University:
Massachusetts Institute of Technology
Course:
18.336 | Numerical Methods for Partial Differential Equations
Academic year:

2024
Views:

401

Pages:

3
Author:

carranzasails1ue

18.336 spring 2009 lecture 2 02/05/09 Well-Posedness Def.: A PDE is called well-posed (in the sense of Hadamard), if (1) a solution exists (2) the solution is unique (3) the solution depends continuously on the data (initial conditions, boundary conditions, right hand side) Careful: Existence and uniqueness involves boundary conditions Ex.: uxx + u = 0 a) u(0) = 0, u( π2 ) = 1 ⇒ unique solution u(x) = sin(x) b) u(0) = 0, u(π) = 1 ⇒ no solution c) u(0) = 0, u(π) = 0 ⇒ inﬁnitely many solutions: u(x) = A sin(x) Continuous dependence depends on considered metric/norm. We typically consider || · ||L∞ , || · ||L2 , || · ||L1 . Ex.: ⎧ heat equation ⎨ ut = uxx u(0, t) = u(1, t) = 0 boundary conditions ⎩ u(x, 0) = u0 (x) initial conditions ⎧ backwards heat equation ⎨ ut = −uxx u(0, t) = u(1, t) boundary conditions ⎩ u(x, 0) = u0 (x) initial conditions Notions of Solutions Classical solution k th order PDE ⇒ u ∈ C k Ex.: �2 u = 0 ⇒ u ∈ C ∞ � � ut + ux = 0 ⇒ u(x, t) ∈ C 1 u(x, 0) ∈ C 1 Weak solution k th order PDE, but u ∈ / Ck. 1 ⎫ ⎬ well-posed ⎭ ⎫ ⎬ no continuous dependence ⎭ on initial data [later] Ex.: Discontinuous coeﬃcients ⎧ ⎫ (b(x)u ) = 0 ⎪ ⎪ x x ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ � 4 u(0) = 0 ⎨ ⎬ x 3 u(1) = � 1 ⇒ u(x) = 2 � x+ ⎪ ⎪ 3 ⎪ ⎪ 1 x < 12 ⎪ ⎪ ⎪ ⎪ ⎩ b(x) = ⎭ 2 x ≥ 12 1 3 x< x≥ 1 2 1 2 Ex.: Conservation laws ut + ( 12 u2 )x = 0 Burgers’ equation Image by MIT OpenCourseWare. Fourier Methods for Linear IVP IVP = initial value problem advection equation ut = ux ut = uxx heat equation ut = uxxx Airy’s equation ut = uxxxx � w=+∞ a) on whole real axis: u(x, t) = eiwx û(w, t)dw Fourier transform w=−∞ +∞ � b) periodic case x ∈ [−π, π[: u(x, t) = ûk (t)eikx Fourier series (FS) k=−∞ Here case b). ∂u ∂ nu PDE: (x, t) − n (x, t) = 0 ∂t ∂x � +∞ � � dûk n insert FS: (t) − (ik) ûk (t) eikx = 0 dt k=−∞ Since (eikx )k∈Z linearly independent: dûk = (ik)n ûk (t) ODE for each Fourier coeﬃcient dt 2 n ûk (t) = e(ik) t ûk (0) � �� Solution: Fourier coeﬃcient of initial conditions: ûk (0)= 21π +∞ � ⇒ u(x, t) = �π −π u0 (x)e−ikx dx nt ûk (0)eikx e(ik) k=−∞ n = 1: u(x, t) = n = 2: u(x, t) = n = 3: u(x, t) = � k � k � ûk (0)eik(x+t) ûk (0)eikx e−k ûk (0)eik(x−k all waves travel to left with velocity 1 2t 2 t) k n = 4: u(x, t) = � ûk (0)eikx ek 4t k frequency k decays with e−k 2t frequency k travels to right with velocity k 2 → dispersion all frequencies are ampliﬁed → unstable Message: For linear PDE IVP, study behavior of waves eikx . The ansatz u(x, t) = e−iwt eikx yields a dispersion relation of w to k. The wave eikx is transformed by the growth factor e−iw(k)t . Ex.: wave equation: heat equation: conv.-diﬀusion: Schrödinger: Airy equation: utt = c2 uxx ut = duxx ut = cux +duxx iut = uxx ut = uxxx w w w w w = ±ck = −idk 2 = −ck−idk 2 = −k 2 = k3 3 conservative dissipative dissipative dispersive dispersive |e±ickt | = 1 2 |e−dk t | → 0 2 |eickt e−dk t | → 0 2 |eik t | = 1 3 |e−ik t | = 1

Well-Posedness

Related Documents

Recommended Documents

Description

Related Documents

Free up your schedule!

Take 5 seconds to unlock