77.4. MEASURABLE APPROXIMATE SOLUTIONS 2615

Proof: Let J−1 be the duality map mentioned above and define Hε : X → X ′ by

⟨Hε (u) ,v⟩= ε⟨Lv,J−1Lu⟩+ ⟨Fu,v⟩+ ⟨Ku,v⟩

for all v ∈ X . Then Hε is pseudo monotone because it is monotone, bounded, and hemi-continuous. This follows from Theorem 77.3.2, and 77.3.3. It is also easy to see that Hε iscoercive.

⟨Hε (u) ,u⟩∥u∥X

= ε∥Lu∥2

V ′

∥u∥X+∥u∥p

V

∥u∥X+

12[⟨Bu(T ) ,u(T )⟩+ ⟨Bu,u⟩(0)] 1

∥u∥X

If not, then there is ∥un∥X → ∞ but for some M,

ε∥Lun∥2

V ′

∥un∥X+∥un∥p

V

∥un∥X+

12[⟨Bun (T ) ,un (T )⟩+ ⟨Bun,un⟩(0)]

1∥un∥X

≤M

Then one of ∥un∥V or ∥Lun∥V ′ is unbounded. Either way, a contradiction is obtained. ThusHε is coercive bounded, and pseudomonotone. It follows that it maps onto X ′.

There exists uε ∈ X such that for all v ∈ X ,

ε⟨Lv,J−1Luε⟩+ ⟨Fuε ,v⟩+ ⟨Kuε ,v⟩= ⟨ f ,v⟩+ ⟨Bv(0) ,u0⟩. (77.4.10)

In 77.4.10, let v = uε . Using the inequality,

|⟨Bv(0) ,u0⟩| ≤ ⟨Bv,v⟩1/2 (0)⟨Bu0,u0⟩1/2

≤ 12⟨Bv,v⟩(0)+ 1

2⟨Bu0,u0⟩,

it follows that

⟨Fuε ,uε⟩+12[⟨Buε ,uε⟩(T )+ ⟨Buε ,uε⟩(0)]

≤ ∥ f∥V ′ ∥uε∥V +12⟨Buε ,uε⟩(0)+

12⟨Bu0,u0⟩

Thus∥uε∥p

V +12⟨Buε ,uε⟩(T )≤

12⟨Bu0,u0⟩+ || f ||V ′ ||uε ||V ,

which implies that there exists a constant C independent of ε such that

||uε ||V ≤C. (77.4.11)

Now let v ∈D(Λ) . Thus v ∈ X and Bv(0) = 0 so the last term of 77.4.10 equals 0. Theterm, ⟨Buε ,v⟩(0) found in the definition of ⟨Kuε ,v⟩ also equals 0. This follows from

⟨Buε ,v⟩(0) = limn→∞⟨Buε ,vn⟩(0) = 0.

where vn = 0 near 0 and converges to v in X by Proposition 77.3.7. Therefore, for v∈D(Λ) ,a dense subset of V ,

ε⟨Λv,J−1Luε⟩+ ⟨Fuε ,v⟩+ ⟨Luε ,v⟩= ⟨ f ,v⟩.