2 orthogonal collocation technique applicable to parabolic partial differential equations
Partial Differential Equations part 2
... differencing of equation (19.1 .20 ) is n+1 n rj+1 /2 − rj+1 /2 n+1 /2 sj n+1 /2 = ∆t n−1 /2 − sj =v ∆t n+1 /2 − sj ∆x n n rj+1 /2 − rj−1 /2 sj+1 (19.1.35) ∆x If you substitute equation (19.1 .22 ) in equation (19.1.35), ... shows that the amplification factor for equation (19 .2. 4) is ξ =1− 4D∆t sin2 (∆x )2 k∆x (19 .2. 5) The requirement |ξ| ≤ leads to the stability criterion 2D∆t ≤1 (∆x )2 (19 .2. 6) Sample page from NUMERICAL ... values uj+1 /2 at the half timesteps tn+1 /2 and the half mesh points xj+1 /2 These are calculated by the Lax scheme: n+1 /2 uj+1 /2 = n ∆t (u (F n − Fjn ) + un ) − j j+1 2 x j+1 (19.1.37) n+1 /2 Using...
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Tài liệu AN INTRODUCTION TO PARTIAL DIFFERENTIAL EQUATIONS ppt
... Index ix 22 6 22 6 22 6 22 8 23 4 24 2 25 8 25 9 26 1 26 3 26 6 26 9 27 5 27 9 28 2 28 2 29 6 306 309 309 311 3 12 318 322 324 329 334 337 361 3 62 3 62 363 363 364 366 Preface This book presents an introduction to the ... 173 173 178 181 1 82 184 187 20 1 20 4 20 8 20 8 20 9 21 9 22 1 22 3 Contents Equations in high dimensions 9.1 Introduction 9 .2 First-order equations 9.3 Classification of second-order equations 9.4 The ... form of hyperbolic equations 3.4 Canonical form of parabolic equations 3.5 Canonical form of elliptic equations 3.6 Exercises vii page xi 1 3 17 20 21 23 23 24 25 30 36 39 41 50 52 58 64 64 64 67...
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introduction to partial differential equations - a computational approach - a. tveito, r. winther
... 20 9 20 9 21 2 21 3 21 6 21 7 21 8 22 2 22 5 22 5 22 8 23 0 23 0 23 1 23 4 23 6 23 7 Orthogonality and General Fourier Series 8.1 The Full Fourier Series 8.1.1 Even and Odd Functions 8.1 .2 Differentiation ... 117 119 122 123 127 129 130 131 1 32 133 134 137 140 143 145 148 2. 3 2. 4 2. 5 2. 6 The 3.1 3 .2 3.3 3.4 3.5 3.6 2. 2.5 Positive Definite Matrices Continuous and Discrete Solutions 2. 3.1 Difference ... 25 6 25 7 25 7 26 1 26 4 26 7 27 3 27 6 28 5 28 5 29 0 29 6 300 3 02 307 10 The Heat Equation Revisited 10.1 Compatibility Conditions 10 .2 Fourier’s...
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Introduction to Partial Differential Equations: A Computational ApproachAslak Tveito Ragnar potx
... 20 9 20 9 21 2 21 3 21 6 21 7 21 8 22 2 22 5 22 5 22 8 23 0 23 0 23 1 23 4 23 6 23 7 Orthogonality and General Fourier Series 8.1 The Full Fourier Series 8.1.1 Even and Odd Functions 8.1 .2 Differentiation ... 117 119 122 123 127 129 130 131 1 32 133 134 137 140 143 145 148 2. 3 2. 4 2. 5 2. 6 The 3.1 3 .2 3.3 3.4 3.5 3.6 2. 2.5 Positive Definite Matrices Continuous and Discrete Solutions 2. 3.1 Difference ... 25 6 25 7 25 7 26 1 26 4 26 7 27 3 27 6 28 5 28 5 29 0 29 6 300 3 02 307 10 The Heat Equation Revisited 10.1 Compatibility Conditions 10 .2 Fourier’s...
