em] &= \frac{2\big((-1)^n - 1\big)}{\pi n^2}, \end{aligned}

which is $-4/(\pi n^2)$ for odd $n$ and $0$ for even $n$ . So

\begin{aligned} u &= \frac{\pi}{2} \\[0.4em] &\quad - \frac{4}{\pi}\sum_{n \text{ odd}} \frac{e^{-n^2 t}\cos nx}{n^2}. \end{aligned}

Every term of the sum decays, and the rod approaches the mean of $f$ , which is $\pi/2$ , in agreement with the conserved mean temperature. The data have $f' = 1$ at both ends, which does not match the insulated ends at $t = 0$ , but the series still solves the problem for every $t > 0$ .

Take $u_t = u_{xx}$ on $0 \le x \le \pi$ with $u(0, t) = 0$ , $u(\pi, t) = t$ and $u(x, 0) = 0$ . With the lift $\ell = xt/\pi$ , which problem does $v = u - \ell$ solve? What goes wrong if the extra term is dropped, and which profile does $u$ approach for large $t$ ?

Show the answer

The lift has $\ell_t = x/\pi$ and $\ell_{xx} = 0$ , so

v_t = v_{xx} - \frac{x}{\pi},

with $v(0, t) = v(\pi, t) = 0$ and $v(x, 0) = 0$ . The data agree at the corners, since both end temperatures start at $0$ . Dropping the term $-x/\pi$ would give $v = 0$ and $u = xt/\pi$ , but that function has $u_t = x/\pi$ and $u_{xx} = 0$ , so it does not solve the heat equation. A steady solution $w$ of the problem for $v$ needs $w'' = x/\pi$ with $w(0) = w(\pi) = 0$ , which gives

w(x) = \frac{x^3 - \pi^2 x}{6\pi}.

The difference $v - w$ solves the homogeneous Dirichlet problem with initial value $-w$ , so it decays like $e^{-t}$ . For large $t$ , therefore,

u \approx \frac{xt}{\pi} + \frac{x^3 - \pi^2 x}{6\pi}.

This profile satisfies $u_t = x/\pi = u_{xx}$ and the end conditions exactly. The full solution adds $-\sum_n \tfrac{2(-1)^n}{\pi n^3}e^{-n^2 t}\sin nx$ , which cancels the profile at $t = 0$ .

The Heat Equation and Separation of Variables

Key terms

Heat equation

Separation of variables for PDEs

Dirichlet and Neumann modes

Conserved mean temperature

Boundary lifting

Diffusive decay

Check yourself

How this lesson connects

Builds on

Used later in