Question

Texts: Prove Proposition 1 - 4 Then u is the weak derivative of Du and we write u = D(Du). By definition, Du = -uVEC0e. Since E C0E C0, and hence (Du = -u). The result is that = VEC0 (11.6) Definition: Du = D(Dw) is called the second order weak derivative of u. Proposition 1: The function D2u is uniquely defined by (11.6). Proof: Exercise. Denote the subset of functions in (20,) with weak derivative up to order 2 by H0. Proposition 2: Suppose the functions u and v are in H20, and c is any real number. Then u+v and cu are in H20, and Du+v = Du + D(Dv); D2cw) = cD2u. Proof: Exercise. Definition: uv2 = uv + DwDv + DrDv Proposition 3 is an inner product for H20. Proof: Exercise. Notation: Norm for the vector space H0: |u|2 = /uw Remark 1: |u|3 = 1|u|2 + ||Du|2 + ||D2u||2 Proposition 4: The space H20 is complete. Proof: Exercise.

          Texts: Prove Proposition 1 - 4
Then u is the weak derivative of Du and we write u = D(Du). By definition, Du = -uVEC0e.
Since E C0E C0, and hence (Du = -u). The result is that
= VEC0
(11.6)
Definition: Du = D(Dw) is called the second order weak derivative of u.
Proposition 1: The function D2u is uniquely defined by (11.6). Proof: Exercise.
Denote the subset of functions in (20,) with weak derivative up to order 2 by H0.
Proposition 2: Suppose the functions u and v are in H20, and c is any real number. Then u+v and cu are in H20, and
Du+v = Du + D(Dv); D2cw) = cD2u.
Proof: Exercise.
Definition:
uv2 = uv + DwDv + DrDv
Proposition 3 is an inner product for H20. Proof: Exercise.
Notation: Norm for the vector space H0: |u|2 = /uw
Remark 1: |u|3 = 1|u|2 + ||Du|2 + ||D2u||2
Proposition 4: The space H20 is complete. Proof: Exercise.

texts prove proposition 1 4 then u is the weak derivative of du and we write u ddu by definition du uvec0e since e c0e c0 and hence du u the result is that vec0 116 definition du ddw is call 45754

Added by Thomas B.

Question

Please give Ace some feedback