6.4.2 The fundamental solution

A bell curve that solves the heat equation

Put one unit of heat at the point $x = 0$ of an endless rod at time 0, and let it spread under $u_t = u_{xx}$ . The temperature it makes is a bell curve, called the fundamental solution or heat kernel.

Rule: Fundamental solution

\Phi(x, t) = \frac{1}{\sqrt{4\pi t}}\,e^{-x^2/(4t)}, \quad t > 0

It solves $u_t = u_{xx}$ at every $x$ and every $t > 0$ . The capital $\Phi$ (phi) is this bell; the normal distribution function keeps its name $N$ , as in lesson 2.6.3.

Check it by computing partials (lesson 4.1.2). The chain rule gives $\Phi_x$ ; the product rule then gives $\Phi_{xx}$ ; and differentiating both factors in $t$ gives $\Phi_t$ :

\begin{aligned} \Phi_x &= -\frac{x}{2t}\,\Phi \\ \Phi_{xx} &= \left(\frac{x^2}{4t^2} - \frac{1}{2t}\right)\Phi \\ \Phi_t &= \left(\frac{x^2}{4t^2} - \frac{1}{2t}\right)\Phi \end{aligned}

The last two agree. Both are 0 at $x = \pm\sqrt{2t}$ , where the bell changes from bending down to bending up (lesson 5.4.3). By lesson 6.4.1 those two points are still for a moment; between them the bell sinks, and outside them it rises.

time $t$	0.01	0.05	0.25	1
height $\Phi(0, t) = \frac{1}{\sqrt{4\pi t}}$	2.820948	1.261566	0.56419	0.282095
width $\sqrt{2t}$	0.141421	0.316228	0.707107	1.414214

The bell is a normal density

Lesson 5.4.3's density of $\mathcal{N}(\mu, \sigma^2)$ is $\frac{1}{\sigma\sqrt{2\pi}}e^{-(x - \mu)^2/(2\sigma^2)}$ . With $\mu = 0$ and $\sigma^2 = 2t$ it is exactly $\Phi(x, t)$ . So $\Phi(\cdot, t)$ is the density of $\mathcal{N}(0, 2t)$ , and three facts follow.

Its area is 1 for every $t$ (lesson 2.6.3): the heat is conserved, as lesson 6.4.1 said.
Its variance is $2t$ and its width $\sqrt{2t}$ : lesson 6.4.1's spreading, now exact.
As $t$ shrinks to 0 it narrows into a spike of area 1 at 0, which is how the heat started.

Heat reaches everywhere at once

$\Phi(x, t) > 0$ for every $x$ as soon as $t > 0$ . At $t = 0.01$ the heat at distance 1 is $\Phi(1, 0.01) \approx 3.9 \times 10^{-11}$ : positive, but far too small to notice. So heat travels at infinite speed, in amounts that fall off like $e^{-x^2/(4t)}$ .

The heat from one point at three times. Each bell has area 1; as $t$ grows four times, the width $\sqrt{2t}$ doubles and the height halves.

\Phi_t

import math import sympy as sp from scipy.stats import norm x = sp.symbols("x", real=True) t = sp.symbols("t", positive=True) Phi = sp.exp(-x**2 / (4 * t)) / sp.sqrt(4 * sp.pi * t) print([sp.simplify(sp.diff(Phi, v, k) / Phi) for v, k in [(t, 1), (x, 1), (x, 2)]]) print(sp.simplify(sp.diff(Phi, t) - sp.diff(Phi, x, 2)), sp.integrate(Phi, (x, -sp.oo, sp.oo)), sp.simplify(sp.integrate(x**2 * Phi, (x, -sp.oo, sp.oo)))) def heat(x, t): return math.exp(-x**2 / (4 * t)) / math.sqrt(4 * math.pi * t) h = 1e-3 for x0, t0 in [(0.5, 0.25), (1, 1)]: ut = (heat(x0, t0 + h) - heat(x0, t0 - h)) / (2 * h) uxx = (heat(x0 + h, t0) - 2 * heat(x0, t0) + heat(x0 - h, t0)) / h**2 print(x0, t0, round(ut, 5), round(uxx, 5)) for t0 in [0.01, 0.05, 0.25, 1]: print(t0, round(heat(0, t0), 6), round(math.sqrt(2 * t0), 6), abs(norm.pdf(0.3, 0, math.sqrt(2 * t0)) - heat(0.3, t0)) < 1e-12) print(f"{heat(1, 0.01):.2g}")

