62 2. FOUR IMPORTANT LINEAR PDE Proof. 1. Fix some point in UT . Upon shifting the coordinates, we may as well assume the point is (0, 0). Suppose first that the cylinder C(1) := C(0, 0 1) lies in UT . Let C ( 1 2 ) := C ( 0, 0 1 2 ) . Then, as in the proof of Theorem 8, u(x, t) = C(1) K(x, t, y, s)u(y, s) dyds ((x, t) ∈ C( 1 2 )) for some smooth function K. Consequently (39) |DxDtu(x, k l t)| ≤ C(1) |DtDxK(x, l k t, y, s)||u(y, s)| dyds ≤ Ckl u L1(C(1)) for some constant Ckl. 2. Now suppose the cylinder C(r) := C(0, 0 r) lies in UT . Let C(r/2) = C(0, 0 r/2). We rescale by defining v(x, t) := u(rx, r2t). Then vt − Δv = 0 in the cylinder C(1). According to (39), |DxDtv(x, k l t)| ≤ Ckl v L1(C(1)) ((x, t) ∈ C( 1 2 )). But DxDtv(x, k l t) = r2l+kDxDtu(rx, k l r2t) and v L1(C(1)) = 1 rn+2 u L1(C(r)). Therefore max C(r/2) |DxDtu| k l ≤ Ckl r2l+k+n+2 u L1(C(r)). If u solves the heat equation within UT , then for each time 0 t ≤ T , the mapping x → u(x, t) is analytic. (See Mikhailov [M].) However the mapping t → u(x, t) is not in general analytic. 2.3.4. Energy methods. a. Uniqueness. We investigate again the initial/boundary-value problem (40) ut − Δu = f in UT u = g on ΓT . We earlier invoked the maximum principle to show uniqueness and now— by analogy with §2.2.5—provide an alternative argument based upon inte- gration by parts. We assume as usual that U ⊂ Rn is open and bounded and that ∂U is C1. The terminal time T 0 is given.

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