June 4, 2020 APM346 – Week 6 Justin Ko
1 Properties of the Heat Equation on R
Recall that the solution to (
ut − kuxx = f (x, t) x ∈ R, t > 0,
(1)
u|t=0 = g(x) x ∈ R.
is given by
Z ∞ Z tZ ∞ (x−y)2
1 (x−y)2 1
u(x, t) = √ e− 4kt g(y) dy + p e− 4k(t−s) f (y, s) dyds. (2)
4πkt −∞ 0 −∞ 4πk(t − s)
We can use this formula to derive several nice properties satisfied by the solutions to (2).
1.1 Well-Posed
Given some minor integrability assumptions on g (bounded and continuous), we can prove the existence
of a C ∞ (R) solution to (1) is obvious from (2). We can show the solutions are also unique and stable.
Proposition 1 (Uniqueness of Solutions that Decay at Infinity )
If u and its derivatives decay at infinity, then (1) has a unique solution.
Proof. We use an energy argument.
Difference of Solutions: Suppose u1 and u2 are C 2 (R) solutions to (1) that decay at infinity. By
linearity, v = u1 − u2 solves (
ut − kuxx = 0 x ∈ R, t > 0,
(3)
u|t=0 = 0 x ∈ R.
To prove uniqueness, it suffices to show that v ≡ 0 on the domain of the solution.
Show the Energy is Zero: We consider the energy of the solution v to (3),
1 ∞ 2
Z
E(t) = v dx.
2 −∞
By the assumptions on the decay of u, we can differentiate under the integral sign with respect to t to
conclude that
Z ∞
E 0 (t) = vt v dx
−∞
Z ∞
=k vxx v dx vt − kvxx = 0
−∞
Z ∞ x=∞
= −k vx2 dx + (vx v) Integrate by Parts
−∞ x=−∞
Z ∞
= −k vx2 dx lim u = 0
−∞ x→±∞
≤ 0.
Since E 0 (t) ≤ 0, we can conclude that E(t) is decreasing by the mean value theorem. Furthermore,
the initial conditions imply
1 ∞
Z
E(0) = v(x, 0)2 dx = 0
2 −∞
Page 1 of 7
, June 4, 2020 APM346 – Week 6 Justin Ko
because v(x, 0) = 0. This implies that E(t) ≤ 0. Combined with the fact E(t) ≥ 0 since it is the
integral of non-negative functions, this implies
0 ≤ E(t) ≤ 0 =⇒ E(t) = 0 for all t.
Show the Difference is Zero: Since E(t) is the integral of a sum of squares of continuous functions,
each term in the integrand must be 0 so
v 2 (x, t) = 0 for all x ∈ R and t ≥ 0 =⇒ v(x, t) ≡ 0.
Therefore, u1 = u2 , so the solution to (3) is unique.
Remark 1. We can also prove uniqueness for the homogeneous heat equation using by applying the
maximum principle covered in the next section and taking limits. We need to assume some integrability
on g to ensure that this limiting procedure is valid.
Proposition 2 (Stability of Homogeneous Solutions that Decay at Infinity )
If f = 0 and u and its derivatives decay at infinity, then (1) is stable.
Proof. Let u1 be the solution to the homogeneous version of (1) with initial data g1 and u2 be the
solution to the homogeneous version of (1) with and initial data g2 . We consider the energy of the
solution v to (3),
1 ∞ 2
Z
E(t) = v dx.
2 −∞
The computations in the proof of uniqueness imply that E 0 (t) ≤ 0, so E(t) is decreasing. Furthermore,
we have
1 ∞ 1 ∞
Z Z
E(0) = v(x, 0)2 dx = (g1 (x) − g2 (x))2 dx = 0
2 −∞ 2 −∞
because v1 (x, 0) = g1 (x) and v2 (x, 0) = g2 (x). We define the L2 norm of f as
Z ∞
2
kf k2 = f (x) dx .
−∞
1
√ ≤ E(0) =
Therefore, E(t) 2 kg1 − g2 k22 by the mean value theorem. For every > 0, if we take
kg1 − g2 k ≤ , then Z ∞
1
E(t) = (u1 − u2 )2 dx ≤ .
2 −∞
for all t. This implies stability for all t in terms of the “square error”.
2
Remark 2. The heat equation is not well-posed for t < 0. For example, take un = 1
n sin(nx)e−n kt
.
1.2 Symmetry
It is easy to check (2) implies that the solution u(x, t) inherits the symmetry properties of the initial
conditions and inhomogeneous term,
Proposition 3 (Symmetry )
Let u(x, t) be the solution to (1).
(i) If f and g are even in x then u(x, t) is even in x.
(ii) If f and g are odd in x then u(x, t) is odd in x.
This means we can use odd or even reflections to solve the heat equation on the half line, in exactly
the same way as for the half line wave equation.
