Heat Equation

Last updated: 2026-04-12

Definition (Heat Equation)

Consider a region

U \subset \mathbb{R}^n

that is open and a time variable

t > 0

, then we can define the heat equation:

\begin{equation*}\frac{\partial u}{\partial t} = \alpha \Delta u\end{equation*}

Here, the unknown function

u: \overline{U} \times [0,\infty) \to \mathbb{R}

represents the temperature distribution, where

u(x,t)

gives the temperature at position

x

and time

t

. The Laplacian operator

\Delta

acts on the spatial variables

x = (x_1,\ldots,x_n)

and is defined as:

\begin{equation*}\Delta u = \sum_{i=1}^n \frac{\partial^2 u}{\partial x_i^2} = \frac{\partial^2 u}{\partial x_1^2} + \frac{\partial^2 u}{\partial x_2^2} + \cdots + \frac{\partial^2 u}{\partial x_n^2}\end{equation*}

The constant

\alpha

represents the thermal diffusivity of the medium, which measures how quickly heat spreads through the material.

Motivation

The foolowing animations are taken from 3blue1brown's video "But what is a partial differential equation? | DE2"

Let's explore the heat equation in the one-dimensional case. Consider a function

f: [0, L] \times [0,\infty) \to \mathbb{R}

that represents initial temperature distribution.

We study a thin metal rod of length

L

and analyze how heat flows through it. Initially, the rod is heated non-uniformly, and at time

t=0

, the temperature at position

x

corresponds to

f(x)

. The function

u: [0,L] \times [0,\infty) \to \mathbb{R}

describes the temperature of the rod at time

t

and position

x

with the initial condition:

\begin{equation*}u(0, x) = f(x) \quad \text{for} \quad 0 \leq x \leq L\end{equation*}

The Intuition Behind Heat Equation

Physical intuition tells us that over time, the rod should reach a uniform temperature distribution. We want to describe this behavior mathematically. For simplicity, let's consider three adjacent points

x_1 < x_2 < x_3

with their temperatures at some time

t

\begin{equation*}T_1 = u(t, x_1), \quad T_2 = u(t, x_2), \quad T_3 = u(t, x_3).\end{equation*}

Let's investigate the temperature change

\text{d}T_2

over a small time interval

\text{d}t

relative to the temperatures at points

T_1

and

T_3

. The fundamental principle is that

T_2

should approach the average of its neighboring points,

\frac{T_1 + T_3}{2}

, as the system moves toward thermal equilibrium. This can be approximated as:

\begin{equation*}\frac{\text{d}T_2}{\text{d}t} = \alpha \left(\frac{T_1 + T_3}{2} - T_2\right) = \frac{\alpha}{2}\left((T_3 - T_2) - (T_2 - T_1)\right)\end{equation*}

The last term represents a finite difference approximation of the Laplacian operator

\Delta

The changing of

T_2

depends on

\frac{T_1 + T_3}{2}

While this explanation is not mathematically rigorous, it provides an intuitive understanding of the heat equation's behavior.

Modeling Temperature Distribution for Rod

To visualize the evolution of temperature in our one-dimensional rod, we need to consider the heat equation with the given initial function

f(x)

. The temperature function

u(t,x)

requires a three-dimensional coordinate system

(x, t, u(t,x))

for complete visualization, where:

$x$ represents the position along the rod
$t$ represents time
$u(t,x)$ gives the temperature at position $x$ and time $t$

The 3D surface of the solution

u(x,t)

to the one-dimensional heat equation with initial distribution

u(x, 0) = f(x)

. The solution was found using finite difference method.

