Existence and Uniqueness of Weak Solutions to Parabolic IVP’s
For T > 0, fixed, and U ⊂ R n a bounded open set, let U T =
Then consider the following initial boundary value problem,
∂ t ux, t + Lux, t = fx, t in U T
ux, 0 = u 0 x
x ∈ U,
ux, t = 0,
x ∈ ∂U, 0 < t < T.
n
n
i,j=1
j=1
x, t : x ∈ U, 0 < t < T .
1
Lux, t = − ∑ ∂ i a ij x, t ∂ j u + ∑ b j x, t ∂ j u + cx, tux, t
where
= − divax, t ∇u + ⃗
bx, t ⋅ ∇u + cx, tux, t
with
a ij , b j , c ∈ L ∞ U T 
for a 0 > 0, ⃗z ⋅ a ij x, t ⃗z ≥ a 0 | ⃗z | 2
and
∀z⃗ ∀x, t ∈ U T .
Then L is said to be uniformly elliptic on U T . We recall that the assumptions on the
coefficients a ij , b j , c, imply the existence of positive constants, a 1 , a 0 and μ 0 such that for all
u, v ∈ H 1 U, and μ > μ 0 ,
|Bu, v, t| ≤ a 1 ‖u‖ 1 ‖v‖ 1
a. e. t ∈ 0, T
Bu, u, t + μu, u 0 ≥ a 0 ‖u‖ 21
a. e. t ∈ 0, T
where
Bu, v, t = ∫
Then the mapping
U
n
n
i,j=1
j=1
∑ ∂ i va ij x, t ∂ j u + v ∑ b j x, t ∂ j u + cx, t
dx
L μ = L + μI : H 10 U → H −1 U
is an isomorphism, and the IBVP (1) is said to be a parabolic problem. For convenience, we
will let
V = H 10 U,
H = H 0 U, and
V ′ = H −1 U.
Then V ⊂ H = H ′ ⊂ V ′ and ⟨f, v⟩ V ′ ×V = f, v H for f ∈ V ′ , v ∈ V.
We recall that
W0, T = u ∈ L 2 0, T : V : ∂ t u ∈ L 2 0, T : V ′ 
and that
W0, T ⊂ C0, T : H.
Then we define a weak solution of (1) to be a function u ∈ W0, T satisfying
i
∂ t u, v H + Bu, v, t = f, v H
ii
u0 = u 0 .
∀v ∈ V a. e. t ∈ 0, T,
2
Equivalently, a weak solution satisfies
T
T
T
i
∫ 0 ∂ t u, v H dt + ∫ 0 Bu, v, tdt = ∫ 0 f, v H dt
ii
u0 = u 0 .
∀v ∈ W0, T
We point out that in a parabolic problem, we can without loss of generality assume that
1
the bilinear form Bu, v is coercive. To see this, we simply observe that when L is uniformly
elliptic, the associated bilinear form is at least V − H − coercive; i.e.,
B μ u, v = Bu, v + μu, v H is coercive for μ > μ 0 > 0. If u is a weak solution for (2) then it is
easy to show that Ux, t = e μt ux, t solves
i
∂ t U, v H + B μ U, v, t = e −μt f, v H
ii
U0 = u 0 .
∀v ∈ V a. e. t ∈ 0, T,
Then existence of U is equivalent to existence of u and the problem for U involves a
coercive bilinear form. Now we have,
Theorem (Existence-uniqueness-continuous dependence) Suppose L is uniformly elliptic on
U T with coefficients in L ∞ U T . Then for every f ∈ L 2 0, T : H −1 U and each u 0 ∈ H 0 U,
there exists a unique u ∈ W0, T satisfying (2).Moreover, the linear mapping
L 2 0, T : H −1 U × H 0 U : f, u 0   u ∈ W0, T
is continuous.
Proof- (uniqueness)
Suppose u ∈ W0, T solves (2.2) in the case f = u 0 = 0. Then
T
T
∫ 0 ∂ t u, u H dt + ∫ 0 Bu, u, tdt = 0.
