June 2, 2020 APM346 – Week 5 Justin Ko
1 The Heat Equation on R
The one dimensional heat equation models the temperature in a rod.
Definition 1. For parameter k ∈ R+ , the homogeneous heat equation on R × R+ is
ut − kuxx = 0. (1)
The corresponding IVP for the inhomogeneous wave equation is
(
ut − kuxx = f (x, t) x ∈ R, t > 0,
(2)
u|t=0 = g(x) x ∈ R.
The solution to this equation is derived using the method of self similar solutions.
Theorem 1 (Solution to the Heat Equation)
(a) The fundamental solution to (1) is
1 x2
G(x, t) = √ e− 4kt . (3)
4πkt
(b) The particular solution to (2) is given by,
Z ∞ Z tZ ∞
u(x, t) = G(x − y, t)g(y) dy + G(x − y, t − s)f (y, s) dyds
−∞ 0 −∞
Z ∞ Z t Z ∞ (x−y)2
1 (x−y)2 1
=√ e− 4kt g(y) dy + p e− 4k(t−s) f (y, s) dyds. (4)
4πkt −∞ 0 −∞ 4πk(t − s)
Remark 1. The general solution does not have simple form in this problem. Instead, we will find the
fundamental solution which is defined as the derivative in x of the solution to the problem
(
ut − kuxx = f (x, t) x ∈ R, t > 0,
(5)
u|t=0 = 1[0,∞) (x) x ∈ R.
By linearity, the solution to (2) can be pieced together from the fundamental solution.
Remark 2. We can think of (4) as the analogue of the solution for linear odes. Consider the ODE
y 0 − y = f (t), y(0) = g.
The solution when f (t) = 0 is solved by the integrating factor et , giving
Z t
y(t) = exp(t) · g + exp(t − s) · f (s) ds.
0
It turns out that (2) can be solved using operators on function spaces. The corresponding solution (4)
can be written in terms of convolutions as
Z t
u(x, t) = G(x, t) ∗ g(x) + G(x, t − s) ∗ f (x, s) ds.
0
This operator approach to PDEs and the similarity between G(x, t) and the exponential function is a
topic called semigroup theory, which appears when we treat PDEs as an ODE for Banach space valued
functions. This theory allows us to use ODEs to study stochastic differential equations.
Page 1 of 8
, June 2, 2020 APM346 – Week 5 Justin Ko
1.1 Derivation of the Fundamental Solution
We will reduce the heat equation into a single variable problem by looking for a self similar solution.
We will first solve (2) for a special choice of initial condition that will allow us to recover more general
initial conditions using linearity. We want to find a solution v to the IVP,
(
vt − kvxx = 0 x ∈ R, t > 0,
(6)
v|t=0 = 1[0,∞) (x) x ∈ R.
Restriction to Self Similar Solutions: Our goal is to find α, β so that the solution v of the inhomoge-
neous heat equation is invariant under the dilation scaling, that is
v(x, t) = λα v(λβ x, λt) (7)
for all λ > 0, x ∈ R and t > 0. To simplify notation, we define w(x, t; λ) := λα v(λβ x, λt). To find such
α and β, we plug w into the left hand sides of the equations in (6) to see that
wt − kwxx = λα+1 vt (λβ x, λt) − kλα+2β vxx (λβ x, λt)
where vx and vt denote the partial derivatives with respect to the first and second coordinates of v,
and
w|t=0 = λα 1[0,∞) (λβ x) = λα 1[0,∞) (x)
because for λ > 0, λβ x ≥ 0 if and only if x ≥ 0. If we take α = 0 and β = 21 , then the computations
above implies w also solves (6) for all λ. Since both v and w solve the same IVP, we can take solutions
such that v = w, which implies (7) holds for
1
v(x, t) = v(λ 2 x, λt) for all λ > 0, x ∈ R, t > 0. (8)
1
Reduction to an ODE: Since (8) holds for all λ > 0, we can define λ = t to conclude that
1 1
v(x, t) = v(t− 2 x, 1) = V (t− 2 x) where V (s) = v(s, 1) (9)
solves the heat equation (1). Notice that V is a single variable function, so we can now solve for V
using ODEs. Notice that
x 1 k 00 − 1 x 1 1
vt − kvxx = − 3 V 0 (t− 2 x) − V (t 2 x) = 0 =⇒ − 1 V 0 (t− 2 x) − kV 00 (t− 2 x) = 0.
2t 2 t 2t 2
1
If we define z = t− 2 x, then the above equation simplifies to
z
kV 00 (z) + V 0 (z) = 0.
2
Finding the General Solution: If we define G̃(z) = V 0 (z), then we have the first order equation
z z2
G̃0 (z) + G̃(z) = 0 =⇒ G̃(z) = V 0 (z) = Ce− 4k
2k
z2
after using an integrating factor of the form I(z) = e 4k . Integrating this implies that
Z z x Z √x
s2 t s2
V (z) = C e− 4k ds + D =⇒ v(x, t) = V √ = C e− 4k ds + D
0 t 0
for some unknown coefficients C and D.
Finding the Particular Solution: We now plug our general solution to solve for C and D. Since
(
x ∞ x>0
lim √ =
t→0+ t −∞ x < 0.
