MEC702_Wk_13_Lect_2

# MEC702_Wk_13_Lect_2 - MEC702 Wk 13 Lect 2 D'Alembert's...

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Unformatted text preview: MEC702 Wk 13 Lect 2 D'Alembert's Solution of the Wave Equation. First, we aim to transform the two partial derivatives in the wave equation, uxx and utt . Let w = x − ct . v = x + ct, Then vx = 1, wx = 1 . Since u(v, w) is a function of both v and w, then ux = uv vx + uw wx = uv + uw . We now apply the Extended Chain Rule to the right of the above equation. By assuming that all the partial derivatives are continuous, we have as a consequence, uwv = uvw .Then uxx = (ux )x = (uv + uw )x = (uv + uw )x = (uv + uw )v vx + (uv + uw )w wx = uvv + 2uvw + uww . Using the same procedure as above, we transform the other derivative as utt = c2 (uvv − 2uvw + uww ) . Inserting the two results, we obtain uvw = 0 Solving, we obtain u(x, t) = Φ(v) + Ψ(w), thus u(x, t) = Φ(x + ct) + Ψ(x − ct) This is known as d'Alembert's solution 1 of the wave equation. Method of Characteristics. Types and Normal Forms of PDEs. The idea of d'Alembert's solution is just a special instance of the Characteristics. This concerns PDEs of the form Method of A uxx + 2B uxy + C uyy = F (x, y, u, ux , uy ) The table below shows the three types of PDEs. Type Hyperbolic Parabolic Elliptic Condition 2 AC − B < 0 AC − B 2 = 0 AC − B 2 > 0 Example of this type Wave equation Heat Equation Laplace equation Transformation to Normal Form. The normal forms and the corresponding transformations depend on the type of PDE. These are obtain by solving the ODE Ay 02 − 2By 0 + C = 0 which is called the characteristic equation. The solutions of this ODE are called the characteristics of the PDE and are written in the form Φ(x, y) = constant, Ψ(x, y) = constant . Then the transformations which gives new variables v, w replacing x, y and the normal forms of the PDE are given in the table below. Type Hyperbolic Parabolic Elliptic New Variables Normal Form v = Φ(x, y), w = Ψ(x, y) v = x, w = Φ(x, y) = Ψ(x, y) v = 12 (Φ(x, y) + Ψ(x, y)), 1 w = 2i (Φ(x, y) + Ψ(x, y)) uvw (v, w) = F uww (v, w) = F uvv (v, w) + uww (v, w) = F Example. D'Alembert's Solution Obtained Systematically. Find the general solution of the one-dimensional wave equation utt − c2 uxx = 0. Solution. Set y = ct. Then ut = uy yt = cuy and utt = c2 uyy . 2 Then substituting, we have uxx − uyy = 0. Note that F = 0. Calculating AC − B 2 = (1)(−1) − (0)2 = −1 < 0 which shows that the wave equation is of type hyperbolic. Solving the characteristic equation Ay 02 − 2By 0 + C = 0 02 = 0 0 (y + 1)(y − 1) = 0 y0 + 1 dy dx dy ˆ dy 0 y −1 0 which gives y Φ(x, y) = y + x = = −1 = −1dx ˆ = −1 dx = −x + c1 = c1 and also Ψ(x, y) = x − y = c2 . Since it is of type hyperbolic, we obtain the new variables u and v , u = Φ(x, y) = y+x = ct + x and = x−y v = Ψ(x, y) = x − ct. Thus the normal form is uvw = F = 0 which agrees with the form we found when nding D'Alembert's solution. Solving the normal form, we obtain u(v, w) = f1 (u) + f2 (v) = f1 (x + ct) + f2 (x − ct). Find the type, transform to normal form, and solve. Show your work in detail. Example. 3 1. uxx + 4uyy = 0. 2. uxx − 16uyy = 0. 3. ut = c2 uxx (One-dimensional heat equation) 4. uxx + uyy = 0 (Two-dimensional Laplace equation). 4 ...
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