Maximum Principle
TRẦN NGUYỄN NHẬT HUY
April 2025
1.1. INTRODUCTION
1.1
Introduction
Theorem (Maximum principle) Let Ω denote an open set in Rn . If a smooth function f : Ω → R attains its
maxima/minima at a point x0 ∈ Ω, then:


∇f (x0 ) = 0
min

f
(x
)
≤
0
≥
0
, ∀ 0 ≤ i, j ≤ n
 xi xj 0
Consider the Hamilton - Jacobi equation:
(
L[u] = ∂t u − ∆u + |∇u|q = 0 in Rn × [0, T ]
u(x, 0) = u0 (x) ≥ 0
2,1
Assume that u ∈ Cx,t
(Rn × [0, T ]) is a solution of (1.1). Question: Is u(x, t) ≥ 0 in QT ? (QT = Rn × [0, T ])
The answer is yes! But to prove this, we must apply the following lemma:
Lemma (Comparison principle) If u and v are a subsolution and a supersolution of (1), then u(x, t) ≤ v(x, t) in
QT , provided that u0 (x) ≤ v0 (x) for all x ∈ Rn .
2,1
So, what can we obtain from this lemma? We can show that u is bounded in Cx,t
(QT ), that is:
0 ≤ u0 ≤ u ≤∥ u0 ∥L∞
Remark.
u is a subsolution of (1) if L[u] ≤ 0 in Rn × [0, T ].
v is a supersolution of (1) if L[v] ≥ 0 in Rn × [0, T ].
Proof.
Let w = u − v
Suppose that the function w has a maximum value at (x0 , t0 ) ∈ Rn × (0, T ) .
From the maximum principle, we have:

→
−


∇w(x0 , t0 ) = 0 ,
∂t u(x0 , t0 ) = 0,


∆w(x , t ) < 0
0
0
Since L[u] ≤ 0, L[v] ≥ 0, we have:
(x0 ,t0 )
0 ≥ L[u] − L[v] ====== −(∆u − ∆v) > 0. (Contradition)
This implies that w can not attain its maximum at (x0 , t0 ) ∈ Rn × [0, T ].
Now, we suppose (x0 ∈ Rn , t = T ) is a maximum point of w:

→
−

∇w(x0 , T ) = 0 ,



w(x0 , T + h) − w(x0 , T )
≥ 0,
∂t u(x0 , T ) = lim−

h
h→0


∆w(x , T ) < 0
0
Similarly, we have:
(x0 ,T )
0 ≥ L[u] − L[v] ====== ∂t w(x0 , T ) − ∆w(x0 , T ) > 0. (Contradition)
Therefore, w = u − v must achieve its max at (x0 ∈ Rn , t = 0) (initial value).
⇒ max w(x, t) = maxn w(x0 , t = 0) = u0 (x0 , 0) − v0 (x0 , 0) ≤ 0
QT
x0 ∈R
⇒ u(x, t) ≤ v(x, t).
Now, for any ε > 0, let w = u − v − εt, t ∈ [0, T ]
From the maximum principle, we have:

→
−


∇w(x0 , t0 ) = 0 ,
∂t u(x0 , t0 ) = 0,


∆w(x , t ) ≤ 0
0
0
2
1.2. GENERAL COMPARISON PRINCIPLE
(x0 ,t0 )
⇒ 0 ≥ L[u] − L[v] ====== ∂t (u − v)(x0 , t0 ) − ∆(u − v)(x0 , t0 ) + |∇u|q − |∇v|q ≥ ε > 0 (Contradition)
Similarly, we obtain the same contradition term for the case t = T.
Therefore, once again, w must achieve its max at (x0 , t = 0)
Then, w(x, t) ≤ 0 ∀(x, t) ∈ QT
⇔ u − v ≤ εt ∀(x, t) ∈ QT
Let ε −→ 0, we complete the proof.
What if w does not have a maximum value? Then, we consider the following auxiliary function Φ(x, t) = w(x, t)−ε|x|2 .
Since u is bouded, it is easy to see that if |x| → ∞ then Φ(x, t) → −∞.
Thus, Φ(x, t) must achieve its maximum at some point (x0 , t0 ) in QT .
2,1
As mentioned at the beginning, we assume the solution of the Hamilton - Jacobi equation is in Cx,t
(Rn × [0, T ]),
this is the unique solution in the viscosity sense to the given problem.
(HW) We can apply the above result for the following equation, where Ω is a open bounded set in Rn :



Lu = 0
in Ω × [0, T ],
u ∂Ω = 0,


u(x, 0) = u (x)
0
Example:
Consider the equation:


Lu = ∂t u − ∆u + |u|q−1 u = 0 in Ω × [0, T ]

on ∂Ω × [0, T ]
u=0


u(x, 0) = u (x)
0
(2)
in Ω
2,1
Assume that u ∈ (Cx,t
(Ω × [0, T ]) ∩ C(Ω × [0, T ]). We continue to check if u ≥ 0 for given u0 (x) ≥ 0.
Put min u(x, t) = u(x0 , t0 ).
Ω×[0,T]
· (x0 , t0 ) ∈ Ω × [0, T ]:
(
∂t u(x0 , t0 ) = 0
∆u(x0 , t0 ) ≥ 0
Thus, L[u](x0 , t0 ) ≤ 0 (Contradition since u < 0, because we are trying to prove u ≥ 0).
· (x0 , t0 ) ∈ ∂Ω × [0, T ]: It is obvious that u = 0 on this domain, which satisfies u ≥ 0.
· x0 ∈ Ω, t = T :
u(x0 , T + h) − u(x0 , T )
< 0.
h
q−1
Since L[u] = ∂t u − ∆u + |u| u < 0, this is again a contradition, with the same argument in the first case.
∂t u(x0 , t0 ) = lim−
h→0
· x0 ∈ Ω, t = 0: min u(x, t) = u(x0 , t0 ) ≥ 0.
Ω×[0,T]
Therefore, the proof for u > 0 in this equation is complete.
1.2
General Comparison Principle
Lemma (General comparison principle) If u and v are a subsolution and a supersolution of (2) , then u(x, t) ≤
(
u0 (x) ≤ v0 (x)
v(x, t) in Ω × [0, T ], provided that:
u ∂Ω = g1 (x, t) ≤ v ∂Ω = g2 (x, t) on ∂Ω × [0, T ]
Proof.
Let (a − b)+ = max{a − b, 0}.
3
1.2. GENERAL COMPARISON PRINCIPLE
We have:
ˆ
(L[u] − L[v])(u − v)+ dx ≤ 0
ˆ
ˆ
ˆ
⇔
∂t (u − v)(u − v)+ dx +
∆(v − u)(u − v)+ dx + (|u|q−1 u − |v|q−1 v)(u − v)+ dx ≤ 0
Ω
Ω
Ω
ˆ
ˆ
ˆ
1 d
2
2
q−1
(u − v)+ dx +
⇔
|∇(u − v)+ | dx + (|u| u − |v|q−1 v)(u − v)+ dx ≤ 0
2 dt Ω
Ω
Ω
ˆ
d
(u − v)2+ dx ≤ 0 (Because the next 2 terms are non-negative)
⇒
dt Ω
ˆ
ˆ
⇒ (u(x, t) − v(x, t))2+ dx ≤ (u0 (x) − v0 (x))2+ dx
Ω
Ω
Ω
⇒ u(x, t) ≤ v(x, t) a.e in Ω × [0, T ].
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