Answer:
Step-by-step explanation:
a) To prove that ∫R2 u(t, x, y) dxdy is constant in time, we will integrate by parts on a large rectangle. Let's consider a rectangle R with sides parallel to the coordinate axes and large enough to contain the support of u(t, x, y) for all t.
Using the divergence theorem, we have:
∫∫R (∇⋅(bu)) dxdy = ∫∫∂R (b⋅n) u dS
where ∂R represents the boundary of the rectangle R and n is the outward unit normal vector on ∂R.
Since div(b) = 0 (given in the problem), the left-hand side becomes:
∫∫R (∇⋅(bu)) dxdy = ∫∫R (0) dxdy = 0
The right-hand side can be further simplified. The boundary of the rectangle R consists of four line segments, and on each segment, the unit normal vector points outward, either parallel or antiparallel to the b vector. Thus, (b⋅n) is either positive or negative but constant on each segment.
Therefore, the integral over the boundary can be written as:
∫∫∂R (b⋅n) u dS = ∫(b⋅n)1 ∫∫R u dxdy - ∫(b⋅n)2 ∫∫R u dxdy
where (b⋅n)1 corresponds to the segments with outward normal vectors pointing parallel to b and (b⋅n)2 corresponds to the segments with outward normal vectors pointing antiparallel to b.
Since u(t, x, y) vanishes outside some bounded subset of R2 for all t∈I (as given in the problem), the integral over the entire rectangle R (∫∫R u dxdy) is zero. Therefore, the right-hand side becomes zero:
0 = ∫∫∂R (b⋅n)1 ∫∫R u dxdy - ∫∫∂R (b⋅n)2 ∫∫R u dxdy
= ∫∫∂R (b⋅n)1 ∫∫R u dxdy - ∫∫∂R (b⋅n)2 ∫∫R u dxdy
Since the rectangle R was chosen to be large enough to contain the support of u(t, x, y) for all t, we can let the sides of the rectangle extend to infinity. This means that the integrals over the boundary of the rectangle tend to zero as the sides of the rectangle become large. Therefore:
0 = ∫∫∂R (b⋅n)1 ∫∫R u dxdy - ∫∫∂R (b⋅n)2 ∫∫R u dxdy
= 0 - 0
= 0
Hence, we conclude that ∫R2 u(t, x, y) dxdy is constant in time.
(b) Let h: R → R be a function such that h(0) = 0. We need to show that h∘u is also a solution to the equation satisfied by u.
To do this, we differentiate (h∘u) with respect to t and x. Applying the chain rule, we have:
∂t (h∘u) = h'(u) ∂t u
∂x (h∘u) = h'(u) ∂x u
Since u satisfies the transport equation ∂t u + b⋅∇u = 0, we can substitute ∂t u = -b⋅∇u in the expression for ∂t (h∘u):
∂t (h∘u) = h'(u) (-b⋅∇u)
Similarly, using the chain rule and the fact that div(b) = 0, we can substitute ∂x u = -(∇⋅(bu)) in the expression for ∂x (h∘u):
∂x (h∘u) = h'(u) (-(∇⋅(bu)))
Combining these results, we have:
∂t (h∘u) + b⋅∇(h∘u)
= h'(u) (-b⋅∇u) + b⋅∇(h∘u)
= h'(u) (-b⋅∇u) + b⋅(h'(u) ∇u)
= h'(u) (-b⋅∇u + b⋅∇u)
= 0
Therefore, h∘u satisfies the transport equation.
Now, using part (a), we know that ∫R2 u(t, x, y) dxdy is constant in time. Let's denote this constant as C. Then, for any t, we have:
∫R2 (h∘u)(t, x, y) dxdy = ∫R2 h(u(t, x, y)) dxdy = C
So, we deduce that ∫R2 h(u(t, x, y)) dxdy is also constant in time.
(c) To prove that the Lp-norm ∥u(t, ⋅)∥Lp is constant in time for all p ∈ (1, ∞), we will use the result from part (b).
Let's consider the function h(u) = |u|^p. Since h(0) = 0, it satisfies the condition h(0) = 0 mentioned in part (b).
By part (b), we know that ∫R2 h(u(t, x, y)) dxdy is constant in time. Substituting h(u) = |u|^p, we have:
∫R2 |u(t, x, y)|^p dxdy = C
This implies that the Lp-norm ∥u(t, ⋅)∥Lp is constant in time for all p ∈ (1, ∞).
Note that the result also holds for p = 1 or p = ∞, although the proof in this case requires additional arguments.
Therefore, we have shown that the Lp-norm of u(t, ⋅) is constant in time for all p ∈ (1, ∞).