Chapter 5: The Heat Equation and Parabolic Smoothing

Date: Oct 2, 2026

Touch a long cold iron bar with a soldering iron for an instant, at one point. A finite amount of heat has been deposited at a single place. A moment later, what is the temperature along the bar? The initial state is as rough as a state can be: zero everywhere except at one point, where it is in effect infinite. Yet the mode experiment of Section 1.5 predicts that the heat equation destroys roughness, the faster the finer it is. So the temperature a moment later should be a smooth hump. How wide is it, how does the width grow, and does it have an edge?

The answers come from a single explicit function, the heat kernel, which plays for \(u_t=ku_{xx}\) the part that the fundamental solution played for the Poisson equation. With it we solve the initial value problem on the whole line. For a finite rod, where no such formula is at hand until Chapter 7, the qualitative tools of Chapter 3, maxima and energy, reappear in a form that respects the direction of time.

5.1 Scaling and the heat kernel

The equation \(u_t=ku_{xx}\) is unchanged when \(x\) is scaled by \(\lambda\) and \(t\) by \(\lambda^2\) (Exercise 1.13). A point release of heat has no length scale of its own, so its temperature profile at time \(t\) should be a fixed shape, stretched by the only available length \(\sqrt{kt}\) . If the total heat \(\int u\,dx\) is to stay constant while the width grows like \(\sqrt t\) , the height must fall like \(1/\sqrt t\) . This suggests

\[ u(x,t)=\frac{1}{\sqrt t}\,v\!\left(\frac{x}{\sqrt t}\right). \]

Writing \(z=x/\sqrt t\) , we compute \(u_t=-\tfrac12t^{-3/2}\bigl(v+zv'\bigr)\) and \(u_{xx}=t^{-3/2}v''\) . The heat equation becomes the ordinary differential equation \(kv''+\tfrac12(zv)'=0\) . Integrating once, \(kv'+\tfrac12zv\) is constant, and the constant is zero for a profile that decays with its derivative at infinity. Then \(v'/v=-z/(2k)\) , so \(v=Ae^{-z^2/(4k)}\) . Choosing \(A\) to make the total heat equal to one gives the function we want.

Definition 5.1 (heat kernel). The heat kernel on the line, for diffusivity \(k>0\) , is

\[ S(x,t)=\frac{1}{\sqrt{4\pi kt}}\,\exp\!\left(-\frac{x^2}{4kt}\right)\qquad(x\in\mathbb R,\ t>0). \]

In \(\mathbb R^n\) it is \(S(x,t)=(4\pi kt)^{-n/2}\exp\bigl(-\lvert x\rvert^2/(4kt)\bigr)\) .

Lemma 5.2 (properties of the heat kernel). (a) \(S\) is infinitely differentiable for \(t>0\) and satisfies \(S_t=kS_{xx}\) . (b) \(S>0\) and \(\int_{-\infty}^{\infty}S(x,t)\,dx=1\) for every \(t>0\) . (c) For every \(\delta>0\) , \(\int_{\lvert x\rvert\ge\delta}S(x,t)\,dx\to 0\) as \(t\to 0\) .

Proof. (a) Taking logarithms, \(\log S=-\tfrac12\log(4\pi kt)-x^2/(4kt)\) , so \(S_t=S\bigl(-\tfrac{1}{2t}+\tfrac{x^2}{4kt^2}\bigr)\) and \(S_x=-\tfrac{x}{2kt}S\) , whence \(S_{xx}=S\bigl(-\tfrac{1}{2kt}+\tfrac{x^2}{4k^2t^2}\bigr)\) and \(kS_{xx}=S_t\) . (b) and (c) The substitution \(x=\sqrt{4kt}\,z\) gives

\[ \int_{\lvert x\rvert\ge\delta}S(x,t)\,dx=\frac{1}{\sqrt\pi}\int_{\lvert z\rvert\ge\delta/\sqrt{4kt}}e^{-z^2}\,dz . \]

With \(\delta=0\) this is \(\tfrac{1}{\sqrt\pi}\int_{-\infty}^{\infty}e^{-z^2}dz=1\) . With \(\delta>0\) the lower limit of integration tends to infinity as \(t\to 0\) , and the tail of a convergent integral tends to zero.

So \(S(\cdot,t)\) is a Gaussian bell of unit area and width proportional to \(\sqrt{kt}\) . As \(t\) decreases to zero it grows tall and narrow while keeping its area, concentrating all its mass at the origin: it is the temperature produced by a unit of heat released at \(x=0\) at time zero, which answers the opening question. It is smooth, and it is positive at every \(x\) . There is no edge.

5.2 The initial value problem on the line

A general initial temperature \(g\) is a superposition of point releases, an amount \(g(y)\,dy\) at each \(y\) , and the equation is linear. The solution should therefore be the same superposition of translated kernels.

Theorem 5.3 (solution of the Cauchy problem). Let \(g\) be bounded and continuous on \(\mathbb R\) and define, for \(t>0\) ,

\[ u(x,t)=\int_{-\infty}^{\infty}S(x-y,t)\,g(y)\,dy . \]

Then (a) \(u\) is infinitely differentiable on \(\mathbb R\times(0,\infty)\) and \(u_t=ku_{xx}\) there; (b) \(\inf g\le u(x,t)\le\sup g\) ; (c) \(u(x,t)\to g(x_0)\) as \((x,t)\to(x_0,0)\) with \(t>0\) , for every \(x_0\) .

