Multivariable-Calculus · Unit 14 · Video 1 · Interactive Practice

The Heat Equation: A First Partial Differential Equation

IKey Formulas

Formula	Name	What it says
$\dfrac{\partial f}{\partial t} = k\left(\dfrac{\partial^2 f}{\partial x^2} + \dfrac{\partial^2 f}{\partial y^2} + \dfrac{\partial^2 f}{\partial z^2}\right)$	Heat equation	$f(x, y, z, t)$ is the temperature; $k > 0$ is the thermal diffusivity (larger $k$ , faster spreading)
$\dfrac{\partial f}{\partial t} = k\,\dfrac{\partial^2 f}{\partial x^2}$	One space direction	$f_{xx} > 0$ (bends up): the spot warms; $f_{xx} < 0$ (bends down): the spot cools
$f(x, t) = e^{-kt}\sin x$	One solution	$f_t = -k e^{-kt}\sin x = k f_{xx}$ , so both sides agree

Key Insight: The solution of a PDE is a whole function, not a number. A PDE has many solutions; the initial temperature distribution and the conditions at the boundary pick out the one that describes a real situation.

IIVisualization 1 — Bending Decides Warming or Cooling

The sign of $\partial^2 f/\partial x^2$ tells whether a spot warms or cools: each point moves toward its neighbours' average.

probe x 1.05

IIIVisualization 2 — A Solution Flattening Out

In $f(x, t) = e^{-kt}\sin x$ every peak cools and every trough warms, faster when $k$ is larger.

time t 0.00 diffusivity

IVVisualization 3 — Is It a Solution?

A function solves $f_t = k f_{xx}$ only if the two sides agree at every $x$ and at every $t$ .

candidate time t 0.40

VQuiz Questions

Problem 1 · The Right Side

Given: $f(x, t) = e^{-4kt}\sin 2x$ . Find the right side of the heat equation, $k\,\dfrac{\partial^2 f}{\partial x^2}$ .

✅ Correct! Each

x

-derivative brings out a factor

2

, and

\sin \to \cos \to -\sin

. Since

f_t = -4k\,e^{-4kt}\sin 2x

as well, this

f

is a solution.

❌ One chain factor missing. Differentiating twice brings out the inner factor

2

twice:

2 \cdot 2 = 4

, not

2

❌ That is $k f_x$ .

f_x = 2e^{-4kt}\cos 2x

is only the first partial; differentiate once more in

x

❌ Check the sign. The derivative of

\cos 2x

-2\sin 2x

, so two derivatives of

\sin 2x

give

-4\sin 2x

❌ Not quite. Treat

e^{-4kt}

as a constant and differentiate

\sin 2x

twice in

x

Show solution

With $t$ held fixed, $e^{-4kt}$ is a constant factor:

f_x = 2e^{-4kt}\cos 2x, \qquad f_{xx} = -4e^{-4kt}\sin 2x.

k\,f_{xx} = -4k\,e^{-4kt}\sin 2x.

The left side is $f_t = -4k\,e^{-4kt}\sin 2x$ too, so $f$ satisfies $f_t = k f_{xx}$ .

Problem 2 · A Hot Spot

Given: along a rod with $k = 2$ , the temperature above the reference value at time $0$ is $f(x, 0) = x^3 - 3x$ . At $x = -1$ the profile has a peak, $f = 2$ . Find $\dfrac{\partial f}{\partial t}$ at $x = -1$ at that moment.

✅ Correct!

f_{xx} = 6x = -6

at the peak, so

f_t = 2 \cdot (-6) = -12

: the hot spot cools.

❌ That is the slope, not the bending.

f_x = 3x^2 - 3 = 0

at the peak, but the heat equation uses the second derivative,

f_{xx} = -6

❌ Check the sign. A peak bends down, so

f_{xx} < 0

and the spot cools:

f_t

must be negative.

❌ Missing $k$ .

-6

f_{xx}

; the rate is

k f_{xx}

with

k = 2

❌ Not quite. Compute

f_{xx}

x = -1

and multiply by

k

Show solution

$f_x = 3x^2 - 3$ and $f_{xx} = 6x$ . At $x = -1$ : $f_x = 0$ (a peak) and $f_{xx} = -6$ .

