6.4.3 Practice set: the heat equation

Diffusion and the heat equation

Diffusion and the heat equation (lesson 6.4.1)

0 of 7 correct

1.
For which number $a$ does $u = e^{at}\sin 4x$ solve $u_t = u_{xx}$ ?
2.
$u = x^3 + 6xt$ solves the heat equation. Find $u_t$ at $x = 2$ , $t = 1$
3.
Three neighbouring temperatures, 0.2 apart, are 3.0, 2.5 and 2.4. Estimate how fast the middle one is changing
4.
Which halves faster, $\sin 2x$ or $\sin 5x$ , and by what factor? Give the half-life of $\sin 2x$ divided by that of $\sin 5x$
5.
A rod obeys $u_t = 4u_{xx}$ and starts at $\sin x$ . What fraction of the wave is left at $t = 0.25$ ? Give an exact answer
6.
At one instant a rod on $[0, 2]$ has slopes $u_x = 0.5$ at $x = 0$ and $u_x = -1.5$ at $x = 2$ . At what rate is its total heat changing?
7.
An insulated rod of length $\pi$ starts at $u = 1 + \cos x$ . Its total heat stays the same, and in the end the rod is the same temperature everywhere. What temperature?

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.

The fundamental solution

The fundamental solution (lesson 6.4.2)

0 of 8 correct

1.
Find $\Phi(0, 0.25)$ exactly
2.
By what factor is the bell $\Phi(0, t)$ higher at $t = 0.1$ than at $t = 0.4$ ?
3.
Spikes of heat 1 start at $x = 0$ and at $x = 4$ . Find $u(2, 1)$ exactly
4.
A box of height 1 on $[-1, 1]$ spreads. Given $N(1) \approx 0.8413$ , find $u(0, 0.5)$ (4 decimal places)
5.
The profile $2\max(x, 0)$ spreads. Find $u(0, 0.5)$ exactly
6.
Start from $u_0 = e^{-x}$ . Find $u(0, 2)$ exactly
7.
Start from $u_0 = x^2$ . Using $u(x, t) = \mathbb{E}[u_0(x + X)]$ with $X \sim \mathcal{N}(0, 2t)$ , find $u(1, 0.5)$
8.
A step from 1 down to 0 at $x = 0$ (1 on the left, 0 on the right) spreads. Given $N(1) \approx 0.8413$ , find $u(1, 0.5)$ (4 decimal places)

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.

Coding exercise

Exercise

The heat equation from any start

Write heat_solution(u0, x, t). It returns $u(x, t)$ for $u_t = u_{xx}$ on the whole line, started from the profile u0 (a function of one number), by lesson 6.4.2's integral $\int u_0(y)\,\Phi(x - y, t)\,dy$ .

Integrate with quad over a window of 12 widths $\sqrt{2t}$ on each side of $x$ , not over a fixed range. For a tiny $t$ the bell is too narrow for quad to find in a wide range.
At $t = 0$ return u0(x). For $t < 0$ raise ValueError: the equation cannot be run backwards from a start.

The tests try the step, the kink, $e^x$ , $x^2$ (which becomes $x^2 + 2t$ ), a box near its edge, $t = 10^{-6}$ , $t = 0$ and a negative $t$ .

See the tests

import math

from scipy.stats import norm


def _step(y):
    return 1.0 if y > 0 else 0.0


def _kink(y):
    return max(y, 0.0)


def _box(y):
    return 1.0 if -1 < y < 1 else 0.0


def test_step():
    """The step from 0 to 1 gives N(1) = 0.841345 at x = 1, t = 0.5"""
    expect_close(heat_solution(_step, 1, 0.5), norm.cdf(1), what="heat_solution(step, 1, 0.5)", rel=1e-8)


def test_kink():
    """The kink max(x, 0) is lifted to sqrt(t/pi) = 0.398942 at x = 0, t = 0.5"""
    expect_close(heat_solution(_kink, 0, 0.5), math.sqrt(0.5 / math.pi), what="heat_solution(kink, 0, 0.5)", rel=1e-8)


def test_exponential():
    """e^x becomes e^(x + t): e^2 at x = 1, t = 1"""
    expect_close(heat_solution(math.exp, 1, 1), math.exp(2), what="heat_solution(exp, 1, 1)", rel=1e-8)


def test_square():
    """x^2 becomes x^2 + 2t: 2 at x = 1, t = 0.5"""
    expect_close(heat_solution(lambda y: y * y, 1, 0.5), 2.0, what="heat_solution(x^2, 1, 0.5)", rel=1e-8)


def test_box_near_its_edge():
    """A box of height 1 on (-1, 1) gives 0.760250 at x = 0.9, t = 0.01"""
    exact = norm.cdf(1.9 / math.sqrt(0.02)) - norm.cdf(-0.1 / math.sqrt(0.02))
    expect_close(heat_solution(_box, 0.9, 0.01), exact, what="heat_solution(box, 0.9, 0.01)", rel=1e-7)


def test_tiny_time():
    """At t = 0.000001 the kink has barely moved: 0.5 at x = 0.5"""
    expect_close(heat_solution(_kink, 0.5, 1e-6), 0.5, what="heat_solution(kink, 0.5, 1e-6)", rel=1e-6)


def test_time_zero():
    """At t = 0 the answer is the starting profile itself"""
    expect_close(heat_solution(_step, 0.3, 0), 1.0, what="heat_solution(step, 0.3, 0)")


def test_negative_time():
    """A negative time raises ValueError"""
    try:
        heat_solution(_kink, 0, -0.1)
    except ValueError:
        return
    raise AssertionError("expected ValueError: the heat equation cannot be run backwards from u0")

Diffusion and the heat equation

The fundamental solution

Coding exercise

Free further reading