Proof. (a) Every derivative of \(S(x-y,t)\) with respect to \(x\) and \(t\) is a polynomial in \((x-y)\) and \(1/t\) times \(S(x-y,t)\) . On a set where \(t\) ranges over a closed interval inside \((0,\infty)\) and \(x\) over a bounded interval, such an expression is bounded by a fixed integrable function of \(y\) . Hence the integral may be differentiated under the integral sign any number of times, by the dominated convergence theorem (math.measure-theory.dominated-convergence-theorem) or, at the level of the Riemann integral, by uniform convergence of the differentiated integrals. Then \(u_t-ku_{xx}=\int(S_t-kS_{xx})(x-y,t)\,g(y)\,dy=0\) .

(b) By Lemma 5.2(b), \(u(x,t)\) is a weighted average of the values of \(g\) , with positive weights of total mass one.

(c) Fix \(x_0\) and \(\eta>0\) and choose \(\delta>0\) so that \(\lvert g(y)-g(x_0)\rvert<\eta\) when \(\lvert y-x_0\rvert<2\delta\) . Since the kernel has total mass one, \(u(x,t)-g(x_0)=\int S(x-y,t)\bigl(g(y)-g(x_0)\bigr)dy\) . For \(\lvert x-x_0\rvert<\delta\) split the integral at \(\lvert y-x\rvert=\delta\) . Where \(\lvert y-x\rvert<\delta\) we have \(\lvert y-x_0\rvert<2\delta\) , and that part is at most \(\eta\) . The remainder is at most \(2\sup\lvert g\rvert\int_{\lvert z\rvert\ge\delta}S(z,t)\,dz\) , which by Lemma 5.2(c) is less than \(\eta\) for all small \(t\) . This is the approximate identity argument of Theorem 4.8, with time in the role that the height above the boundary played there.

Four features of the solution can be read off the formula, and each distinguishes the heat equation from the other two model equations.

Smoothing. The solution is \(C^\infty\) for every \(t>0\) , even if \(g\) is merely continuous, and the formula still makes sense and still gives a smooth solution if \(g\) is only bounded and piecewise continuous. All derivatives fall on the kernel.

Infinite speed of propagation. If \(g\ge 0\) is positive on some interval, however short, then \(u(x,t)>0\) for every \(x\) and every \(t>0\) , because the kernel is positive everywhere. A disturbance is felt at once at any distance. The effect at a distance \(d\) in a time \(t\) is of the order \(e^{-d^2/(4kt)}\) , which is fantastically small unless \(d\) is at most a few multiples of \(\sqrt{kt}\) ; the practical extent of a disturbance grows like \(\sqrt t\) , not like \(t\) .

Averaging and decay. Extremes are smoothed out: by (b) the solution never exceeds the range of its initial values.

Conservation. If in addition \(g\) is integrable, integrating the formula in \(x\) and exchanging the order of integration, which Tonelli’s theorem permits for the absolutely integrable integrand (math.measure-theory.tonelli-fubini-theorem), gives \(\int u(x,t)\,dx=\int g(y)\,dy\) . Total heat is conserved, as the conservation law of Section 1.2 requires on a line with no ends for heat to escape through.

Example 5.1: A hot segment cools

Let \(k=1\) and let the initial temperature be \(1\) on the segment \(-1\le x\le 1\) and \(0\) elsewhere. The formula gives \(u(x,t)=\int_{-1}^{1}S(x-y,t)\,dy\) . Substituting \(z=(y-x)/\sqrt{4t}\) and writing the result with the error function \(\operatorname{erf}(s)=\tfrac{2}{\sqrt\pi}\int_0^se^{-z^2}dz\) ,

\[ u(x,t)=\frac12\left[\operatorname{erf}\!\left(\frac{x+1}{\sqrt{4t}}\right)-\operatorname{erf}\!\left(\frac{x-1}{\sqrt{4t}}\right)\right]. \]

At the centre, \(u(0,t)=\operatorname{erf}(1/\sqrt{4t})\) , which is \(0.9747\) at \(t=0.1\) , \(0.5205\) at \(t=1\) , and \(0.1769\) at \(t=10\) . At the original edge \(x=1\) the temperature at \(t=0.1\) is \(0.5000\) to four decimals: an edge between a hot and a cold region immediately takes the average of the two and holds it until the opposite edge makes itself felt. At \(x=3\) , well outside the segment, the temperature at \(t=0.1\) is \(3.9\times10^{-6}\) : positive, as infinite propagation speed demands, and negligible, as common sense demands. By \(t=1\) it has risen to \(0.0763\) .

Temperature profiles of an initially hot segment from minus one to one at five times: a square pulse at time zero, then rounded corners, then progressively lower and wider bell-shaped curves, all positive everywhere. — **Figure 5.1.** The heat equation rounds the jumps of the initial profile at once, keeps the value one half at the original edges for small times, and afterwards lowers and widens the profile at the rate of the square root of time while preserving the area under it.

Figure 5.1 shows all four features. The corners are gone at the earliest positive time shown. Every later curve is positive on the whole axis. No curve rises above \(1\) . And the area under each is \(2\) .

Example 5.2: A Gaussian stays Gaussian

Let \(g(x)=e^{-x^2}\) . The convolution of two Gaussians is a Gaussian, and carrying out the integral gives

\[ u(x,t)=\frac{1}{\sqrt{1+4kt}}\,\exp\!\left(-\frac{x^2}{1+4kt}\right). \]

It is quicker to verify this than to derive it. At \(t=0\) it is \(e^{-x^2}\) . With \(q=1+4kt\) we have \(u_t=u\bigl(-\tfrac{2k}{q}+\tfrac{4kx^2}{q^2}\bigr)\) and \(u_{xx}=u\bigl(-\tfrac{2}{q}+\tfrac{4x^2}{q^2}\bigr)\) , so \(u_t=ku_{xx}\) , and \(u\) is bounded. By the uniqueness theorem proved in the next section it is the solution given by the formula. The squared width \(q\) grows linearly in time, the height decays like \(q^{-1/2}\) , and \(\int u\,dx=\sqrt\pi\) for all \(t\) . With \(k=1\) the peak has fallen to \(\tfrac13\) at \(t=2\) .