\frac{\partial f}{\partial t} = k\,\frac{\partial^2 f}{\partial x^2} = 2 \cdot (-6) = -12.

The profile bends down at the peak, so the temperature there falls, at $12$ degrees per unit time at this instant.

Problem 3 · Decay Rate and Half-Life

Given: $f(x, t) = e^{ct}\sin 3x$ and the heat equation $f_t = k f_{xx}$ with $k = \tfrac{1}{2}$ . Find the constant $c$ that makes $f$ a solution, then the time at which the amplitude $e^{ct}$ has fallen to half.

What is c?

When does the amplitude halve?

✅ Correct!

c = -9k = -\tfrac{9}{2}

, and

e^{-4.5t} = \tfrac{1}{2}

gives

t = \tfrac{\ln 2}{4.5} \approx 0.154

❌ Missing a factor of 3. Two

x

-derivatives of

\sin 3x

bring out

3 \cdot 3 = 9

, so

f_{xx} = -9f

, not

-3f

❌ Check the sign.

f_{xx} = -9f

, so

c f = k f_{xx}

forces

c < 0

: the sine profile decays, it does not grow.

❌ Missing $k$ .

c f = k f_{xx} = -9k f

, and

k = \tfrac{1}{2}

, so

c = -\tfrac{9}{2}

❌ Check c. Substitute:

f_t = c f

and

f_{xx} = -9f

, then match

c f = k f_{xx}

❌ Multiplied instead of divided.

4.5\ln 2 \approx 3.12

; solving

4.5t = \ln 2

gives

t = \ln 2 / 4.5

❌ That uses $c = -\tfrac{3}{2}$ .

\ln 2 / 1.5 \approx 0.462

; with the correct

c = -\tfrac{9}{2}

the decay is three times faster.

❌ That uses $c = -9$ .

\ln 2 / 9 \approx 0.077

leaves out

k = \tfrac{1}{2}

; the rate is

4.5

, not

9

❌ Check the time. Solve

e^{-4.5t} = \tfrac{1}{2}

by taking logarithms.

Show solution

Differentiate:

f_t = c\,e^{ct}\sin 3x = c f, \qquad f_{xx} = -9e^{ct}\sin 3x = -9f.

The heat equation needs $c f = k f_{xx} = -9k f$ for every $x$ and $t$ , so $c = -9k = -\tfrac{9}{2}$ .

The amplitude is $e^{-4.5t}$ . Setting it equal to $\tfrac{1}{2}$ :

-4.5t = \ln\tfrac{1}{2} = -\ln 2 \quad\Rightarrow\quad t = \frac{\ln 2}{4.5} \approx 0.154.

Problem 4 · Diffusion in Two Directions

Given: the concentration of dye in a thin layer obeys the diffusion equation $\dfrac{\partial f}{\partial t} = k\left(\dfrac{\partial^2 f}{\partial x^2} + \dfrac{\partial^2 f}{\partial y^2}\right)$ . Find the constant $a$ for which $f(x, y, t) = e^{at}\sin x\cos 2y$ is a solution.

✅ Correct!

f_{xx} = -f

and

f_{yy} = -4f

, so

a f = k(-f - 4f) = -5k f

❌ The factor 2 must be squared. Two

y

-derivatives of

\cos 2y

give

-4\cos 2y

, so

f_{yy} = -4f

, not

-2f

❌ Check the sign. Both second partials are negative multiples of

f

, so the concentration pattern must decay:

a < 0

❌ Only the $x$ -direction.

-k

comes from

f_{xx} = -f

alone; the diffusion equation also adds

f_{yy} = -4f

❌ Not quite. Express

f_t

f_{xx}

and

f_{yy}

as multiples of

f

, then match both sides.

Show solution

f_t = a f, \qquad f_{xx} = -e^{at}\sin x\cos 2y = -f, \qquad f_{yy} = -4e^{at}\sin x\cos 2y = -4f.

The equation requires $a f = k(f_{xx} + f_{yy}) = k(-f - 4f) = -5k f$ at every point, so $a = -5k$ .

The same PDE describes heat (with $f$ a temperature) and diffusion (with $f$ a concentration): the pattern fades at rate $5k$ and the dye blends toward uniform.

Solved: 0 / 4