Fundamental Solution and Cauchy Problem

The foolowing material is taken from [EvansPDE2010] p. 46-47

Now, as we understand the idea behind the heat equation we want to find the solution to it. The key to solving the heat equation lies in understanding the fundamental solution

\Phi(x,t)

. This remarkable function not only solves the heat equation itself but also enables us to construct solutions for any given initial distribution function

f(x)

Definition (Fundamental solution)

The function

\begin{equation*}\Phi(x,t) := \begin{cases}\frac{1}{(4\pi t)^{n/2}} e^{-\frac{|x|^2}{4t}} & (x \in \mathbb{R}^n, \; t > 0) \\0 & (x \in \mathbb{R}^n, \; t < 0)\end{cases}\end{equation*}

is called the fundamental solution of the heat equation. Note that

\Phi

has a singularity at the point

(0,0)

Using

\Phi

, we can construct solutions to the initial-value (or Cauchy) problem:

\begin{align*}u_t &= \Delta u \quad \text{in } \mathbb{R}^n \times (0,\infty) \\u &= g \quad \text{on } \mathbb{R}^n \times \{t=0\}.\end{align*}

A key observation is that for any fixed

y \in \mathbb{R}^n

and

t > 0

, both

(x,t) \mapsto \Phi(x,t)

and

(x,t) \mapsto \Phi(x-y,t)

solve the heat equation away from their respective singularities. This property suggests that we can construct a solution through convolution:

\begin{align*}u(x,t) &= \int_{\mathbb{R}^n} \Phi(x-y,t)g(y)\,\text{d}y \notag\\&= \frac{1}{(4\pi t)^{n/2}} \int_{\mathbb{R}^n} e^{-\frac{|x-y|^2}{4t}}g(y)\,\text{d}y \quad (x \in \mathbb{R}^n, \; t > 0)\end{align*}

The fact that this convolution indeed provides a solution to our initial-value problem will be established in the subsequent theorem.

Solution to Cauchy Problem

The foolowing material is taken from [EvansPDE2010] p. 47-48

Using the definition of the fundamental solution we can establish the following facts:

1. Regularity of the solution

u(x,t)

, i.e. the amount of derivatives.

2. Next, we verify that the convolution with

\Phi(x,t)

indeed yields a solution to the heat equation.

3. Finally, despite the singularity of

\Phi(x,t)

(0,0)

, we will prove that our solution satisfies the initial condition by showing

\begin{equation*}\lim_{t \to 0^+} u(x,t) = g(x)\end{equation*}

for all

x \in \mathbb{R}^n

Theorem (Solution of initial-value problem)

Assume

g \in C(\mathbb{R}^n) \cap L^{\infty}(\mathbb{R}^n)

(continuous and bounded), and define

u

as above. Then

$u \in C^{\infty}(\mathbb{R}^n \times (0,\infty))$ ,
$u_t(x,t) - \Delta u(x,t) = 0 \quad (x \in \mathbb{R}^n, t > 0)$ ,
$\displaystyle\lim_{\substack{(x,t)\to(x^0,0) \\ x\in\mathbb{R}^n, \, t>0}} u(x,t) = g(x^0) \quad \text{for each point } x^0 \in \mathbb{R}^n$ .

\quad

Proof. 1. Since the function

\frac{1}{t^{n/2}}e^{-\frac{|x|^2}{4t}}

is infinitely differentiable, with uniformly bounded derivatives of all orders, on

\mathbb{R}^n \times [\delta,\infty)

for each

\delta > 0

. As a consequence, our solution

u(x,t)

, obtained through convolution with this function, inherits these smoothness properties. Specifically:

\begin{equation*}u \in C^\infty(\mathbb{R}^n \times (0,\infty)).\end{equation*}

This means our solution possesses continuous derivatives of all orders in both space and time variables, making it an extremely well-behaved function from an analytical perspective.

2. Let us now verify that our solution satisfies the heat equation. For each fixed

y \in \mathbb{R}^n

, we know that

\Phi(x-y,t)

solves the heat equation, meaning:

\begin{equation*}\Phi_t(x-y,t) = \Delta_x\Phi(x-y,t)\end{equation*}

Using this property and the linearity of integration, we can show that

u(x,t)

solves the heat equation:

\begin{align*}u_t(x,t) - \Delta u(x,t) &= \int_{\mathbb{R}^n} [\Phi_t - \Delta_x\Phi](x-y,t)g(y)\,\text{d}y \\&= \int_{\mathbb{R}^n} [0]g(y)\,\text{d}y \\&= 0 \quad (x \in \mathbb{R}^n, t > 0)\end{align*}

Thus, we have confirmed that our constructed solution

u(x,t)

is indeed a solution to the heat equation.