5.3 Maximum principle for a finite rod

For a rod \(0\le x\le L\) observed during \(0\le t\le T\) , the natural region is the rectangle

\[ Q_T=\{(x,t):0<x<L,\ 0<t\le T\}. \]

The data of an initial-boundary value problem are prescribed on the bottom \(t=0\) and on the two sides \(x=0\) and \(x=L\) . This set, the boundary of the rectangle with the top removed, is the parabolic boundary \(\Gamma_T\) . The top \(t=T\) is where the solution is being sought, not where it is given.

Theorem 5.4 (weak maximum principle for the heat equation). Let \(u\) be continuous on \(\overline{Q_T}\) with \(u_t\) and \(u_{xx}\) continuous in \(Q_T\) . If \(u_t-ku_{xx}\le 0\) in \(Q_T\) , then the maximum of \(u\) over \(\overline{Q_T}\) is attained on the parabolic boundary \(\Gamma_T\) . If \(u_t-ku_{xx}\ge 0\) , the same holds for the minimum.

Proof. Suppose first that \(u_t-ku_{xx}<0\) in \(Q_T\) , and let the maximum over the compact set \(\overline{Q_T}\) be attained at \((x_0,t_0)\) . If this point were in \(Q_T\) , then \(x\mapsto u(x,t_0)\) would have an interior maximum at \(x_0\) , so \(u_{xx}(x_0,t_0)\le 0\) ; and \(t\mapsto u(x_0,t)\) would have a maximum over \((0,T]\) at \(t_0\) , so \(u_t(x_0,t_0)=0\) if \(t_0<T\) and \(u_t(x_0,T)\ge 0\) if \(t_0=T\) , since the function cannot be decreasing as it arrives at its maximum. In either case \(u_t-ku_{xx}\ge 0\) at the point, a contradiction. So the maximum is on \(\Gamma_T\) .

In general put \(v=u-\varepsilon t\) with \(\varepsilon>0\) . Then \(v_t-kv_{xx}=u_t-ku_{xx}-\varepsilon<0\) , so by the first case

\[ \max_{\overline{Q_T}}u\le\max_{\overline{Q_T}}v+\varepsilon T=\max_{\Gamma_T}v+\varepsilon T\le\max_{\Gamma_T}u+\varepsilon T , \]

and \(\varepsilon\) is arbitrary. For the minimum apply the result to \(-u\) .

The proof is that of Theorem 3.6 with one new twist, the one-sided condition on \(u_t\) at the top edge. It is that asymmetry that removes the top from the boundary. Physically: with no heat sources, the hottest point of a rod over a time interval is found either in the initial state or at an end, never later in the interior.

Theorem 5.5 (uniqueness and stability on a finite rod). Let \(u_1,u_2\) be solutions of \(u_t-ku_{xx}=f\) in \(Q_T\) , continuous on \(\overline{Q_T}\) , with initial values \(g_1,g_2\) and boundary values \(a_1(t),a_2(t)\) at \(x=0\) and \(b_1(t),b_2(t)\) at \(x=L\) . Then

\[ \max_{\overline{Q_T}}\lvert u_1-u_2\rvert\le\max\Bigl\{\max\lvert g_1-g_2\rvert,\ \max\lvert a_1-a_2\rvert,\ \max\lvert b_1-b_2\rvert\Bigr\}. \]

In particular the initial-boundary value problem has at most one solution.

Proof. The difference \(w=u_1-u_2\) satisfies \(w_t-kw_{xx}=0\) , so Theorem 5.4 applies to both \(w\) and \(-w\) , and the values of \(w\) on \(\Gamma_T\) are the differences of the data.

The maximum principle also settles uniqueness on the whole line, within the class of bounded solutions.

Theorem 5.6 (uniqueness of bounded solutions on the line). Let \(u\) be bounded and continuous on \(\mathbb R\times[0,T]\) with \(u_t=ku_{xx}\) for \(0<t\le T\) and \(u(x,0)=0\) . Then \(u=0\) .

Proof. Let \(\lvert u\rvert\le M\) . For \(\ell>0\) the function \(w(x,t)=\dfrac{M}{\ell^2}\,(x^2+2kt)\) solves the heat equation. Compare \(u\) and \(w\) on the rectangle \(-\ell\le x\le\ell\) , \(0\le t\le T\) : on the bottom \(w\ge 0=u\) , and on the sides \(w\ge M\ge u\) . By Theorem 5.4 applied to \(u-w\) on this rectangle, \(u\le w\) inside it. Fix a point \((x,t)\) and let \(\ell\to\infty\) : then \(w(x,t)\to 0\) , so \(u(x,t)\le 0\) . The same argument for \(-u\) gives \(u\ge 0\) .

Hence the function of Theorem 5.3 is the only bounded solution with initial values \(g\) . The word “bounded” matters: Tychonoff constructed nonzero solutions of the heat equation on the line with zero initial values, which grow extremely fast as \(\lvert x\rvert\to\infty\) . They are excluded by any reasonable growth condition, and we shall not meet them again.