Page 2 of 8
1 The Heat Equation on R
The one dimensional heat equation models the temperature in a rod.
Definition 1. For parameter k ∈ R+ , the homogeneous heat equation on R × R+ is
ut − kuxx = 0. (1)
The corresponding IVP for the inhomogeneous wave equation is
(
ut − kuxx = f (x, t) x ∈ R, t > 0,
(2)
u|t=0 = g(x) x ∈ R.
The solution to this equation is derived using the method of self similar solutions.
Theorem 1 (Solution to the Heat Equation)
(a) The fundamental solution to (1) is
1 x2
G(x, t) = √ e− 4kt . (3)
4πkt
(b) The particular solution to (2) is given by,
Z ∞ Z tZ ∞
u(x, t) = G(x − y, t)g(y) dy + G(x − y, t − s)f (y, s) dyds
−∞ 0 −∞
Z ∞ Z t Z ∞ (x−y)2
1 (x−y)2 1
=√ e− 4kt g(y) dy + p e− 4k(t−s) f (y, s) dyds. (4)
4πkt −∞ 0 −∞ 4πk(t − s)
Remark 1. The general solution does not have simple form in this problem. Instead, we will find the
fundamental solution which is defined as the derivative in x of the solution to the problem
(
ut − kuxx = f (x, t) x ∈ R, t > 0,
(5)
u|t=0 = 1[0,∞) (x) x ∈ R.
By linearity, the solution to (2) can be pieced together from the fundamental solution.
Remark 2. We can think of (4) as the analogue of the solution for linear odes. Consider the ODE
y 0 − y = f (t), y(0) = g.
The solution when f (t) = 0 is solved by the integrating factor et , giving
Z t
y(t) = exp(t) · g + exp(t − s) · f (s) ds.
0
It turns out that (2) can be solved using operators on function spaces. The corresponding solution (4)
can be written in terms of convolutions as
Z t
u(x, t) = G(x, t) ∗ g(x) + G(x, t − s) ∗ f (x, s) ds.
0
This operator approach to PDEs and the similarity between G(x, t) and the exponential function is a
topic called semigroup theory, which appears when we treat PDEs as an ODE for Banach space valued
functions. This theory allows us to use ODEs to study stochastic differential equations.
Page 1 of 8
, June 2, 2020 APM346 – Week 5 Justin Ko
1.1 Derivation of the Fundamental Solution
We will reduce the heat equation into a single variable problem by looking for a self similar solution.
We will first solve (2) for a special choice of initial condition that will allow us to recover more general
initial conditions using linearity. We want to find a solution v to the IVP,
(
vt − kvxx = 0 x ∈ R, t > 0,
(6)
v|t=0 = 1[0,∞) (x) x ∈ R.
Restriction to Self Similar Solutions: Our goal is to find α, β so that the solution v of the inhomoge-
neous heat equation is invariant under the dilation scaling, that is
v(x, t) = λα v(λβ x, λt) (7)
for all λ > 0, x ∈ R and t > 0. To simplify notation, we define w(x, t; λ) := λα v(λβ x, λt). To find such
α and β, we plug w into the left hand sides of the equations in (6) to see that
wt − kwxx = λα+1 vt (λβ x, λt) − kλα+2β vxx (λβ x, λt)
where vx and vt denote the partial derivatives with respect to the first and second coordinates of v,
and
w|t=0 = λα 1[0,∞) (λβ x) = λα 1[0,∞) (x)
because for λ > 0, λβ x ≥ 0 if and only if x ≥ 0. If we take α = 0 and β = 21 , then the computations
above implies w also solves (6) for all λ. Since both v and w solve the same IVP, we can take solutions
such that v = w, which implies (7) holds for
1
v(x, t) = v(λ 2 x, λt) for all λ > 0, x ∈ R, t > 0. (8)
1
Reduction to an ODE: Since (8) holds for all λ > 0, we can define λ = t to conclude that
1 1
v(x, t) = v(t− 2 x, 1) = V (t− 2 x) where V (s) = v(s, 1) (9)
solves the heat equation (1). Notice that V is a single variable function, so we can now solve for V
using ODEs. Notice that
x 1 k 00 − 1 x 1 1
vt − kvxx = − 3 V 0 (t− 2 x) − V (t 2 x) = 0 =⇒ − 1 V 0 (t− 2 x) − kV 00 (t− 2 x) = 0.
2t 2 t 2t 2
1
If we define z = t− 2 x, then the above equation simplifies to
z
kV 00 (z) + V 0 (z) = 0.
2
Finding the General Solution: If we define G̃(z) = V 0 (z), then we have the first order equation
z z2
G̃0 (z) + G̃(z) = 0 =⇒ G̃(z) = V 0 (z) = Ce− 4k
2k
z2
after using an integrating factor of the form I(z) = e 4k . Integrating this implies that
Z z x Z √x
s2 t s2
V (z) = C e− 4k ds + D =⇒ v(x, t) = V √ = C e− 4k ds + D
0 t 0
for some unknown coefficients C and D.
Finding the Particular Solution: We now plug our general solution to solve for C and D. Since
(
x ∞ x>0
lim √ =
t→0+ t −∞ x < 0.
Page 2 of 8