But
T
T
T
∫ 0 ∂ t u, u H dt = 1 ∫ 0 ∂ t u, u H dt = 1 d ∫ 0 u, u H dt
2
2 dt
= 1 ‖uT‖ 2H − ‖u0‖ 2H .
2
Then
1 ‖uT‖ 2 + ∫ T Bu, u, tdt = 0,
H
0
2
and, using the fact that B is coercive,
1 ‖uT‖ 2 + a 0 ∫ T ‖ut‖ 2 dt ≤ 1 ‖uT‖ 2 + ∫ T Bu, u, tdt = 0
H
H
H
0
0
2
2
Since u ∈ W0, T is continuous in t with values in H = H 0 U, it follows that ut = 0. This
proves that the solution is unique, if it exists. The proof of existence will be carried out in
steps.
1) Existence of Approximate Solutions
Let w k denote an orthonormal basis for H which is, at the same time, an orthogonal
basis for V. For N a positive integer, let
N
u N t =
∑ C jN t w j
j=1
where the coefficients C jN t are required to satisfy
i
∂ t u N , w k  H + Bu N , w k , t = f, w k  H
k = 1, . . , N a. e. t ∈ 0, T,
ii
u N 0, w k  H = u 0 , w k , . for k = 1, . . , N
3
2
Since w k denotes an orthonormal basis for H, this reduces to,
N
d C t + ∑ C jN tBw j , w k , t = f k t
dt kN
for k = 1, . . , N
j=1
C kN 0 = u 0 , w k , . for k = 1, . . , N
This is a system of linear ordinary differential equations in N unknowns where the coefficient
matrix is positive definite on 0, T hence it follows that for every f ∈ L 2 0, T : H −1 U and
each u 0 ∈ H 0 U, there exists a unique solution C kN t : for k = 1, . . , N ∈ C0, T N for the
system ofODE’s and a corresponding approximate solution u N t for (3).
2) Energy Estimates
We will show that there exist constants C 1 , C 2 , C 3 depending only on U, T and the
coefficients in L, such that for each N,
1.
a.
‖u N ‖ C0,T:H ≤ C 1 ‖u 0 ‖ H + ‖f‖ L 2
0,T:V ′
b.
‖u N ‖ L 2 0,T:V ≤ C 2 ‖u 0 ‖ H + ‖f‖ L 2
0,T:V ′
c.
‖u ′N ‖ L 2 0,T:V ′  ≤ C 3 ‖u 0 ‖ H + ‖f‖ L 2
0,T:V ′
We note first that
N
‖u N 0‖ 2H =
∑u 0 , w j  2H ≤ ‖u 0 ‖ 2H
j=1
and
2|f, u N  H | = 2|⟨f, u N ⟩ V ′ ×V | ≤ 2‖ft‖ V ′ ‖u N t‖ V
1 ‖ft‖ 2 ′ + α‖u N t‖ 2
≤ α
V
V
Now it follows from (3) that
∂ t u N , u N  H + Bu N , u N , t = f, u N  H
and we have that,
∂ t u N , u N  H = 1 d u N , u N  H = 1 d ||u N || 2H
2 dt
2 dt
which leads to,
i.e.,
d ||u || 2 + 2Bu N , u N , t ≤ 1 ‖ft‖ 2 ′ + α‖u N t‖ 2
V
V
α
dt N H
d ||u || 2 + 2α‖u N t‖ 2 ≤ 1 ‖ft‖ 2 ′ + α‖u N t‖ 2
V
V
V
α
dt N H
Then,
d ||u || 2 + α‖u N t‖ 2 ≤ 1 ‖ft‖ 2 ′
V
V
α
dt N H
and since
‖u N t‖ 2H ≤ ‖u N t‖ 2V ,
3
d ||u || 2 + α‖u N t‖ 2 ≤ 1 ‖ft‖ 2 ′
H
V
α
dt N H
We rewrite this as,
d ||u N || 2 e αt e −αt ≤ 1 ‖ft‖ 2 ′
H
V
α
dt
from which it follows by integrating from 0 to t,
1 ∫ T e αt ‖ft‖ 2 ′ dt
‖u N t‖ 2H ≤ e −αt ‖u N 0‖ 2H + α
V
0
Since this holds for every t ∈ 0, T, it follows.that
sup ‖u N t‖ 2H ≤ C 1 ‖u 0 ‖ 2H + ‖f‖ 2L 2
0≤t≤T
0,T:V ′
proving a.
