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### Transcript of MIT Numerical PDE

18.336 spring 2009

lecture 1

02/03/09

18.336 Numerical Methods for Partial Dierential Equations Fundamental ConceptsDomain Rn with boundary PDE in b.c. on

PDE = partial dierential equation b.c. = boundary conditions (if time involved, also i.c. = initial conditions) Def.: An expression of the form F (Dk u(x), Dk1 u(x), ..., Du(x), u(x), x) = 0, x Rn (1)

is called k th order PDE, k k1 where F : Rn Rn ... Rn R R is given, and u : R is the unknown. A function u satisfying (1) is called solution of the PDE. = (ux1 , ..., uxn ) u x1 x1 . . . .. . D2 u = . . . u xn x1 . . . Du gradient (vector) ux1 xn . . Hessian (matrix) . u xn xn . . . etc. 1

Def.: The PDE (1) is called... (i) linear, if ||k

a (x)D u = f (x)

homogeneous, if f = 0 (ii) semilinear, if ||k

a (x)D u + F0 (Dk1 u, ..., Du, u, x) = 0

(iii) quasilinear, if a (Dk1 u, ..., Du, u, x) D u + F0 (Dk1 u, ..., Du, u, x) = 0||k

(iv) fully nonlinear, if neither (i), (ii) nor (iii). Def.: An expression of the form F (Dk u(x), Dk1 u(x), ..., Du(x), u(x), x) = 0, is called k th order system of PDE, where F : Rmn Rmn ... Rmn Rm Rm and u : Rm , u = (u1 , ..., um ). Typically: # equations = # unknowns , i.e. n = m. Some examples: ut + ux = 0 linear advection equation ut = uxx heat equation uxx = f (x) Poisson equation (1D) 2 u = f Poisson equation (nD) ut + cux = Duxx convection diusion equation ut + ( 1 u2 )x = 0 ut + u ux = 0 Burgers equation (quasilinear) 2 2 u = u2 a semilinear PDE utt = uxx wave equation (1D) u 0 1 u wave equation, written as a system = v t 1 0 v x utt = vxt = vtx = uxx ut + uux = uxxx Korteweg-de-Vries equation ut + ( u ) u = p + 2 u incompressible Navier-Stokes equation u=0 [dynamic-algebraic system] shallow water equations ht + (uh)x = 0 ut + uux = ghx [system of hyberbolic conservation laws] |u| = 1 Eikonal equation (nonlinear) 2k k1

x Rn

MIT OpenCourseWare http://ocw.mit.edu

18.336 Numerical Methods for Partial Differential EquationsSpring 2009

18.336 spring 2009

lecture 2

02/05/09

Well-PosednessDef.: 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( ) = 1 unique solution u(x) = sin(x) 2 b) u(0) = 0, u( ) = 1 no solution c) u(0) = 0, u( ) = 0 innitely 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

well-posed

no continuous dependence on initial data [later]

Notions of SolutionsClassical 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

Ex.: Discontinuous coecients ( b ( x ) u ) = 0 x x 4 u (0) = 0 x x< 1 3 2 u(1) = 1 u ( x ) = 2 1 1 x + x 1 3 3 2 1 x < 2 b(x) = 2 x 1 2 Ex.: Conservation laws u2 )x = 0 Burgers equation ut + ( 1 2

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Fourier Methods for Linear IVPIVP = initial value problem advection equation ut = ux ut = uxx heat equation ut = uxxx Airys equation ut = uxxxx w=+ a) on whole real axis: u(x, t) = eiwx u (w, t)dw Fourier transform

w=

b) periodic case x [, [: u(x, t) =

+ k=

u k (t)eikx Fourier series (FS)

Here case b). u nu PDE: (x, t) n (x, t) = 0 t x + du k n insert FS: (t) (ik ) u k (t) eikx = 0 dt k= Since (eikx )kZ linearly independent: du k = (ik )n u k (t) ODE for each Fourier coecient dt 2

Solution:

Fourier coecient of initial conditions: u k (0)= 21 + k=

u k (t) = e(ik) t u (0) k

n

u0 (x)eikx dx

u(x, t) = k k k

u k (0)eikx e(ik)

nt

n = 1: u(x, t) = n = 2: u(x, t) = n = 3: u(x, t) = n = 4: u(x, t) = Message:

u k (0)eik(x+t) u k (0)eikx ek u k (0)eik(xk u k (0)eikx ek2t

all waves travel to left with velocity 1 frequency k decays with ek2 t

2 t)

k

4t

frequency k travels to right with velocity k 2 dispersion all frequencies are amplied unstable

For linear PDE IVP, study behavior of waves eikx . The ansatz u(x, t) = eiwt eikx yields a dispersion relation of w to k . The wave eikx is transformed by the growth factor eiw(k)t . Ex.: wave equation: heat equation: conv.-diusion: Schr odinger: 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 conservative dissipative dissipative dispersive dispersive |eickt | = 1 2 |edk t | 0 2 |eickt edk t | 0 2 |eik t | = 1 3 |eik t | = 1