Example 5.3: Two-sided bounds without solving

Let \(u\) solve \(u_t=u_{xx}\) for \(0<x<1\) with \(u(0,t)=u(1,t)=0\) and \(u(x,0)=x(1-x)\) . Each function \(w_c(x,t)=c\,e^{-\pi^2t}\sin(\pi x)\) solves the same equation with the same boundary values. The initial parabola lies between two multiples of the sine arch,

\[ \tfrac14\sin(\pi x)\le x(1-x)\le\tfrac1\pi\sin(\pi x)\qquad(0\le x\le 1), \]

with equality on the left at \(x=\tfrac12\) and on the right in the limit at the endpoints, where both sides have slope \(1\) (Exercise 8 asks for a proof). By Theorem 5.4 applied to \(u-w_{1/\pi}\) and to \(w_{1/4}-u\) , the ordering of the initial values persists for all time:

\[ \tfrac14e^{-\pi^2t}\sin(\pi x)\le u(x,t)\le\tfrac1\pi e^{-\pi^2t}\sin(\pi x). \]

At \(x=\tfrac12\) and \(t=0.1\) this reads \(0.0932\le u\le 0.1186\) . The exact value, from the series solution of Chapter 7, is \(0.0962\) . The comparison has captured the decay rate \(e^{-\pi^2t}\) exactly and the amplitude to within the gap between \(\tfrac14\) and \(\tfrac1\pi\) .

5.4 Energy and the rate of decay

Theorem 5.7 (energy decay). Let \(u\) be continuous on \([0,L]\times[0,\infty)\) with \(u_t\) , \(u_x\) , \(u_{xx}\) continuous on \([0,L]\times(0,\infty)\) , and let \(u_t=ku_{xx}\) with \(u(0,t)=u(L,t)=0\) . Then \(E(t)=\tfrac12\int_0^Lu(x,t)^2\,dx\) satisfies

\[ E'(t)=-k\int_0^Lu_x^2\,dx\le 0\qquad\text{and}\qquad E(t)\le E(0)\,e^{-4kt/L^2}. \]

Proof. Differentiating under the integral and using the equation, \(E'(t)=\int_0^Luu_t\,dx=k\int_0^Luu_{xx}\,dx\) . Integrating by parts, \(k\int_0^Luu_{xx}\,dx=k\bigl[uu_x\bigr]_0^L-k\int_0^Lu_x^2\,dx\) , and the boundary term vanishes because \(u\) does at both ends.

For the rate we need an inequality between a function and its derivative. If \(w\in C^1[0,L]\) and \(w(0)=0\) , then \(w(x)=\int_0^xw'(s)\,ds\) , and by the Cauchy–Schwarz inequality \(w(x)^2\le x\int_0^L(w')^2ds\) . Integrating in \(x\) ,

\[ \int_0^Lw^2\,dx\le\frac{L^2}{2}\int_0^L(w')^2\,dx . \]

Applying this to \(w=u(\cdot,t)\) gives \(E'(t)=-k\int_0^Lu_x^2dx\le-\dfrac{2k}{L^2}\int_0^Lu^2dx=-\dfrac{4k}{L^2}E(t)\) . Hence \(\bigl(e^{4kt/L^2}E(t)\bigr)'\le 0\) for \(t>0\) , and since \(E\) is continuous at \(t=0\) the bound follows.

The identity \(E'=-k\int u_x^2\) says that the mean square temperature decreases at a rate measured by how uneven the temperature is. It gives a second proof of uniqueness: the difference of two solutions with the same data has \(E(0)=0\) and \(E\ge0\) nonincreasing, so \(E=0\) . Unlike the maximum principle, the argument works unchanged for insulated ends, where \(u_x\) vanishes at the boundary and the boundary term drops out for that reason. The inequality between \(\int w^2\) and \(\int(w')^2\) is the simplest instance of the Poincaré inequality, which becomes a main tool in Chapter 9. The constant \(L^2/2\) is not the best possible. For functions that vanish only at \(x=0\) , which is all the proof above used, the best constant is \(4L^2/\pi^2\) , attained by \(\sin(\pi x/(2L))\) . But \(u(\cdot,t)\) vanishes at both ends, and for such functions Chapter 7 shows that the best constant is \(L^2/\pi^2\) , which improves the exponent \(4k/L^2\) to \(2\pi^2k/L^2\) .

Example 5.4: The energy identity on a two-mode solution

With \(k=1\) and \(L=1\) take \(u(x,t)=e^{-\pi^2t}\sin(\pi x)+\tfrac12e^{-4\pi^2t}\sin(2\pi x)\) , a sum of two solutions that vanish at both ends. Since \(\int_0^1\sin(\pi x)\sin(2\pi x)\,dx=0\) and each squared sine integrates to \(\tfrac12\) ,

\[ E(t)=\tfrac14e^{-2\pi^2t}+\tfrac1{16}e^{-8\pi^2t},\qquad E'(t)=-\tfrac{\pi^2}{2}e^{-2\pi^2t}-\tfrac{\pi^2}{2}e^{-8\pi^2t}. \]

On the other side, \(u_x=\pi e^{-\pi^2t}\cos(\pi x)+\pi e^{-4\pi^2t}\cos(2\pi x)\) and \(\int_0^1u_x^2\,dx=\tfrac{\pi^2}{2}e^{-2\pi^2t}+\tfrac{\pi^2}{2}e^{-8\pi^2t}\) , which is \(-E'(t)\) as the theorem states. Numerically \(E(0)=0.3125\) and \(E(0.1)\approx0.03475\) , while the bound of Theorem 5.7 gives \(E(0.1)\le0.3125\,e^{-0.4}\approx0.2095\) . The bound is true and far from sharp: the actual decay exponent of the slowest mode is \(2\pi^2\approx19.7\) , not \(4\) . Notice also that the second mode, which carried one fifth of the initial energy, carries less than a thousandth of it at \(t=0.1\) . Fine structure disappears first.