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Partial Differential Equations part 1
... 19.0 .2) , uj+1,l − 2uj,l + uj−1,l uj,l+1 − 2uj,l + uj,l−1 + = ρj,l 2 2 (19.0.5) or equivalently uj+1,l + uj−1,l + uj,l+1 + uj,l−1 − 4uj,l = 2 ρj,l (19.0.6) To write this system of linear equations ... vector F is called the conserved flux For example, the prototypical hyperbolic equation, the one-dimensional wave equation with constant velocity of propagation v ∂2u ∂2u = v2 ∂t2 ∂x (19.1 .2) ... Numerical Methods for Partial Differential Equations, 2nd ed (New York: Academic Press) [1] Richtmyer, R.D., and Morton, K.W 1967, Difference Methods for Initial Value Problems, 2nd ed (New York:...
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Tài liệu Partial Differential Equations part 3 pptx
... North America) FTCS 851 19 .2 Diffusive Initial Value Problems and the heuristic stability criterion is ∆t ≤ j (∆x )2 2Dj+1 /2 (19 .2. 21) Dj+1 /2 = D(un ) + D(un ) j+1 j (19 .2. 22) Implicit schemes are ... North America) τ∼ 849 19 .2 Diffusive Initial Value Problems ∂2u =0 ∂x2 (19 .2. 11) What about stability? The amplification factor for equation (19 .2. 8) is ξ= + 4α sin2 k∆x (19 .2. 12) Clearly |ξ| < for ... stability A slight generalization of (19 .2. 8) leads to i =− n+1 n+1 n+1 ψj+1 − 2 j + ψj−1 n+1 + Vj ψj (∆x )2 (19 .2. 27) for which ξ= k∆x 4∆t 1+i sin2 (∆x) (19 .2. 28) + Vj ∆t This is unconditionally stable,...
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Tài liệu Partial Differential Equations part 4 ppt
... going to see all hell break loose! 19.3 Initial Value Problems in Multidimensions 855 The expression for |ξ |2 can be manipulated into the form |ξ |2 = − (sin2 kx∆ + sin2 ky ∆) − ( 2 + 2 ) x y ... The last two terms are negative, and so the stability requirement |ξ |2 ≤ becomes − ( 2 + 2 ) ≥ x y ∆ ∆t ≤ √ 2 2(vx + vy )1 /2 or (19.3.9) (19.3.10) This is an example of the general result for ... ∆ )2 − (αy sin kx ∆ − αx sin ky ∆ )2 856 Chapter 19 Partial Differential Equations (19.3.13) Called the alternating-direction implicit method (ADI), this embodies the powerful concept of operator...
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Tài liệu Partial Differential Equations part 5 ppt
... (19.0.6), we find umn e2πim/J + e 2 im/J + e2πin/L + e 2 in/L − = ρmn 2 or umn = ρmn 2 2πn 2 m + cos 2 cos J L (19.4.4) (19.4.5) Thus the strategy for solving equation (19.0.6) by FFT techniques is: ... ∂x 8 62 Chapter 19 Partial Differential Equations The finite-difference form of equation (19.4 .28 ) can be written as a set of vector equations uj−1 + T · uj + uj+1 = gj 2 (19.4 .29 ) T = B − 21 (19.4.30) ... and then adding the three equations, we get (1) uj 2 + T(1) · uj + uj +2 = gj 2 (19.4. 32) This is an equation of the same form as (19.4 .29 ), with T(1) = 21 − T2 (1) gj = 2 (gj−1 − T · gj + gj+1...
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Tài liệu Partial Differential Equations part 6 doc
... the asymptotic formula for large J turns out to be ρs 1− 2 2J (19.5.11) The number of iterations r required to reduce the error by a factor of 10−p is thus r 2pJ ln 10 2 pJ (19.5. 12) In other ... − 2 Jacobi (19.5 .20 ) + π/J 2 1− J ω ρSOR (19.5 .21 ) for large J (19.5 .22 ) Equation (19.5.10) gives for the number of iterations to reduce the initial error by a factor of 10−p , r pJ ln 10 2 ... Chapter 19 Partial Differential Equations The Gauss-Seidel method, equation (19.5.6), corresponds to the matrix decomposition (L + D) · x(r) = −U · x(r−1) + b (19.5.13) ρs r 1− 2 J2 pJ ln 10 2 (19.5.14)...
