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 onedimensional 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 (Onedimensional heat equation)
4. uxx + uyy = 0 (Twodimensional Laplace equation). 4 ...
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 Winter '16
 Alveen Chand
 Partial differential equation, wave equation, d'Alembert, UVW

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