5.5 Sources, and the arrow of time

Sources. For the inhomogeneous equation \(u_t-ku_{xx}=f(x,t)\) with zero initial values, think of the source during the time interval from \(s\) to \(s+ds\) as depositing the temperature \(f(\cdot,s)\,ds\) , which then evolves freely. Adding the contributions gives Duhamel’s formula

\[ u(x,t)=\int_0^t\!\!\int_{-\infty}^{\infty}S(x-y,t-s)\,f(y,s)\,dy\,ds . \]

If \(f\) and \(f_x\) are bounded and continuous this function is a solution: writing \(u=\int_0^tv(x,t;s)\,ds\) , where \(v(\cdot,\cdot\,;s)\) is the solution of the homogeneous equation for \(t>s\) that equals \(f(\cdot,s)\) at \(t=s\) , the derivative of the integral with respect to its upper limit contributes \(v(x,t;t)=f(x,t)\) and the derivative of the integrand contributes \(\int_0^tkv_{xx}\,ds=ku_{xx}\) . The same principle, with the same proof, converts the solution operator of any linear homogeneous evolution equation into a solution formula for the equation with a source.

Example 5.5: A steady source switched on

Solve \(u_t-ku_{xx}=\sin x\) with \(u(x,0)=0\) . The free evolution of the datum \(\sin x\) from time \(s\) is \(v(x,t;s)=e^{-k(t-s)}\sin x\) , since this solves the heat equation, is bounded, and equals \(\sin x\) at \(t=s\) . Duhamel’s formula gives

\[ u(x,t)=\int_0^te^{-k(t-s)}\sin x\,ds=\frac{1-e^{-kt}}{k}\,\sin x . \]

Check: \(u_t=e^{-kt}\sin x\) and \(-ku_{xx}=(1-e^{-kt})\sin x\) , which add up to \(\sin x\) , and \(u(x,0)=0\) . As \(t\to\infty\) the solution approaches \(\tfrac1k\sin x\) . The bounded solutions of the steady equation \(-ku''=\sin x\) are \(\tfrac1k\sin x+C\) , and the limit is the one with mean value zero over a period, as it must be since \(u(\cdot,t)\) has mean zero for every \(t\) : diffusion and the source come into balance, with time constant \(1/k\) .

The arrow of time. Every property in this chapter has a direction. The kernel exists only for \(t>0\) . The maximum principle excludes the top of the rectangle, not the bottom. The energy decreases. The solution at a positive time is infinitely differentiable, so a state with a corner, such as \(e^{-\lvert x\rvert}\) , cannot be the result of any earlier bounded state: the heat equation cannot in general be solved backward in time at all. And where a backward solution does exist it is violently unstable, as Exercise 1.9 showed. This is the mathematical form of irreversibility. Heat flow forgets its initial state at a rate that depends on scale, and what is forgotten cannot be reconstructed.

Common mistake. Reading the diffusive spread as a speed. Because a disturbance spreads over a width proportional to \(\sqrt{kt}\) , the time needed to affect a point at distance \(d\) is proportional to \(d^2/k\) : doubling the distance quadruples the time. There is no front moving at a constant velocity, and “the speed of heat” is not a meaningful quantity for this equation. A second mistake of direction is to apply the maximum principle with data on the top edge: knowing \(u\) at time \(T\) and on the sides does not bound \(u\) at earlier times, because the top edge is not part of the parabolic boundary (Exercise 9).

The wave equation, the subject of the next chapter, is the opposite in every one of these respects. It has a finite speed, it does not smooth, it conserves its energy, and it runs equally well in both directions of time. What features of its formula are responsible?

Exercises

Verify that the heat kernel in \(\mathbb R^n\) is the product of \(n\) one-dimensional kernels, one in each coordinate, and deduce from Lemma 5.2 that it solves \(S_t=k\Delta S\) and has total integral one.

Hint

\((4\pi kt)^{-n/2}e^{-\lvert x\rvert^2/(4kt)}=\prod_{i=1}^{n}S_1(x_i,t)\) with \(S_1\) the one-dimensional kernel. Differentiate the product in \(t\) by the product rule and use \(\partial_tS_1=k\,\partial_{x_ix_i}S_1\) in each factor; the integral over \(\mathbb R^n\) factors into \(n\) integrals, each equal to one.
A half-line \(x>0\) at temperature \(1\) is joined at time zero to a half-line \(x<0\) at temperature \(0\) . Write the solution with the error function, find the temperature at the junction for \(t>0\) , and show that the time at which the temperature at the point \(x=-d\) reaches \(0.25\) is proportional to \(d^2/k\) , finding the constant numerically.

Answer

\(u(x,t)=\tfrac12\bigl(1+\operatorname{erf}(x/\sqrt{4kt})\bigr)\) , so \(u(0,t)=\tfrac12\) for all \(t>0\) . At \(x=-d\) the condition \(u=0.25\) is \(\operatorname{erf}(d/\sqrt{4kt})=0.5\) , that is, \(d/\sqrt{4kt}=z_0\approx0.4769\) , giving \(t=d^2/(4kz_0^2)\approx1.099\,d^2/k\) .
Show that \(u=e^{-ka^2t}\cos(ax)\) is the bounded solution of the heat equation on the line with initial values \(\cos(ax)\) , and deduce without integration the value of \(\int_{-\infty}^{\infty}S(y,t)\cos(ay)\,dy\) .

Answer

\(u_t=-ka^2u=ku_{xx}\) , \(u\) is bounded and continuous, and \(u(x,0)=\cos(ax)\) , so by Theorem 5.6 it is the function of Theorem 5.3. Evaluating that formula at \(x=0\) and using the evenness of \(S\) gives \(\int_{-\infty}^{\infty}S(y,t)\cos(ay)\,dy=e^{-ka^2t}\) .
Prove the semigroup property \(\int_{-\infty}^{\infty}S(x-y,t)\,S(y,s)\,dy=S(x,t+s)\) for \(s,t>0\) by a uniqueness argument, and interpret it.