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Tài liệu Partial Differential Equations part 7 doc
... + h2 u2 (x, y) + · · · Then, assuming R is of high enough order that we can neglect its effect, equation (19.6. 32) gives τh LH (u + h2 u2 ) − Lh (u + h2 u2 ) = LH (u) − Lh (u) + h2 [LH (u2 ) ... Problems 881 u [2] [2] = -h*h*rhs [2] [2] /4.0; } h=1.0/(n-1); h2=h*h; for (ipass=1;ipass< =2; ipass++,jsw=3-jsw) { Red and black sweeps isw=jsw; for (j =2; j
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Functional analysis sobolev spaces and partial differential equations
... Universitext For other titles in this series, go to www.springer.com/series /22 3 Haim Brezis Functional Analysis, Sobolev Spaces and Partial Differential Equations 1C Haim Brezis Distinguished Professor ... interesting application to PDEs, due toY Pinchover, is presented in S Agmon [2] For a proof of the Krein–Milman theorem, see Problem (b) In the theory of partial differential equations Let us mention, ... E; x2n = ∀n ≥ 1} and Y = y = (yn )n≥1 ∈ E; y2n = y2n−1 ∀n ≥ 2n Check that X and Y are closed linear spaces and that X + Y = E Let c ∈ E be defined by 24 The Hahn–Banach Theorems Introduction to...
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Tài liệu Boundary Value Problems, Sixth Edition: and Partial Differential Equations pptx
... References 124 Chapter Review 125 Miscellaneous Exercises 125 v vi CHAPTER Contents The Heat Equation 135 2. 1 2. 2 2. 3 2. 4 2. 5 2. 6 2. 7 2. 8 2. 9 2. 10 2. 11 2. 12 2.13 CHAPTER The Wave Equation 21 5 3.1 3 .2 3.3 ... equation Be careful to identify the dependent and independent variables d2 φ + 2 φ = dx2 d2 φ − 2 φ = dx2 d2 u = dt dT = − 2 kT dt d dw r r dr dr − 2 w = r2 d2 R dR − n(n + 1)R = + 2 dρ dρ In Exercises ... v1 u1 + v2 u2 = v1 u1 + v2 u2 , (13) up = v1 u1 + v2 u2 + v1 u1 + v2 u2 , (14) then we find that and the equation that results from substituting Eq (11) into Eq (9) becomes v1 u1 + v2 u2 + v1 (u1...
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Partial Differential Equations and Fluid Mechanics doc
... 29 2 A quantum groups primer, S MAJID 21 6 21 7 21 8 22 0 22 1 22 2 22 3 22 4 22 5 22 6 29 3 29 4 29 5 29 6 29 7 29 8 29 9 300 301 3 02 303 304 305 306 307 308 309 310 311 3 12 313 314 315 316 317 318 319 320 321 ... 318 319 320 321 322 323 324 325 326 327 328 329 330 331 3 32 333 334 335 336 337 338 339 340 341 3 42 343 344 345 346 347 348 349 350 351 3 52 353 354 355 356 357 358 359 360 361 3 62 363 364 365 366 ... m j=1 m j=1 m j=1 ⎞1 ⎛ 2 ⎠ ⎝ j ⎞ m j=1 ∂ϕj ∂x2 2 ⎠ μ4 x2 dx + j 2 (C0 )2 16 ⎠ 2 x2 dx m C0 Ω j=1 2 2 ⎠ dx + j ⎞1 ∂ϕj ∂x2 μ8 x4 dx + Ω C0 dx m Ω j=1 ∂ϕj ∂x2 dx 12 M Boukrouche & G Lukaszewicz...
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Entropy and partial differential equations evans l c
... imply V2 V1 Let T1 → T2 = T∗ : T2 ΛV (V, T2 )dV = T2 − T1 V2 T2 T1 V2 ΛV (V, T∗ )dV = T∗ V1 V1 V2 (T ) V1 (T ) ∂P dV dT ∂T ∂P (V, T∗ )dV ∂T Divide by V2 − V1 and let V2 → V1 = V∗ , to deduce (22 ) ... ∂2F ∂T ∂2F ∂V (17) ∂S = − ∂T = ∂ E ∂S ∂V ∂S ∂V + ∂2E ∂V Next differentiate (16): 1= 0= Thus (17) gives: ∂2F ∂T =− ∂2F ∂V = ∂2E ∂V ∂2E ∂S ∂2E ∂S ∂2E ∂S − ∂S ∂T ∂S ∂V + ∂2E ∂S∂V −1 ∂2E ... and (22 ) says: ΛV = T ∂P ∂T Thus ∂P ∂2S = (P − ΛV )2 < 0, − ∂V T ∂V CV T ( 32) since ∂P ∂V < 0, CV > Lastly, ∂T − ∂V ∂2S = ∂E∂V T2 39 E = ΛV − P T CV Consequently (31), ( 32) imply ∂2S ∂E ∂2S ∂V...