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Building any profile from spikes

A spike of heat 1 at $y$ instead of 0 spreads into $\Phi(x - y, t)$ : the same bell, centred at $y$ . A spike of heat $q$ gives $q\,\Phi(x - y, t)$ . The equation is linear (lesson 6.3.1), so several spikes spread independently and their bells add up.

Two spikes. Put heat 1 at $x = -1$ and heat 2 at $x = 1$ . Then $u(x, t) = \Phi(x + 1, t) + 2\,\Phi(x - 1, t)$ , with total heat 3 at every time.

$u(x, t)$	$x = -1$	$x = 0$	$x = 1$	$x = 2$
$t = 0.05$	1.2616	0.0255	2.5231	0.017
$t = 0.25$	0.5849	0.6227	1.1387	0.4152
$t = 1$	0.4896	0.6591	0.668	0.4691

At $t = 0.05$ the two bumps are separate, with almost nothing between them. By $t = 1$ they have merged into one broad hump, slightly higher on the right.

A whole profile

Any starting profile $u_0(x)$ can be cut into thin strips of width $\Delta y$ . The strip at $y$ holds heat $u_0(y)\,\Delta y$ , and spreads into a bell. Adding the bells is a Riemann sum (lesson 2.5.1), and as $\Delta y$ shrinks it becomes an integral.

Rule: Solution by the fundamental solution

u(x, t) = \int_{-\infty}^{\infty} u_0(y)\,\Phi(x - y, t)\,dy

$u_0$ is the profile at $t = 0$ . For $t > 0$ , $u(x, t)$ is an average of $u_0$ near $x$ , weighted by a bell centred at $x$ with width $\sqrt{2t}$ . As $t$ shrinks to 0 it tends to $u_0(x)$ wherever $u_0$ is continuous.

The next section's step profile shows the sum settling. At $x = 1$ , $t = 0.5$ , strips on $[0, 10]$ give 0.843903 for $\Delta y = 0.5$ , 0.841446 for 0.1 and 0.841346 for 0.01. The integral itself is 0.841345.

\Delta y \cdot \Phi(1 - y, 0.5)

import math from scipy.integrate import quad def heat(x, t): return math.exp(-x**2 / (4 * t)) / math.sqrt(4 * math.pi * t) def two(x, t): return heat(x + 1, t) + 2 * heat(x - 1, t) for t in [0.05, 0.25, 1]: print(t, [round(two(x, t), 4) for x in [-1, 0, 1, 2]], round(quad(lambda x: two(x, t), -math.inf, math.inf)[0], 6)) for dy in [0.5, 0.1, 0.01]: n = round(10 / dy) total = sum(dy * heat(1 - (i + 0.5) * dy, 0.5) for i in range(n)) print(dy, round(total, 6))

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The step

Start with a rod that is cold on the left and hot on the right: $u_0 = 0$ for $x < 0$ and $u_0 = 1$ for $x > 0$ . The integral keeps only $y > 0$ :

u(x, t) = \int_0^{\infty} \Phi(x - y, t)\,dy

Substitute $z = \frac{y - x}{\sqrt{2t}}$ , a linear inner function with $dy = \sqrt{2t}\,dz$ (lesson 2.6.1). The integrand becomes $\varphi(z)$ , the standard normal density, and the lower limit $y = 0$ becomes $z = -\frac{x}{\sqrt{2t}}$ . The area under $\varphi$ above $-w$ is $N(w)$ , by symmetry (lesson 5.4.3).