Page 2 of 7
1 Properties of the Heat Equation on R
Recall that the solution to (
ut − kuxx = f (x, t) x ∈ R, t > 0,
(1)
u|t=0 = g(x) x ∈ R.
is given by
Z ∞ Z tZ ∞ (x−y)2
1 (x−y)2 1
u(x, t) = √ e− 4kt g(y) dy + p e− 4k(t−s) f (y, s) dyds. (2)
4πkt −∞ 0 −∞ 4πk(t − s)
We can use this formula to derive several nice properties satisfied by the solutions to (2).
1.1 Well-Posed
Given some minor integrability assumptions on g (bounded and continuous), we can prove the existence
of a C ∞ (R) solution to (1) is obvious from (2). We can show the solutions are also unique and stable.
Proposition 1 (Uniqueness of Solutions that Decay at Infinity )
If u and its derivatives decay at infinity, then (1) has a unique solution.
Proof. We use an energy argument.
Difference of Solutions: Suppose u1 and u2 are C 2 (R) solutions to (1) that decay at infinity. By
linearity, v = u1 − u2 solves (
ut − kuxx = 0 x ∈ R, t > 0,
(3)
u|t=0 = 0 x ∈ R.
To prove uniqueness, it suffices to show that v ≡ 0 on the domain of the solution.
Show the Energy is Zero: We consider the energy of the solution v to (3),
1 ∞ 2
Z
E(t) = v dx.
2 −∞
By the assumptions on the decay of u, we can differentiate under the integral sign with respect to t to
conclude that
Z ∞
E 0 (t) = vt v dx
−∞
Z ∞
=k vxx v dx vt − kvxx = 0
−∞
Z ∞ x=∞
= −k vx2 dx + (vx v) Integrate by Parts
−∞ x=−∞
Z ∞
= −k vx2 dx lim u = 0
−∞ x→±∞
≤ 0.
Since E 0 (t) ≤ 0, we can conclude that E(t) is decreasing by the mean value theorem. Furthermore,
the initial conditions imply
1 ∞
Z
E(0) = v(x, 0)2 dx = 0
2 −∞
Page 1 of 7
, June 4, 2020 APM346 – Week 6 Justin Ko
because v(x, 0) = 0. This implies that E(t) ≤ 0. Combined with the fact E(t) ≥ 0 since it is the
integral of non-negative functions, this implies
0 ≤ E(t) ≤ 0 =⇒ E(t) = 0 for all t.
Show the Difference is Zero: Since E(t) is the integral of a sum of squares of continuous functions,
each term in the integrand must be 0 so
v 2 (x, t) = 0 for all x ∈ R and t ≥ 0 =⇒ v(x, t) ≡ 0.
Therefore, u1 = u2 , so the solution to (3) is unique.
Remark 1. We can also prove uniqueness for the homogeneous heat equation using by applying the
maximum principle covered in the next section and taking limits. We need to assume some integrability
on g to ensure that this limiting procedure is valid.
Proposition 2 (Stability of Homogeneous Solutions that Decay at Infinity )
If f = 0 and u and its derivatives decay at infinity, then (1) is stable.
Proof. Let u1 be the solution to the homogeneous version of (1) with initial data g1 and u2 be the
solution to the homogeneous version of (1) with and initial data g2 . We consider the energy of the
solution v to (3),
1 ∞ 2
Z
E(t) = v dx.
2 −∞
The computations in the proof of uniqueness imply that E 0 (t) ≤ 0, so E(t) is decreasing. Furthermore,
we have
1 ∞ 1 ∞
Z Z
E(0) = v(x, 0)2 dx = (g1 (x) − g2 (x))2 dx = 0
2 −∞ 2 −∞
because v1 (x, 0) = g1 (x) and v2 (x, 0) = g2 (x). We define the L2 norm of f as
Z ∞
2
kf k2 = f (x) dx .
−∞
1
√ ≤ E(0) =
Therefore, E(t) 2 kg1 − g2 k22 by the mean value theorem. For every > 0, if we take
kg1 − g2 k ≤ , then Z ∞
1
E(t) = (u1 − u2 )2 dx ≤ .
2 −∞
for all t. This implies stability for all t in terms of the “square error”.
2
Remark 2. The heat equation is not well-posed for t < 0. For example, take un = 1
n sin(nx)e−n kt
.
1.2 Symmetry
It is easy to check (2) implies that the solution u(x, t) inherits the symmetry properties of the initial
conditions and inhomogeneous term,
Proposition 3 (Symmetry )
Let u(x, t) be the solution to (1).
(i) If f and g are even in x then u(x, t) is even in x.
(ii) If f and g are odd in x then u(x, t) is odd in x.
This means we can use odd or even reflections to solve the heat equation on the half line, in exactly
the same way as for the half line wave equation.
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