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harmonic analysis and partial differential equations - b. dahlberg, c. kenig
... We want to show jh 1; T 2ij C k 1k2k 2k2 for all 1; 2 S (Rn): We note that h 1; Pt2TPt2 2i ! as t ! and hence it is enough to prove Z d j dt h 1; Pt2TPt2 2idtj C k 1k2k 2k2 for all 1; 2 S (Rn) ... it is enough to show that Z j h 1; RtQtTPt2 2i dt j C k 1k2k 2k2 for all 1; 2 S (Rn): t Hence it is enough to show Z j h 1; RtQtTPt2 2i dt j C (k 1k2 + k 2k2) for all 1; 2 S (Rn): 2 t But Z Z ... 1; QtTPt2 2ij dt t 0Z dt + Z kQ TP k2 dt I + II: krt 1k2 t t t 2 t 0 33 Here I= Z ^ k ' t(jtjj j) ^1( )k2 dt C k ^1k2 = C k 1k2: t 2 To cope with II, we rewirte QtTPt2 = LtPt as ((LtPt) 2) (x) =...
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hilbert space methods for partial differential equations - r. showalter
... then (a) j(x y)j2 (x x) (y y) x y2V (b) kxk (x x)1 =2 de nes a norm k k on V for which kx + yk2 + kx ; yk2 = 2( kxk2 + kyk2 ) x y2V (c) the scalar product is continuous from V and V to K Proof : ... A show \fM : Ag V 2. 1 Prove parts (d) and (e) of Lemma 2. 1 2. 2 If V1 p1 and V2 p2 are seminormed spaces, show p(x) p1 (x1 ) + p2(x2 ) is a seminorm on the product V1 V2 2. 3 Let V p be a seminormed ... r1 ) Similarly, there is an integer n2 > n1 and s(y2 r2 ) s(y1 r1 ) such that r2 < 1 =2 UNIFORM BOUNDEDNESS WEAK COMPACTNESS 23 and j(xn2 y)H j > for y s(y2 r2 ) We inductively de ne s(yj rj ) s(yj;1...
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partial differential equations and the finite element method - pave1 solin
... 194 197 197 20 0 20 2 20 5 20 9 21 0 21 0 21 2 21 4 21 4 21 6 21 6 21 8 22 0 22 0 22 2 22 5 22 6 22 8 22 8 23 23 3 23 6 23 6 23 8 24 0 24 2 24 2 24 3 24 6 24 8 25 0 25 4 xii CONTENTS 6.6 6.7 Discretization by H2-conforming ... forms to the reference domain 7.5.5 Interpolation on edge elements 25 5 25 5 25 6 25 7 25 9 26 0 26 5 26 6 26 9 27 0 27 0 21 2 27 3 27 4 27 5 27 5 27 6 27 7 27 9 27 9 28 28 28 3 28 3 28 4 28 5 28 5 28 6 28 7 28 8 28 9 28 9 29 0 ... elements 2. 2.1 Model problem Finite-dimensional subspace V,, C 2. 2 .2 Piecewise-affine basis functions 2. 2.3 The system of linear algebraic equations 2. 2.4 v 2. 3 2. 4 24 26 29 30 30 30 31 32 33 33...
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partial differential equations, (ma3132 lecture notes) - b. neta (
... 27 7 27 7 27 8 28 0 28 1 28 2 28 2 28 3 28 4 28 6 28 6 28 7 29 1 29 3 29 3 29 4 29 6 29 6 301 301 3 02 3 02 303 306 307 311 3 12 13.9 Hyperbolic Equations 13.9.1 Stability 13.9 .2 Euler ... factor modulus for upstream differencing vi 101 1 02 103 103 103 104 104 104 127 128 160 161 167 167 173 174 191 196 197 20 8 22 7 22 7 28 0 28 0 28 6 28 6 29 1 29 2 29 2 29 3 29 4 29 5 29 7 29 7 29 8 ... ∗ = 0) B ∗ = 2y 2( −x)x + 2( −x2 )y · y = −2x2 y − 2x2 y = −4x2 y D ∗ = y 2( −1) + (−x2 ) · = −x2 − y 21 E ∗ = y · + (−x2 ) · = y − x2 Now solve (2. 3.1 .20 ) - (2. 3.1 .21 ) for x, y x2 = η − ξ, y...
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