Rule: The step spreads into N

u(x, t) = N\!\left(\frac{x}{\sqrt{2t}}\right)

For the step that is 0 on the left of 0 and 1 on the right. At $x = 0$ the value is $\frac{1}{2}$ for every $t > 0$ , halfway up the jump. It tends to 1 far to the right and to 0 far to the left.

At $t = 0.5$ the width $\sqrt{2t}$ is 1, so $u(x, 0.5) = N(x)$ : 0.02275, 0.158655, 0.5, 0.841345 and 0.97725 at $x$ = $-2$ , $-1$ , 0, 1 and 2. Now follow the fixed point $x = 1$ as the jump spreads past it. Its value falls from 1.0 (to 4 decimals) at $t = 0.01$ to 0.9214 at 0.25, 0.7602 at 1 and 0.6915 at 2.

N

import math from scipy.integrate import quad from scipy.stats import norm def heat(x, t): return math.exp(-x**2 / (4 * t)) / math.sqrt(4 * math.pi * t) t = 0.5 for x in [-2, -1, 0, 1, 2]: print(x, round(norm.cdf(x / math.sqrt(2 * t)), 6), round(quad(lambda y: heat(x - y, t), 0, math.inf)[0], 6)) for t in [0.01, 0.25, 1, 2]: print(t, round(float(norm.cdf(1 / math.sqrt(2 * t))), 4))

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The kink

Now start from the kink $u_0 = \max(x, 0)$ : zero on the left, the line $y = x$ on the right. Then $u(x, t) = \int_0^\infty y\,\Phi(x - y, t)\,dy$ . Split the $y$ in front as $x + (y - x)$ .

The $x$ part is $x$ times the step's integral, $x\,N\!\left(\frac{x}{\sqrt{2t}}\right)$ . The same substitution turns the $(y - x)$ part into $\sqrt{2t}$ times the integral of $z\,\varphi(z)$ above $-\frac{x}{\sqrt{2t}}$ . Since $-\varphi(z)$ is an antiderivative of $z\,\varphi(z)$ (lesson 2.6.1), that integral is $\varphi\!\left(\frac{x}{\sqrt{2t}}\right)$ .

Rule: The kink spreads

\begin{gathered} u_0 = \max(x, 0) \\[4pt] u = x\,N\!\left(\tfrac{x}{\sqrt{2t}}\right) + \sqrt{2t}\,\varphi\!\left(\tfrac{x}{\sqrt{2t}}\right) \end{gathered}

The corner is rounded and lifted: at $x = 0$ the value is $\sqrt{2t}\,\varphi(0) = \sqrt{t/\pi}$ . Far from 0 the solution hugs the two straight pieces.

At $t = 0.5$ the values at $x$ = $-2$ , $-1$ , 0, 1 and 2 are 0.008491, 0.083315, 0.398942, 1.083315 and 2.008491. Each is above $\max(x, 0)$ , by 0.008491, 0.083315, 0.398942, 0.083315 and 0.008491. That fits lesson 6.4.1: the kink never bends down, so it can only rise.

Its slope is the step's solution: $u_x = N\!\left(\frac{x}{\sqrt{2t}}\right)$ , since the slope of the kink is the step. Case study 6.4.4 turns a call option's payoff into a profile very like this kink.

The step and the kink, with their solutions at three times. The step's curves all pass through one half at 0. The kink's corner is rounded and lifted by $\sqrt{t/\pi}$ , while far from 0 the straight pieces barely move.

t = 0.5

import math from scipy.integrate import quad from scipy.stats import norm def heat(x, t): return math.exp(-x**2 / (4 * t)) / math.sqrt(4 * math.pi * t) def kink(x, t): w = math.sqrt(2 * t) return x * norm.cdf(x / w) + w * norm.pdf(x / w) t = 0.5 for x in [-2, -1, 0, 1, 2]: by_quad = quad(lambda y: y * heat(x - y, t), 0, math.inf)[0] print(x, round(kink(x, t), 6), round(by_quad, 6), round(kink(x, t) - max(x, 0), 6)) h = 1e-5 for x in [-1, 0, 1]: print(x, round((kink(x + h, t) - kink(x - h, t)) / (2 * h), 6), round(float(norm.cdf(x)), 6)) print([round(math.sqrt(t / math.pi), 4) for t in [0.01, 0.1, 0.5]])