3. Let us now prove that our solution satisfies the initial condition by showing that

u(x,t) \to g(x)

t \to 0^+

. Fix any

x^0 \in \mathbb{R}^n

and

\varepsilon > 0

. Since

g

is continuous, there exists

\delta > 0

such that

\begin{equation*}|g(y) - g(x^0)| < \varepsilon \quad \text{whenever } |y-x^0| < \delta, \; y \in \mathbb{R}^n.\end{equation*}

For

|x-x^0| < \frac{\delta}{2}

, we can decompose the difference between

u(x,t)

and

g(x^0)

as follows:

\begin{align*}|u(x,t) - g(x^0)| &= \left|\int_{\mathbb{R}^n} \Phi(x-y,t)[g(y) - g(x^0)]\,\text{d}y\right| \\&\leq \int_{B(x^0,\delta)} \Phi(x-y,t)|g(y) - g(x^0)|\,\text{d}y \\&\quad + \int_{\mathbb{R}^n-B(x^0,\delta)} \Phi(x-y,t)|g(y) - g(x^0)|\,\text{d}y \\&=: I + J.\end{align*}

For the first term

I

, since

\Phi(x-y,t)

is the density of a normal distribution with mean

x

and variance

t

\begin{equation*}I \leq \varepsilon \int_{\mathbb{R}^n} \Phi(x-y,t)\,\text{d}y = \varepsilon\end{equation*}

For the second term

J

, observe that when

|x-x^0| \leq \frac{\delta}{2}

and

|y-x^0| \geq \delta

\begin{equation*}|y-x^0| \leq |y-x| + \frac{\delta}{2} \leq |y-x| + \frac{1}{2}|y-x^0|\end{equation*}

which implies

|y-x| \geq \frac{1}{2}|y-x^0|

. This leads to:

\begin{align*}J &\leq 2\|g\|_{L^\infty} \int_{\mathbb{R}^n-B(x^0,\delta)} \Phi(x-y,t)\,\text{d}y \\&\leq \frac{C}{t^{n/2}} \int_{\mathbb{R}^n-B(x^0,\delta)} e^{-\frac{|x-y|^2}{4t}}\,\text{d}y \\&\leq \frac{C}{t^{n/2}} \int_{\mathbb{R}^n-B(x^0,\delta)} e^{-\frac{|y-x^0|^2}{16t}}\,\text{d}y \\&= C\int_{\mathbb{R}^n-B(x^0,\delta/\sqrt{t})} e^{-\frac{|z|^2}{16}}\,\text{d}z\end{align*}

The final integral approaches zero as

t \to 0^+

. Therefore, for sufficiently small

t > 0

and

|x-x^0| < \frac{\delta}{2}

, we have:

\begin{align*}|u(x,t) - g(x^0)| < 2\varepsilon\end{align*}

So, the theorem is proved

\Box

Mean-value property of Heat Equation

The foolowing material is taken from [EvansPDE2010] p. 51-54

Illustration of

\Gamma_T

and

U

Definition

Assume

U \subset \mathbb{R^n}

is open and bounded region, and fix a time

T > 0

We define the parabolic cylinder $U_T := U \times [0,T].$
The parabolic boundary of $U_T$ is $\Gamma_T := \overline{U_T} - U_T.$

We want next to derive a kind of analogue to the mean-value property for harmonic functions. There is no such simple formula. However let us observe that for fixed

x

the spheres

\partial B(x,r)

are level sets of the fundamental solution

\Phi(x-y)

for Laplace's equation. This suggests that perhaps for fixed

(x,t)

the level sets of fundamental solution

\Phi(x - y, t - s)

for the heat equation may be relevant.