Solution

Fix \(s>0\) and put \(g(y)=S(y,s)\) , a bounded continuous function. By Theorem 5.3 the function

\[ u_1(x,t)=\int_{-\infty}^{\infty}S(x-y,t)\,S(y,s)\,dy\qquad(t>0) \]

solves the heat equation, is bounded by \(\sup g=(4\pi ks)^{-1/2}\) , and extends continuously to \(t=0\) with the values \(g\) . On the other hand \(u_2(x,t)=S(x,t+s)\) solves the heat equation for \(t>-s\) , hence for \(t\ge 0\) ; it is continuous there, bounded by \((4\pi ks)^{-1/2}\) , and equals \(g\) at \(t=0\) . The difference \(u_1-u_2\) is bounded and continuous on \(\mathbb R\times[0,T]\) for every \(T\) , solves the heat equation, and vanishes at \(t=0\) . By Theorem 5.6 it is zero, which is the claimed identity.

Interpretation: letting a unit point release diffuse for a time \(s\) and then letting the result diffuse for a further time \(t\) is the same as letting the point release diffuse for the time \(t+s\) . In terms of Gaussians, the squared widths \(2ks\) and \(2kt\) add under convolution.
For the half-line \(x>0\) with \(u(0,t)=0\) and \(u(x,0)=g(x)\) , extend \(g\) to an odd function on the line to show that \(u(x,t)=\int_0^\infty\bigl(S(x-y,t)-S(x+y,t)\bigr)g(y)\,dy\) . For \(g=1\) evaluate this as \(\operatorname{erf}\bigl(x/\sqrt{4kt}\bigr)\) and find the heat flux into the wall at \(x=0\) .

Hint

The odd extension \(g_{\mathrm{odd}}\) gives, by Theorem 5.3, a solution on the line that is odd in \(x\) for each \(t\) (replace \(x\) by \(-x\) and \(y\) by \(-y\) in the integral), hence zero at \(x=0\) . Fold the integral over \(y<0\) onto \(y>0\) . For \(g=1\) the two pieces are \(\tfrac12\bigl(1+\operatorname{erf}(x/\sqrt{4kt})\bigr)\) and \(\tfrac12\bigl(1-\operatorname{erf}(x/\sqrt{4kt})\bigr)\) , whose difference is \(\operatorname{erf}(x/\sqrt{4kt})\) . Then \(u_x(0,t)=1/\sqrt{\pi kt}\) and heat enters the wall at the rate \(k\,u_x(0,t)=\sqrt{k/(\pi t)}\) .
Treat the half-line \(x>0\) with an insulated end, \(u_x(0,t)=0\) , by an even extension, and write the kernel. Compare its sign and its integral over the half-line with those of the kernel in Exercise 5, and interpret the difference.

Answer

The kernel is \(S(x-y,t)+S(x+y,t)\) . It is positive and its integral over \(y>0\) is \(1\) : no heat is lost through an insulated end. The Dirichlet kernel \(S(x-y,t)-S(x+y,t)\) is also positive for \(x,y>0\) , but its integral is \(\operatorname{erf}(x/\sqrt{4kt})<1\) : the deficit is the heat that has flowed out through the cold end.
Let \(g\) be bounded, continuous, and integrable with \(\int x^2\lvert g(x)\rvert\,dx\) finite, and let \(u\) be given by Theorem 5.3. Show that \(\int x^2u(x,t)\,dx=\int x^2g(x)\,dx+2kt\int g(x)\,dx\) , and interpret the result as a statement about the width of the temperature distribution.

Hint

Exchange the order of integration and use \(\int x^2S(x-y,t)\,dx=y^2+2kt\) , which follows from the substitution \(x=y+z\) and the Gaussian moments \(\int zS(z,t)\,dz=0\) , \(\int z^2S(z,t)\,dz=2kt\) . If \(g\ge0\) has unit mass, the variance of the temperature distribution grows by exactly \(2kt\) .
Prove the inequalities \(\tfrac14\sin(\pi x)\le x(1-x)\le\tfrac1\pi\sin(\pi x)\) on \([0,1]\) used in Example 5.3, by examining the convexity of the difference on suitable subintervals.

Solution

Upper inequality. Let \(d(x)=\tfrac1\pi\sin(\pi x)-x(1-x)\) . Then \(d(0)=0\) , \(d'(x)=\cos(\pi x)-1+2x\) so \(d'(0)=0\) , and \(d''(x)=2-\pi\sin(\pi x)\) . Let \(x_1\in(0,\tfrac12)\) be the point with \(\sin(\pi x_1)=2/\pi\) . On \([0,x_1]\) we have \(d''\ge0\) , so \(d\) is convex there and lies above its tangent line at \(0\) , which is the zero function: \(d\ge0\) on \([0,x_1]\) . By the symmetry \(d(x)=d(1-x)\) , also \(d\ge0\) on \([1-x_1,1]\) . On the middle interval \([x_1,1-x_1]\) we have \(d''\le0\) , so \(d\) is concave and lies above its chord, whose endpoint values \(d(x_1)=d(1-x_1)\) are nonnegative by what was just shown. Hence \(d\ge0\) on \([0,1]\) .