To prove b), we return to the estimate,
d ||u || 2 + α‖u N t‖ 2 ≤ 1 ‖ft‖ 2 ′
V
V
α
dt N H
and we integrate with respect to t over 0, T to get
T
1 ∫ T ‖ft‖ 2 ′ dt
‖u N T‖ 2H − ‖u N 0‖ 2H + α ∫ ‖u N t‖ 2V dt ≤ α
V
0
0
Then
T
T
∫ 0 ‖u N t‖ 2V d ≤ α1 ‖u N 0‖ 2H + 12 ∫ 0 ‖ft‖ 2V ′ dt
α
which asserts that
‖u N ‖ L2 2 0,T:V ≤ C 2 ‖u 0 ‖ 2H + ‖f‖ 2L 2
0,T:V ′
Finally, to prove c), let V N = spanw 1 , … , w N . Then, for each v ∈ V with ‖v‖ V = 1, we can
use the fact that w j is an ON basis in H to write v = v 1 + v 2 where v 1 ∈ V N and v 2 , w j  H = 0
for j = 1, … , N. Since w j  is also an orthogonal basis in V, we have
N
‖v 1 ‖ 2V = v 1 , v 1  V =
∑v 1 , w j  2V ‖w j ‖ 2V ≤ ‖v‖ 2V ≤ 1
j=1
Now, it follows from (3) that
∂ t u N , v 1  H + Bu N , v 1 , t = f, v 1  H
and
∂ t u N , v H = ∂ t u N , v 1 + v 2  H = ∂ t u N , v 1  H
since
N
∂ t u N t =
∑ C ′jN t w j
and v 2 , w j  H = 0 for j = 1, … , N.
j=1
Then,
4
∂ t u N , v H = ∂ t u N , v 1  H = f, v 1  H − Bu N , v 1 , t
and
|∂ t u N , v H | ≤ ‖f‖ V ′ ‖v 1 ‖ V + a 1 ‖u N ‖ V ‖v 1 ‖ V
≤ ‖f‖ V ′ + a 1 ‖u N ‖ V
since ‖v 1 ‖ V ≤ 1
Then
sup |⟨∂ t u N , v⟩ V ′ ×V | = ‖u ′N t‖ V ′ ≤ ‖f‖ V ′ + a 1 ‖u N ‖ V
‖v‖ V ≤1
Integrating this expression from 0 to T, and using b leads to c.
3) Existence of Weak Solutions
u N t is bounded in L 2 0, T : V
We see that b) implies
c) implies
∂ t u N t is bounded in L 2 0, T : V ′ 
Then there exists a subsequence, u M t ⊂ u N t such that
u M t ⇀ ut
weakly in L 2 0, T : V
weakly in L 2 0, T : V ′ 
∂ t u M t ⇀ vt
But then v must equal u ′ t since weak convergence in in L 2 0, T : V implies convergence in
D ′ 0, T : V, which implies convergence in D ′ 0, T : V ′  and u M t ⇀ ut in D ′ 0, T : V ′ 
implies ∂ t u M t ⇀ u ′ t in D ′ 0, T : V ′ . Similarly, weak convergence in L 2 0, T : V ′  implies
convergence in D ′ 0, T : V ′  so ∂ t u M t ⇀ vt in D ′ 0, T : V ′  and v = u ′ . Now we have,
u M t ⇀ ut
weakly in L 2 0, T : V
∂ t u M t ⇀ ∂ t ut
weakly in L 2 0, T : V ′ 
which implies that u = w − lim u M belongs to W0, T. To see that this weak limit is a weak
solution to the IVP, let
p
Vp =
vt =
∑ d j t w j : d j ∈ C 1 0, T, 1 ≤ j ≤ p
.