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Growing profiles: $e^{cx}$

The integral also works for profiles that grow, if they grow slowly enough. Take $u_0 = e^{cx}$ for a number $c$ . Completing the square in the exponent (lesson 1.2.2), as case study 2.6.5 did, moves the bell:

\begin{aligned} &e^{cy}\,\Phi(x - y, t) \\ &= e^{cx + c^2 t}\,\Phi(x + 2ct - y, t) \end{aligned}

The bell is now centred at $x + 2ct$ and scaled by $e^{cx + c^2 t}$ . Its area is 1, so the whole integral is that scale.

Rule: Exponential profiles

u_0 = e^{cx} \quad \Rightarrow \quad u = e^{cx + c^2 t}

Check: $u_t = c^2 u$ and $u_{xx} = c^2 u$ . The profile bends up everywhere, so it rises everywhere. If $u_0 = e^{cx}$ only for $x > 0$ , and 0 elsewhere, only the moved bell's area above 0 counts; by the step's rule, $u = e^{cx + c^2 t}\,N\!\left(\frac{x + 2ct}{\sqrt{2t}}\right)$ .

For $c = 1$ at $x = 0$ , $t = 1$ , the value is $e \approx 2.718282$ ; for $c = 2$ at $t = 0.5$ it is $e^2 \approx 7.389056$ . With the cut at 0, $c = 1$ at $x = 0$ , $t = 0.5$ gives $e^{0.5}N(1) \approx 1.387143$ . Case study 6.4.4 needs exactly this cut version.

Where the integral fails

A profile that grows like $e^{x^2}$ is too fast. For $u_0 = e^{x^2}$ at $x = 0$ the integral is $\frac{1}{\sqrt{1 - 4t}}$ , which is 2.236068 at $t = 0.2$ ; from $t = \frac{1}{4}$ on it diverges, and there is no solution. The kink, polynomials and every $e^{cx}$ are fine.

e^{cy}\,\Phi(x - y, t)

import math from scipy.integrate import quad from scipy.stats import norm def heat(x, t): return math.exp(-x**2 / (4 * t)) / math.sqrt(4 * math.pi * t) for c, x, t in [(1, 0, 1), (2, 0, 0.5)]: centre, w = x + 2 * c * t, 12 * math.sqrt(2 * t) by_quad = quad(lambda y: math.exp(c * y) * heat(x - y, t), centre - w, centre + w)[0] print(c, x, t, round(by_quad, 6), round(math.exp(c * x + c**2 * t), 6)) cut = quad(lambda y: math.exp(y) * heat(0 - y, 0.5), 0, 13)[0] print(round(cut, 6), round(math.exp(0.5) * norm.cdf(1), 6)) for t in [0.2, 0.3]: print(t, [f"{quad(lambda y: math.exp(y**2) * heat(-y, t), -L, L)[0]:.6g}" for L in [5, 10, 20]])

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The same answer from random draws

$\Phi(\cdot, t)$ is the density of a normal variable $X \sim \mathcal{N}(0, 2t)$ , so the rule reads $u(x, t) = \mathbb{E}[u_0(x + X)]$ : the starting profile, averaged over a random normal shift (module 5.4). That shift is like the end of a random walk of many small steps (lessons 5.11.1 and 5.7.3). Heat spreads the way a crowd of random walkers does.