Definition

For fixed

x \in \mathbb{R}^n

t \in \mathbb{R}

r > 0

, we define

\begin{equation*}E(x,t;r) := \left\{(y,s) \in \mathbb{R}^{n+1} \,|\, s \leq t, \,\Phi(x-y,t-s) \geq \frac{1}{r^n} \right\}.\end{equation*}

Theorem (Mean-value property of Heat Equation)

Let

u \in C_1^2(U_T)

solve the heat equation. Then

\begin{equation*}u(x,t) = \frac{1}{4r^n} \iint_{E(x,t;r)} u(y,s) \frac{|x-y|^2}{(t-s)^2} \,dy\,ds\end{equation*}

for each

E(x,t;r) \subset U_T

Formula above is a sort of analogue for the heat equation of the mean-value formulas for Laplace's equation. Observe that the right-hand side involves only

u(y,s)

for times

s \leq t

. This is reasonable, as the value

u(x,t)

should not depend upon future times.

\quad

Proof. Shift the space and time coordinates so that

x = 0

and

t = 0

. Upon mollifying if necessary, we may assume

u

is smooth. Write

E(r) = E(0,0;r)

and set

\begin{equation*}\phi(r) := \frac{1}{r^n} \iint_{E(r)} u(y,s) \frac{|y|^2}{s^2} \,dy\,ds= \iint_{E(1)} u(ry,r^2s) \frac{|y|^2}{s^2} \,dy\,ds.\end{equation*}

We compute

\begin{align*}\phi'(r) &= \iint_{E(1)} \sum_{i=1}^n u_{y_i}y_i\frac{|y|^2}{s^2} + 2ru_s\frac{|y|^2}{s} \,dy\,ds \\&= \frac{1}{r^{n+1}} \iint_{E(r)} \sum_{i=1}^n u_{y_i}y_i\frac{|y|^2}{s^2} + 2u_s\frac{|y|^2}{s} \,dy\,ds \\&=: A + B.\end{align*}

Also, let us introduce the useful function

\begin{equation*}\psi := -\frac{n}{2}\log(-4\pi s) + \frac{|y|^2}{4s} + n\log r\end{equation*}

and observe

\psi = 0

\partial E(r)

, since

\Phi(y,-s) = r^{-n}

\partial E(r)

. We utilize

\psi

to write

\begin{align*}B &= \frac{1}{r^{n+1}} \iint_{E(r)} 4u_s \sum_{i=1}^n y_i\psi_{y_i} \,dy\,ds \\&= -\frac{1}{r^{n+1}} \iint_{E(r)} 4nu_s\psi + 4\sum_{i=1}^n u_{sy_i}\psi \,dy\,ds;\end{align*}

there is no boundary term since

\psi = 0

\partial E(r)

. Integrating by parts with respect to

s

, we discover

\begin{align*}B &= \frac{1}{r^{n+1}} \iint_{E(r)} -4n u_s\psi + 4\sum_{i=1}^n u_{y_i}\psi_s \,dy\,ds \\&= \frac{1}{r^{n+1}} \iint_{E(r)} -4n u_s\psi + 4\sum_{i=1}^n u_{y_i}y_i\left(-\frac{n}{2s} - \frac{|y|^2}{4s^2}\right) \,dy\,ds \\&= \frac{1}{r^{n+1}} \iint_{E(r)} -4n u_s\psi - \frac{2n}{s}\sum_{i=1}^n u_{y_i}y_i \,dy\,ds - A.\end{align*}

Consequently, since

u

solves the heat equation,

\begin{align*}\phi'(r) &= A + B \\&= \frac{1}{r^{n+1}} \iint_{E(r)} -4n\Delta u\psi - \frac{2n}{s}\sum_{i=1}^n u_{y_i}y_i \,dy\,ds \\&= \sum_{i=1}^n \frac{1}{r^{n+1}} \iint_{E(r)} 4nu_{y_i}\psi_{y_i} - \frac{2n}{s}u_{y_i}y_i \,dy\,ds \\&= 0.\end{align*}

Thus

\phi

is constant, and therefore

\begin{equation*}\phi(r) = \lim_{t\to 0}\phi(t) = u(0,0)\left(\lim_{t\to 0}\frac{1}{t^n}\iint_{E(t)} \frac{|y|^2}{s^2}\,dy\,ds\right) = 4u(0,0),\end{equation*}

\begin{equation*}\frac{1}{t^n}\iint_{E(t)} \frac{|y|^2}{s^2}\,dy\,ds = \iint_{E(1)} \frac{|y|^2}{s^2}\,dy\,ds = 4.\end{equation*}

We omit the details of this last computation

\Box

Strong Maximum Principle and Uniqueness

The foolowing material is taken from [EvansPDE2010] p. 55-57

We employ the mean-value property to give a quick proof of the strong maximum principle.