Lower inequality. Let \(e(x)=x(1-x)-\tfrac14\sin(\pi x)\) . Then \(e(\tfrac12)=\tfrac14-\tfrac14=0\) , \(e'(x)=1-2x-\tfrac\pi4\cos(\pi x)\) so \(e'(\tfrac12)=0\) , and \(e''(x)=-2+\tfrac{\pi^2}{4}\sin(\pi x)\) . Let \(x_2\in(0,\tfrac12)\) satisfy \(\sin(\pi x_2)=8/\pi^2\) . On \([x_2,1-x_2]\) , \(e''\ge0\) : \(e\) is convex and lies above its tangent at \(\tfrac12\) , which is zero, so \(e\ge0\) there. On \([0,x_2]\) , \(e''\le0\) : \(e\) is concave and lies above the chord joining \(e(0)=0\) to \(e(x_2)\ge0\) , so \(e\ge0\) ; the interval \([1-x_2,1]\) follows by symmetry.

Numerically the ratio \(x(1-x)/\sin(\pi x)\) decreases from \(1/\pi\approx0.3183\) at the ends to \(0.25\) at the middle, consistent with both bounds.
A student is given the temperature \(h(x)\) of a rod at time \(T\) and zero boundary values, and argues: “By the maximum principle \(\lvert u\rvert\le\max\lvert h\rvert\) on the whole rectangle, so the earlier temperatures are no larger than the final one.” Find the error, and exhibit a solution with \(k=1\) and \(L=1\) for which \(\max\lvert u(\cdot,T)\rvert=1\) while \(\max\lvert u(\cdot,0)\rvert=e^{\pi^2T}\) .

Solution

Theorem 5.4 places the maximum of a solution on the parabolic boundary: the initial line and the two sides. The top edge \(t=T\) is not part of it. The student has data on the top and sides and none on the bottom, so the theorem gives no bound at all: the maximum over the rectangle may sit on the bottom edge, where nothing is known. Knowledge of the final state does not control earlier states.

For the example take

\[ u(x,t)=e^{\pi^2(T-t)}\sin(\pi x). \]

Then \(u_t=-\pi^2u=u_{xx}\) , \(u(0,t)=u(1,t)=0\) , and \(u(x,T)=\sin(\pi x)\) with maximum \(1\) , while \(u(x,0)=e^{\pi^2T}\sin(\pi x)\) has maximum \(e^{\pi^2T}\) ; for \(T=1\) that is about \(1.93\times10^{4}\) . The maximum principle, correctly applied, says \(\max u=e^{\pi^2T}\) , attained on the initial line. The correct statement in the student’s setting is the reverse of the claim: final temperatures are no larger than initial ones.
Let \(u\) solve \(u_t=ku_{xx}\) on \(0\le x\le L\) with insulated ends. Prove that the heat content is constant and that \(E(t)=\tfrac12\int_0^Lu^2dx\) is nonincreasing, and deduce uniqueness for the Neumann initial-boundary value problem.

Hint

\(\tfrac{d}{dt}\int_0^Lu\,dx=k\bigl[u_x\bigr]_0^L=0\) . For the energy, \(E'(t)=k\bigl[uu_x\bigr]_0^L-k\int_0^Lu_x^2dx\) , and the boundary term vanishes now because \(u_x\) does. Apply this to the difference of two solutions.
Prove by the energy method that the problem \(u_t=ku_{xx}\) on \(0<x<L\) with \(u(x,0)=g(x)\) , \(u_x(0,t)=\alpha u(0,t)\) , \(u_x(L,t)=-\alpha u(L,t)\) , where \(\alpha>0\) , has at most one solution, and explain why these signs correspond to cooling at both ends.

Solution

Let \(w\) be the difference of two solutions, regular enough for the computation of Theorem 5.7. It satisfies the same equation and boundary conditions with zero initial values. With \(E(t)=\tfrac12\int_0^Lw^2dx\) ,

\[ E'(t)=k\int_0^Lww_{xx}\,dx=k\,w(L,t)w_x(L,t)-k\,w(0,t)w_x(0,t)-k\int_0^Lw_x^2\,dx . \]

Insert the boundary conditions \(w_x(L,t)=-\alpha w(L,t)\) and \(w_x(0,t)=\alpha w(0,t)\) :

\[ E'(t)=-k\alpha\,w(L,t)^2-k\alpha\,w(0,t)^2-k\int_0^Lw_x^2\,dx\le 0 , \]

since \(\alpha>0\) . So \(E\) is nonincreasing, nonnegative, and zero at \(t=0\) ; hence \(E=0\) and \(w=0\) .

The signs. The outward normal derivative is \(u_x\) at \(x=L\) and \(-u_x\) at \(x=0\) , so both conditions read \(\partial u/\partial\nu=-\alpha u\) . The outward heat flux is \(-k\,\partial u/\partial\nu=k\alpha u\) : each end loses heat at a rate proportional to its own temperature, which is Newton’s law of cooling into surroundings at temperature zero, as in Exercise 1.17. With \(\alpha<0\) the ends would gain heat in proportion to their temperature, the boundary terms would have the wrong sign, and the argument would fail.
For \(u=e^{-k\pi^2t/L^2}\sin(\pi x/L)\) compute \(E(t)\) exactly and compare its decay exponent with that guaranteed by Theorem 5.7.

Answer

\(E(t)=\tfrac12e^{-2k\pi^2t/L^2}\int_0^L\sin^2(\pi x/L)\,dx=\tfrac{L}{4}e^{-2k\pi^2t/L^2}\) . The true exponent \(2\pi^2k/L^2\approx19.7\,k/L^2\) is about five times the guaranteed \(4k/L^2\) ; the theorem’s bound holds but is not sharp.
Use Duhamel’s formula to solve \(u_t-ku_{xx}=e^{-t}\cos x\) on the line with \(u(x,0)=0\) , assuming \(k\ne 1\) , and verify the result by substitution. What happens to the formula as \(k\to1\) ?

