MEC702_Wk_13_Lect_1

# MEC702_Wk_13_Lect_1 - MEC702 Wk 13 Lect 1 Partial...

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Unformatted text preview: MEC702 Wk 13 Lect 1 Partial Dierential Equations (PDEs) Denition. • A partial dierential equation (PDE) is an equation involving one or more partial derivatives of an (unknown) function, call it u, that depends on two or more variables, often time t and one or several variables in space. • Just as ODEs, we have that a PDE is linear if the function u and all its partial derivatives are of the rst degree. • But the denition of homogeneous PDE is dierent from homogeneous ODE. A PDE is homogeneous if all the terms of the PDE contain either u or one of its partial derivatives. Otherwise, the PDE is nonhomogeneous. Example. Important PDEs. ∂2u ∂2u = c2 2 2 ∂t ∂x 2 ∂u ∂ u = c2 2 ∂t ∂x ∂2u ∂2u + 2 =0 ∂x2 ∂y ∂2u ∂2u + 2 = f (x, y) ∂x2 ∂y  2  ∂2u ∂ u ∂2u 2 = c + ∂t2 ∂x2 ∂y 2 ∂2u ∂2u ∂2u + 2 + 2 =0 ∂x2 ∂y ∂z One-dimensional wave eqn One-dimensional heat eqn Two-dimensional Laplace eqn Two-dimensional Poisson eqn Two-dimensional wave eqn Three-dimensional Laplace eqn where c > 0 is a constant, t is the time variable, x, y , z are the Cartesian coordinates, and dimension is the number of these coordinates in the equation. Example. Verify that the following functions are all solutions to the twodimensional Laplace equation. 1. u = x2 − y 2 2. u = ex cos y 1 3. u = ln(x2 + y 2 ) The above example shows that number of solutions that a PDE can have is very large. To obtain a unique solution corresponding to a physical problem, we make use of additional conditions which maybe • Initial Conditions - when time t is one of the variables, the value of the function u may be given for t = 0 • Boundary Conditions - for the region R where the function u is dened, values of u may be given at the boundaries of R. Remark. A partial derivative ∂f ∂x can also be written as fx , hence ∂2f ∂x2 = fxx . Theorem. (Fundamental Theorem on Superposition) If u1 and u2 are solutions of a homogeneous linear PDE in some region R, then u = c1 u1 + c2 u2 with any constants c1 and c2 is also a solution of that PDE in the region R. Example. Solving the PDE uxx − u = 0 like an ODE. Find the solutions u of the PDE uxx − u = 0 depending on x and y . Solution. Since no y-derivatives occur, we can solve this PDE like the ODE u00 − u = 0. For the ODE, we have the solution u = Aex + Be−x with constant A and B . For the PDE, we have the solution u(x, y) = A(y)ex + B(y)e−x with arbitrary functions A(y) and B(y), thus obtaining a variety of solutions. Example. Solving the PDE uxy = ux Find solutions u = u(x, y) of this PDE. like an ODE. Solution. Setting ux = p, we have py = −p. We have py = p ln |p| = p = −1 −y + ∂(x) c(x)e−y and by integration with respect to x, where u(x, y) = f (x)e−y + g(y) where f (x) and g(y) are arbitrary functions. 2 f (x) = ´ c(x)dx Modeling: Vibrating String, Wave Equation The model of a vibrating elastic string (a violin string, for instance) consists of the one-dimensional wave equation ∂2u ∂2u = c2 2 2 ∂t ∂x where c2 = Tρ (the term c2 indicates that the constant is positive). Since the string is fastened at the ends x = 0 and x = L, we have the boundary conditions u(0, t) = 0 u(L, t) = 0 for all t ≥ 0. Furthermore, the form of the motion of the string will depend on its initial deection (deection at time t = 0), call it f (x), and on its initial velocity (velocity at t = 0), call it g(x). Thus, we have the two initial conditions u(x, 0) = f (x) ut (x, 0) = g(x) for all 0 ≤ x ≤ L. Solution by Separating Variables. Use of Fourier Series. We will now solve the one-dimensional wave equation. Step 1: Two ODEs from the Wave Equation In the method of separating variables, or product method, we determine solutions of the wave equation of the form u(x, t) = F (x) · G(t) which are product of two functions. This is a powerful general method that has various applications in engineering mathematics. Dierentiating the two functions, ∂2u ∂2u 00 00 and ∂t2 = F G ∂x2 = F G. Inserting into the wave equation gives F G00 = c2 F 00 G. 