Theorem (Strong maximum principle for the Heat Equation)

Assume

u \in C_1^2(U_T) \cap C(\overline{U_T})

solves the heat equation in

U_T

1. Then

\begin{equation*}\max_{U_T} u = \max_{\Gamma_T} u.\end{equation*}

2. Furthermore, if

U

is connected and there exists a point

(x_0,t_0) \in U_T

such that

\begin{equation*}u(x_0,t_0) = \max_{U_T} u,\end{equation*}

then

\begin{equation*}u \text{ is constant in } \overline{U}_{t_0}.\end{equation*}

Assertion (1) is the maximum principle for the heat equation and (2) is the strong maximum principle. Similar assertions are valid with "min" replacing "max."

\quad

Proof. 1. Suppose there exists a point

(x_0,t_0) \in U_T

with

\begin{equation*}u(x_0,t_0) = M := \max_{U_T} u.\end{equation*}

Then for all sufficiently small

r > 0

E(x_0,t_0;r) \subset U_T

; and we employ the mean-value property to deduce

\begin{equation*}M = u(x_0,t_0) = \frac{1}{4r^n} \iint_{E(x_0,t_0;r)} u(y,s)\frac{|x_0-y|^2}{(t_0-s)^2} \,dy\,ds \leq M,\end{equation*}

since

\begin{equation*}1 = \frac{1}{4r^n} \iint_{E(x_0,t_0;r)} \frac{|x_0-y|^2}{(t_0-s)^2} \,dy\,ds.\end{equation*}

Equality holds only if

u

is identically equal to

M

within

E(x_0,t_0;r)

. Consequently

\begin{equation*}u(y,s) = M \quad \text{for all } (y,s) \in E(x_0,t_0;r).\end{equation*}

Draw any line segment

L

U_T

connecting

(x_0,t_0)

with some other point

(y_0,s_0) \in U_T

, with

s_0 < t_0

. Consider

\begin{equation*}r_0 := \min\{s \geq s_0 \,|\, u(x,t) = M \text{ for all points } (x,t) \in L, \,s \leq t \leq t_0\}.\end{equation*}

Since

u

is continuous, the minimum is attained. Assume

r_0 > s_0

. Then

u(z_0,r_0) = M

for some point

(z_0,r_0)

L \cap U_T

and so

u \equiv M

E(z_0,r_0;r)

for all sufficiently small

r > 0

. Since

E(z_0,r_0;r)

contains

L \cap \{r_0-\sigma \leq t \leq r_0\}

for some small

\sigma > 0

, we have a contradiction. Thus

r_0 = s_0

, and hence

u \equiv M

L

2. Now fix any point

x \in U

and any time

0 \leq t < t_0

. There exist points

\{x_0,x_1,\ldots,x_m = x\}

such that the line segments in

\mathbb{R}^n

connecting

x_{i-1}

x_i

lie in

U

for

i = 1,\ldots,m

. (This follows since the set of points in

U

which can be so connected to

x_0

by a polygonal path is nonempty, open and relatively closed in

U

.) Select times

t_0 > t_1 > \cdots > t_m = t

. Then the line segments in

\mathbb{R}^{n+1}

connecting

(x_{i-1},t_{i-1})

(x_i,t_i)

(i = 1,\ldots,m)

lie in

U_T

. According to step 1,

u \equiv M

on each such segment and so

u(x,t) = M

\Box

References

[EvansPDE2010]
Lawrence C. Evans. Partial Differential Equations, 2010.