By Chin-Yuan Lin
This quantity is on initial-boundary price difficulties for parabolic partial differential equations of moment order. It rewrites the issues as summary Cauchy difficulties or evolution equations, after which solves them by way of the means of basic distinction equations. due to this, the amount assumes much less heritage and offers a simple strategy for readers to understand.
Readership: Mathematical graduate scholars and researchers within the zone of study and Differential Equations. it's also strong for engineering graduate scholars and researchers who're drawn to parabolic partial differential equations.
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Additional resources for An Exponential Function Approach to Parabolic Equations
1, Chapter 1 that F (t) satisﬁes the dissipativity condition (H1). Step 2. 13, Chapter 4], the range of (I − λF (t)), λ > 0, equals C[0, 1], so F (t) satisﬁes the range condition (H2). Step 3. ) Let gi (x) ∈ C[0, 1], i = 1, 2, and let v1 = (I − λF (t))−1 g1 ; v2 = (I − λF (τ ))−1 g2 ; page 42 July 9, 2014 17:2 9229 - An Exponential Finction Approach to Parabolic Equations 3. EXAMPLES main4 43 where λ > 0 and 0 ≤ t, τ ≤ T . Then (v1 − v2 ) − λ(v1 − v2 ) = λ[f0 (x, t) − f0 (x, τ )] + (g1 − g2 ), and so v1 − v2 ∞ ≤ g1 − g 2 ∞ + λ max |f0 (x, t) − f0 (x, τ )| ≤ g1 − g2 ∞ + λ|ζ(t) − ζ(τ )|, x∈[0,1] proving the condition (HA).
N ˆ Here 0 < μ < 1, is a constant. It will be shown  that G(t) satisﬁes the four conditions, namely, the dissipativity condition (H1), the weaker range condition (H2) , the time-regulating condition (HA), and the embedding condition (HB). 1. 2. 1). Under additional assumptions on u0 and f0 (x, t), we will make further estimates, so that u(t) for u0 ∈ D(G(0)) with G(0)u0 = ( u0 + f0 (x, 0)) ∈ D(G(0)) is, in fact, a unique classical solution. We now begin the proof, which is composed of eight steps.
6). 3. The Linear Nonhomogeneous Case. 2: Jμn x = (I − μB)−n x; λ ≥ μ > 0, λω < 1; Jλm x = (I − λB)−m x; α, β > 0, α + β = 1. Here x ∈ D(B). 7: Proof. We divide the proof into ﬁve steps. ˜ satisﬁes the range condition (A1). For, Step 1. As a nonlinear operator, B the equation ˜ = w, u − λBu ˜ = D(B), is the same as the equation where w is a given element in D(B) u − λBu = w + λf0 , and f0 satisﬁes the condition (F0). Here λ < λ0 . Step 2. Since the linear operator B satisﬁes the dissipativity condition (B2) or ˜ −1 w is single-valued the mixture condition (B3), it follows from Step 1 that (I − λB) for w ∈ D(B), and that ˜ −1 w1 ˜ −1 w2 − (I − λB) (I − λB) = (I − λB)−1 (w2 − w1 ) for w1 , w2 ∈ D(B).
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