j=1
Then it follows from (3) that for each M and all p ≤ M,
T
T
T
∫ 0 ∂ t u M , vt H dt + ∫ 0 Bu M , vt, tdt = ∫ 0 f, vt H dt
∀v ∈ V p
Now, the weak convergence results imply that we can let M tend to infinity to get,
T
T
T
∫ 0 ∂ t u, vt H dt + ∫ 0 Bu, vt, tdt = ∫ 0 f, vt H dt
and this holds for all v ∈ V p , for arbitrarily large p. But it is evident that
⋃ Vp
is dense in L 2 0, T : V
p>0
which means that the last result extends to all v ∈ L 2 0, T : V. Then u is a weak solution of
the equation in (3). To see that u0 = u 0 , note that for arbitrary v ∈ C 1 0, T : V such that
vT = 0,
5
T
T
T
∫ 0 −u, ∂ t vt H dt + ∫ 0 Bu, vt, tdt = ∫ 0 f, vt H dt + u0, v0 H
Similarly, for each M,
T
T
T
∫ 0 −u M , ∂ t vt H dt + ∫ 0 Bu M , vt, tdt = ∫ 0 f, vt H dt + u M 0, v0 H
and if we subtract the latter equation from the former, we get
T
T
∫ 0 −u − u M , ∂ t vt H dt + ∫ 0 Bu − u M , vt, tdt = u0 − u M 0, v0 H .
Now we let M tend to infinity, and obtain
0=
u0 − lim u M 0, v0
M
∀v ∈ C 1 0, T : V.
H
But lim u M 0 = u 0 and C 1 0, T : V ⊂ C0, T : H and it follows that u0 = u 0 . Then u is a
M
weak solution of the IVP. Note that it now follows that every weakly convergent
subsequence must converge to a weak solution of the IVP and since the weak solution has
been shown to be unique, all subsequences must have the same weak limit. But in that
case, the sequence u N  itself must converge weakly to u ∈ W0, T.
By using a v ∈ C 1 0, T : V such that v0 = 0, we can show in the same way that
lim u M T = uT in H. Then we can show that u N  converges strongly to u in L 2 0, T : V.
M
To see this, we write,
T
T
T
∫ 0 ∂ t u, ut H dt + ∫ 0 Bu, ut, tdt = ∫ 0 f, ut H dt
and
T
T
∫ 0 ∂ t u, ut H dt = 1 ∫ 0 d ‖ut‖ 2H dt = 1 ‖uT‖ 2H − 1 ‖u0‖ 2H
2
2
dt
2
Then
T
T
∫ 0 f, ut H dt − ∫ 0 But, ut, tdt = 1 ‖uT‖ 2H − 1 ‖u0‖ 2H
2
2
4
Next, we observe that
T
a 0 ‖u N − u‖ L2 2 0,T:V ≤ ∫ Bu N − u, u N − ut, tdt
0
T
T
T
0
0
0
≤ ∫ Bu N , u N , tdt − ∫ Bu N , u, tdt − ∫ Bu, u N − u, tdt
But,
T
T
T
∫ 0 Bu N , u N , tdt = ∫ 0 f, u N t H dt − ∫ 0 ∂ t u N , u N  H dt
T
= ∫ f, u N t H dt − 1 ‖u N T‖ 2H − ‖u N 0‖ 2H
0
2
and this leads to,
T
a 0 ‖u N − u‖ L2 2 0,T:V ≤ ∫ f, u N t H − Bu N , u, tdt
0
T
− ∫ Bu, u N − u, tdt − 1 ‖u N T‖ 2H − ‖u N 0‖ 2H
0
2
Now, the weak convergence results imply we can allow N to tend to infinity on the right side
6
of this last expression and make use of (4) to obtain
T
T
RHS  ∫ f, ut H dt − ∫ But, ut, tdt − 1 ‖uT‖ 2H + 1 ‖u0‖ 2H = 0
0
0
2
2
from which it follows that ‖u N − u‖ L2 2 0,T:V  0 as N → ∞.
Note it now follows that the energy inequalities b) and c) apply to the weak solution u and
this implies that the solution depends continuously on the data. That is,
u 0 , f ∈ H × L 2 0, T; V ′   u ∈ W0, T
is continuous,
where we recall that
‖u‖ 2W0,T = ∫
T
0
‖ut‖ 2V + ‖∂ t ut‖ 2V ′ dt.
7