3 Dividing by c2 F G and simplifying gives 1 G00 F 00 = . c2 G F The variables are now separated; the left-side depending only on t and the right-side only on x. Hence both sides must be constant. Thus, 1 G00 F 00 = = k. c2 G F Multiplying by the denominators gives immediately two ODEs F 00 − kF = 0 and G00 − c2 kG = 0 where the separation constant k is still arbitrary. Step 2: Satisfying the Boundary Conditions We now determine solutions F (x) and G(t) of the wave equation so that u = F (x)G(t) satises the boundary conditions, that is, u(0, t) = F (0)G(t) = 0 u(L, t) = F (L)G(t) = 0 for all t. We now solve for F (x). If G ≡ 0, then u = F G = 0, which is of no interest. Hence consider G 6= 0 and then by the boundary conditions, F (0) = 0, F (L) = 0 . We must show that k must be negative. • Assume that k = 0, the general solution of F 00 − kF = 0 is F = ax + b, and from the conditions above, a = b = 0, so that F ≡ 0 and u = F G ≡ 0, which is of no interest. • Assume that positive k = µ2 , a general solution of F 00 − µ2 F = 0 is F (x) = Aeµx + Be−µx , and from the conditions above, we obtain F ≡ 0 as before. 4 • Hence we are left with the possibility of choosing k to be negative, say k = −p2 . Then F 00 + p2 F = 0 and has a general solution F (x) = A cos px + B sin px. From this and the boundary conditions, we have F (0) = A=0 F (L) = B sin pL = 0. We must assume B 6= 0 otherwise F ≡ 0. Then we have sin pL = 0, thus so that pL = nπ p= nπ L . Setting B = 1, we thus obtain innitely many solutions F (x) = Fn (x), where   Fn (x) = sin nπ x L for n = 1, 2, . . .. We next solve for G(t). We have k = −p2 = − then G00 + λ2 G = 0 where  nπ 2 L λn = cp = cnπ L . We obtain the general solution of Gn (t) is Gn (t) = Bn cos (λn t) + Bn∗ sin (λn t) Therefore the general solution of the one-dimensional wave equation h  nπ i un (x, t) = [Bn cos (λn t) + Bn∗ sin (λn t)] sin x L where n = 1, 2, . . .. • For each n = 1, 2, . . ., the functions un (x, t) are called the eigenfunctions, or characteristics functions. • The values λn = cnπ L the vibrating string. are called eigenvalues, or characteristic values, of • The set [λ1 , λ2 , . . .] is called the spectrum. 5 Step 3: Solution of the Entire Problem. Fourier Series. By the Fundamental Theorem on Superposition, the complete general solution of the one-dimensional wave equation is u(x, t) = ∞ X un (x, t) = n=1 where λn = ∞ h  nπ i X x (Bn cos (λn t) + Bn∗ sin (λn t)) sin L n=1 which is an innite series. xnπ L Applying Initial Conditions. • Applying the rst initial condition u(x, 0) = f (x) (initial displacement), u(x, 0) = ∞ X Bn sin n=1  nπ  x = f (x). L Hence we must choose the terms Bn so that u(x, 0) becomes the Fourier sine series of f (x), thus 2 Bn = L ˆ L f (x) sin  nπx  0 L dx • Applying the second initial condition ut (x, 0) = g(x) (initial velocity), "∞ #  nπ o Xn ∂u ∗ = x [−Bn λn sin (λn t) + Bn λn cos (λn t)] sin ∂t t=0 L n=1 t=0 = ∞ X  nπ  Bn∗ λn sin x = g(x). L n=1 This gives a Fourier sine series of g(x). Then the Fourier coecients are Bn∗ λn Since λn = cnπ L 2 = L ˆ L g(x) sin 0  nπ  x dx L , we obtain by division Bn∗ 2 = cnπ ˆ L g(x) sin 0 6  nπ  x dx L Establishing the solution. The general solution is purely a formal expression. For the sake of simplicity, we consider only the case when initial velocity g(x) = 0. Then the terms Bn∗ = 0, thus u(x, t) = P∞  n=1 Bn cos (λn t) sin nπx L  , λn = cnπ L . Using the formula, cos  nπ   nπ  1 h  nπ   nπ i sin ct sin x = (x − ct) + sin (x + ct) , L L 2 L L we have u(x, t) = = = ∞ X Bn cos (λn t) sin  nπx  n=1 ∞ X L h  nπ   nπ i 1 Bn sin (x − ct) + sin (x + ct) 2 n=1 L L "∞ # ∞ n nπ o X n nπ o 1 X Bn sin (x − ct) + Bn sin (x + ct) . 2 n=1 L L n=1 Notice the two Fourier sine series. Denote f ∗ (x) = Bn ∞ X n=1 then u(x, t) = sin  nπ  x , L 1 ∗ [f (x − ct) + f ∗ (x + ct)] 2 where f ∗ (x) is the odd extension of f (x) with the period 2L. 7 ...
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• Winter '16
• Alveen Chand
• Sin, Partial differential equation, wave equation, one-dimensional wave equation

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