By Michael Renardy Robert C. Rogers
Partial differential equations are basic to the modeling of typical phenomena. the need to appreciate the strategies of those equations has regularly had a favourite position within the efforts of mathematicians and has encouraged such diversified fields as advanced functionality concept, practical research, and algebraic topology. This booklet, intended for a starting graduate viewers, offers a radical advent to partial differential equations.
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Extra info for An Introduction to Partial Differential Equations, 2nd edition
Let u satisfy the wave equation. 139) d d 2 (u + u2x )/2 − (ux ut ). 140) = dt t dx We now use this in an integration over the rectangle (x, t) ∈ [0, 1] × [t0 , t1 ], in which we change the order of integration at will, and we obtain the following: 0 1 = 1 u2t (x, t1 ) + u2x (x, t1 ) dx − 0 u2t (x, t0 ) + u2x (x, t0 ) dx 0 1 t1 = 0 t0 t1 d 2 (u + u2x ) dt dx dt t 1 = 2 t0 t1 = 0 d (ux ut ) dx dt dx 2ux (1, t)ut (1, t) dt − t0 t1 2ux (0, t)ut (0, t) dt. 138). Note that the quantity we call the energy for solutions of the wave equation and the quantity we call the energy for solutions of the heat equation seem very diﬀerent mathematically.
38) In both cases, we have det Lp (iξ) = |ξ|4 . 4 Nonlinear Equations For nonlinear equations and systems, the type can depend not only on the point in space but on the solution itself. We simply linearize the equation at a given solution and deﬁne the type to be that of the linearized equation. Characteristic surfaces are similarly deﬁned as the characteristic surfaces of the linearized equation. For future use we give the deﬁnition of quasilinear and semilinear: usual way as a sum of products, then the degree of each of these products as a polynomial in ξ does not exceed the degree of the determinant.
107) 0 In the language of Chapter 6, for any solution of the heat equation satisfying the given boundary conditions, the L2 norm (in space) decreases with time. Proof. We ﬁrst use the heat equation to derive the following diﬀerential identity for u. 108) 2u2x . Integrating both sides of this identity with respect to x gives us 1 1 2 1 (u2 (x, t))t dx = u(1, t)ux (1, t) − u(0, t)ux (0, t) − 0 u2x (x, t) dx. 109) We now use the boundary conditions to eliminate the boundary terms in the equation above and integrate the result with respect to time.
An Introduction to Partial Differential Equations, 2nd edition by Michael Renardy Robert C. Rogers