Solution

The free evolution of the datum \(e^{-s}\cos x\) from time \(s\) is \(v(x,t;s)=e^{-s}e^{-k(t-s)}\cos x\) : it solves the heat equation, is bounded, and equals \(e^{-s}\cos x\) at \(t=s\) . Duhamel’s formula gives

\[ u(x,t)=\cos x\;e^{-kt}\int_0^te^{(k-1)s}\,ds=\frac{e^{-t}-e^{-kt}}{k-1}\,\cos x . \]

Verification: \(u(x,0)=0\) ; \(u_t=\dfrac{-e^{-t}+ke^{-kt}}{k-1}\cos x\) and \(-ku_{xx}=\dfrac{ke^{-t}-ke^{-kt}}{k-1}\cos x\) , and the sum is \(\dfrac{(k-1)e^{-t}}{k-1}\cos x=e^{-t}\cos x\) .

As \(k\to1\) the quotient \((e^{-t}-e^{-kt})/(k-1)\) tends to the derivative of \(-e^{-kt}\) with respect to \(k\) at \(k=1\) , namely \(te^{-t}\) , so the solution becomes \(u=te^{-t}\cos x\) . This is resonance: the source decays at exactly the rate at which the mode \(\cos x\) decays on its own, and the response acquires a factor \(t\) .
Explain why \(g(x)=e^{-\lvert x\rvert}\) cannot be the temperature at time \(t=1\) of any solution of the heat equation that is bounded and continuous on \(\mathbb R\times[0,1]\) . State exactly which theorems are used.

Solution

Suppose \(u\) is bounded and continuous on \(\mathbb R\times[0,1]\) , solves the heat equation for \(0<t\le1\) , and has \(u(x,1)=e^{-\lvert x\rvert}\) . Let \(g_0=u(\cdot,0)\) , which is bounded and continuous. By Theorem 5.3 the function \(\tilde u(x,t)=\int S(x-y,t)g_0(y)\,dy\) is a bounded solution for \(t>0\) that extends continuously to \(t=0\) with values \(g_0\) , by part (c), and is infinitely differentiable for \(t>0\) , by part (a). The difference \(u-\tilde u\) is bounded and continuous on \(\mathbb R\times[0,1]\) , solves the heat equation, and vanishes at \(t=0\) ; by Theorem 5.6 it is zero. Hence \(u(\cdot,1)=\tilde u(\cdot,1)\) is infinitely differentiable in \(x\) . But \(e^{-\lvert x\rvert}\) has one-sided derivatives \(-1\) and \(+1\) at \(x=0\) and is not differentiable there. This contradiction shows no such \(u\) exists.

The theorems used are Theorem 5.3(a) and (c) and the uniqueness Theorem 5.6. The conclusion is that the set of states reachable at a positive time is a very small part of the bounded continuous functions: the backward problem fails already at the level of existence.
Let \(u\) solve \(u_t=ku_{xx}\) on \(0<x<L\) with constant boundary temperatures \(u(0,t)=a\) , \(u(L,t)=b\) and initial temperature \(g\) . Show that \(u\) converges in mean square, as \(t\to\infty\) , to the one-dimensional harmonic function with the same boundary values, and give a rate.

Solution

In one dimension the harmonic function with \(\ell(0)=a\) , \(\ell(L)=b\) is the straight line \(\ell(x)=a+(b-a)x/L\) . Put \(w=u-\ell\) . Since \(\ell\) does not depend on \(t\) and \(\ell''=0\) , \(w_t=u_t=ku_{xx}=kw_{xx}\) , and \(w(0,t)=w(L,t)=0\) , with \(w(x,0)=g(x)-\ell(x)\) . Assuming \(u\) has the regularity required in Theorem 5.7, that theorem applies to \(w\) and gives

\[ \int_0^L\bigl(u(x,t)-\ell(x)\bigr)^2dx\le e^{-4kt/L^2}\int_0^L\bigl(g(x)-\ell(x)\bigr)^2dx . \]

So \(u(\cdot,t)\to\ell\) in mean square at least at the exponential rate \(4k/L^2\) , whatever the initial temperature. The steady state depends only on the boundary values: the equilibrium equation of Chapter 3 describes where the evolution equation of this chapter is heading. With the sharp constant from Chapter 7 the rate improves to \(2\pi^2k/L^2\) .
Show that \(w=S_x\) is a solution of the heat equation for \(t>0\) , compute \(\int w\,dx\) and \(\int xw\,dx\) , sketch \(w(\cdot,t)\) , and describe what initial state it represents as \(t\to0\) .

Answer

Differentiating \(S_t=kS_{xx}\) in \(x\) shows \(w=S_x=-\tfrac{x}{2kt}S\) is a solution. It is odd, positive for \(x<0\) and negative for \(x>0\) , so \(\int w\,dx=0\) ; and \(\int xw\,dx=-\int S\,dx=-1\) by integration by parts. As \(t\to0\) it concentrates at the origin with equal and opposite lobes: a hot spot immediately to the left of a cold spot, a dipole of zero total heat.
For the hot segment of Example 5.1, use the formula to show that \(0<u(x,t)\le\tfrac12\bigl(1-\operatorname{erf}((x-1)/\sqrt{4t})\bigr)\) for \(x>1\) , and evaluate the bound at \(x=3\) for \(t=0.1\) and \(t=1\) .

Answer

For \(x>1\) both arguments of the error function are positive, \(\operatorname{erf}\le1\) , and the difference is positive because \(\operatorname{erf}\) is increasing. At \(x=3\) the bound is \(\tfrac12(1-\operatorname{erf}(2/\sqrt{0.4}))\approx3.9\times10^{-6}\) for \(t=0.1\) and \(\tfrac12(1-\operatorname{erf}(1))\approx0.0786\) for \(t=1\) , close to the values of Example 5.1.