X

import math import numpy as np from scipy.stats import norm rng = np.random.default_rng(3) X = rng.normal(0, math.sqrt(2 * 0.5), 200_000) step = (1 + X > 0).astype(float) kink = np.maximum(1 + X, 0) print(round(step.mean(), 4), round(norm.cdf(1), 6)) print(round(kink.mean(), 4), round(norm.cdf(1) + norm.pdf(1), 6)) print(round(step.std() / math.sqrt(X.size), 4), round(kink.std() / math.sqrt(X.size), 4))

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Common mistakes

The variance is $2t$ , not $t$ . $\Phi(\cdot, t)$ is the $\mathcal{N}(0, 2t)$ density, with width $\sqrt{2t}$ .
$\Phi$ here is the heat kernel, a density. The normal distribution function is $N$ .
Centre the bell where you want the answer. $u(x, t)$ weights $u_0(y)$ by $\Phi(x - y, t)$ , a bell centred at $x$ .
Only for $t > 0$ . $\Phi(x, 0)$ divides by 0; at $t = 0$ the answer is $u_0$ itself.
The step stays at $\frac{1}{2}$ at its jump, for every $t > 0$ .
The kink's solution lies above the kink, by $\sqrt{t/\pi}$ at the corner.
Mind the growth of $u_0$ . $e^{cx}$ is fine; $e^{x^2}$ has a solution only until $t = \frac{1}{4}$ .

The fundamental solution

0 of 13 correct

1.
Find $\Phi(0, 1)$ exactly
2.
At what time $t$ is $\Phi(0, t) = 1$ ? Give an exact answer
3.
$\Phi(\cdot, 3)$ is a normal density. What is its variance?
4.
At $t = 2$ , at which positive $x$ is the bell $\Phi(x, t)$ momentarily still?
5.
Heat 3 starts at the point $y = 2$ . Find $u(2, 1)$ exactly
6.
Spikes of heat 1 and 2 start at $x = -1$ and $x = 1$ . What is the total heat at $t = 5$ ?
7.
A step from 0 to 1 at $x = 0$ spreads. Find $u(0, 7)$
8.
Given $N(1) \approx 0.8413$ , find the step's value $u(2, 2)$ (4 decimal places)
9.
Given $N(1) \approx 0.8413$ , find the step's value $u(-1, 0.5)$ (4 decimal places)
10.
The kink $\max(x, 0)$ spreads. Find $u(0, \pi)$
11.
Given $N(1) \approx 0.8413$ and $\varphi(1) \approx 0.2420$ , find the kink's value $u(1, 0.5)$
12.
Start from $u_0 = e^{2x}$ . Find $u(0, 0.5)$ exactly
13.
Start from $u_0 = e^{x^2}$ . Up to what time does the solution exist?

How to type answers

Powers with ^: x^7, 2^(1/3), a^-2.
Multiply with * or by writing side by side: 3x^2, 2(x+1).
ln(x) is the natural log, log(x, 2) is log base 2, e is Euler's number, sqrt(x) is a square root.
sin(x), cos(x) and tan(x) take radians; write pi for π and sin(x)^2 for sin²x.
Your first answer to each question is saved: it counts towards your score and brings the question back for review.
Exact answers like ln(20)/ln(3) are best. Decimals need about 4 significant figures.

\Phi(x, t) = \frac{1}{\sqrt{4\pi t}}e^{-x^2/(4t)}

The fundamental solution

A bell curve that solves the heat equation

Rule: Fundamental solution

The bell is a normal density

Heat reaches everywhere at once

Building any profile from spikes

A whole profile

Rule: Solution by the fundamental solution

The step

Rule: The step spreads into N

The kink

Rule: The kink spreads

Growing profiles: $e^{cx}$

Rule: Exponential profiles

Where the integral fails

The same answer from random draws

Common mistakes

The fundamental solution

Summary card

Free further reading

A bell curve that solves the heat equation

Rule: Fundamental solution

The bell is a normal density

Heat reaches everywhere at once

Building any profile from spikes

A whole profile

Rule: Solution by the fundamental solution

The step

Rule: The step spreads into N

The kink

Rule: The kink spreads

Growing profiles: ecxe^{cx}ecx

Rule: Exponential profiles

Where the integral fails

The same answer from random draws

Common mistakes

Summary card

Free further reading

Growing profiles: $e^{cx}$