3

MIT OpenCourseWare http://ocw.mit.edu

18.336 Numerical Methods for Partial Differential EquationsSpring 2009

18.336 spring 2009

lecture 3

02/10/09

Four Important Linear PDELaplace/Poisson equation 2 u = f in u = g on 1 Dirichlet boundary condition u = h on 2 Neumann boundary condition n 2 = 1

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f 0 Laplace equation 2 u = 0 u = harmonic function Physical example: Heat equation: ut 2 u = f Dirichlet: prescribe u = g Neumann: prescribe ux u =h nsource

stationary (t ) : ut = 0 2 u = f

1

Fundamental solution of Laplace equation: = Rn no boundary conditions Radially symmetric solution in Rn \{0} : r 2xi xi n 1 2 xi = 2|x| = r 2 r = |x| = xi r 2 2 r 1 r xi x 1 xi i i=1 = = 3 xi 2 r2 r r u(x) = v (r) r uxi = v (r) x i 2 r 2r xi 2 1 xi 2 uxi xi = v (r) + v (r) = v (r) 2 + v (r) 3 xi xi 2 r r r n n 1 2 u = uxi xi = v (r) + v (r) r i=1 Hence: n1 v (r) = 0 r v (r) 1n v =0 (log v (r)) = = r v (r) log v (r) = (1 n) log r + log b 2 u = 0 v (r) +1n v (r) = b r br + c n = 1 b log r + c n = 2 v (r) = b + c n 3 rn2

Def.: The function 1 n = 1 2 |x| log |x| n = 2 (x= 0, (n) = volume of unit ball in Rn ) (x) = 21 1 1 n3 n(n2)(n) |x|n2 is called fundamental solution of the Laplace equation. Rem.: In the sense of distributions, is the solution to 2 (x) =Dirac delta

(x)

2

Poisson equation: Given f: Rn R, u(x) = (x y )f (y ) dy (convolution) solves u(x) = f (x). Motivation: 2 u(x) =R2 Rn 2

x (x y )f (y ) dy =R2

2

(x y )f (y ) dy = f (x).

is a Greens function for the Poisson equation on Rn . Properties of harmonic functions: Mean value property u harmonic u(x) = B (x,r) u ds u(x) = B (x,r) u dy for any ball B (x, r) = {y : ||y x|| r}. Implication: u harmonic u C Proof: u(x) = Rn B (0,r) (x y )u(y ) dy u Ckconvolution average average

=

u C k+1

Maximum principle Domain Rn bounded. (i) u harmonic max u = max u

(weak MP) (strong MP)

(ii) connected; u harmonic If x0 : u(x0 ) = max u, then u constant

Implications u u max min uniqueness of solution of Poisson equation with Dirichlet boundary conditions 2 u = f in u = g on Proof: Let u1 , u2 be two solutions. Then w = u1 u2 satises 2 w = 0 in max principle = w 0 u1 u2 w = 0 on

3

Pure Neumann Boundary Condition: 2 u = f in u = h on h has innitely many solutions (u u + c), if f dx = h dS. Otherwise no solution. Compatibility Condition: 2 f dx = u dx = divf dx = f n dS =

f dS = n

h dS.

4

MIT OpenCourseWare http://ocw.mit.edu

18.336 Numerical Methods for Partial Differential EquationsSpring 2009

18.336 spring 2009 Heat equation ut = 2 u Physics:

lecture 4

02/12/09

Ficks law: ux F = au d mass balance: u dx = b F n dS = b divF dx dt V V V ut = b div(au) = c 2 u

simple: c = 1

Fundamental Solution |x|2 1 4t (x, t) = e (4t)n/2 ut = 2 u in Rn ] 0, [ solves u(x, 0) = (x) t = 0

Superposition: u(x, t) = (x y, t)u0 (y ) dy Rn ut = 2 u in Rn ] 0, [ solves u(x, 0) = u0 (x) Maximum Principle Rn bounded T := [0, T ], T = ( {0}) ( ] 0, T [ )

C u C

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1

If u is the solution to ut = 2 u in T u = u0 on {0} u=g on ] 0, T [ then (i) max u = max uT T

(weak)

(ii) For is connected: If (x0 , t0 ) T : u(x0 , t0 ) = max u, then u = constantT

(strong)

Implications: max min uniqueness (see Poisson equation) innite speed of propagation: ut = 2 u in T u=0 on ] 0, T [ u=g on {0} g 0 = u > 0 in T . Inhomogenous Case: ut 2 u = f in Rn ] 0, [ u=0 on Rn {0} t solution: u(x, t) = (x y, t s)f (y, s) dy ds0 Rn strong max principle

Duhamels Principle (variation of constants): superposition of solutions starting at s with initial conditions f (s). Transport equation ut + b u = 0 in Rn ] 0, [ u = u0 on Rn {0} solution: u(x, t) = u0 (x tb). check: ut = b u0 (x tb) = b u Inhomogenous Case: ut + b u = f in Rn ] 0, [ u = u0 on Rn {0} Duhamels principle yields the solution: t

b = direction vector (eld)

u(x, t) = u0 (x tb) +0