# Lecture6 - ESE 502 Mathematics of Modern Engineering II...

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Unformatted text preview: ESE 502 Mathematics of Modern Engineering II Lecture 6 ”Wave equation continued. D’Alembert’s method”. ◦ We have solved the wave equation with specified boundary/initial conditions using the method of separation of variables. ◦ It is useful to notice that the obtained solution of the wave euqation can be written in a closed form as well (that is, without an infinite series sign) Indeed: If we assume for simplicity that g = 0, then the solution is written as ∞ X nπ u(t, x) = an cos λn t sin x L n=1 or u(t, x) = ∞ X an cos n=1 cnπ nπ t sin x. L L Using the formulas for the sine of a sum and a difference of two variables, we obtain that u(t, x) = ∞ X 1 nπ nπ an [sin (x − ct) + sin (x + ct)] = 2 L L n=1 ∞ ∞ nπ nπ 1X 1X an sin an sin (x − ct) + (x + ct) 2 n=1 L 2 n=1 L where an are the Fourier coefficients of the function f . When we define by f ? the odd periodic extension of the function f on IR with the period 2L, then we can rewrite the solution u(t, x) as 1 1 u(t, x) = f ? (x − ct) + f ? (x + ct). 2 2 (1) Example 1. Let f (x) = Kx2 (1 − x), 0 &lt; x &lt; 1 where K = .01 and g(x) = 0, c2 = 1. Find the solution of the wave equation specified by the given conditions. 1 Solution: The initial deflection function satisfies the conditions needed to allow the representation through Fourier series so that the solution can be written in the form u(t, x) = ∞ X an cos n=1 where 2 an = L L Z f (x) sin 0 cnπ nπ t sin x L L nπ xdx, n = 1, 2, ... L Alternatively, we can use the formula (1) as well. Basically, we have to find only the coefficients an . Z an = 2 1 Kx2 (1 − x) sin nπxdx = 0 −Kx2 (1 − x) cos nπx 1 K 2[ |0 + nπ nπ 1 Z 2K (2x − 3x2 ) sin nπx 1 1 [ |0 − nπ nπ nπ (2x − 3x2 ) cos nπxdx] = 0 Z 1 (2 − 6x) sin nπxdx] = 0 (2 − 6x) cos nπx 1 6 2K [− | − − 0 (nπ)2 nπ nπ − − Z 1 cos nπxdx] = 0 2K 4 cos nπ + 2 [= (nπ)2 nπ 4K [1 + 2 cos nπ], n = 1, 2, ... (nπ)3 Another way of solving the wave equation: change of variables ◦ Change of variables method is a common approach when trying to solve a particular problem. The goal is that by introducing the new variables the given problem becomes the one more easier to solve in the new variables. After the transfromed problem is solved in new variables, one uses the inverse change of variables to express the solution in terms of the original variables. 2 Assume that we have to solve the wave equation utt = c2 uxx . (2) We introduce new variables (τ, y) defined as τ = x + ct and y = x − ct. To rewrite the equation (2) in new variables (τ, y), we first use the chain rule to find that: ux = uτ τx + uy yx = uτ + uy ; uxx = (uτ + uy )x = uτ τ τx + uτ y yx + uτ y τx + uyy yx = uτ τ + 2uτ y + uyy where we assume that the partial derivatives of u are continuous functions thus the mixed partial derivatives coincide. ut = uτ τt + uy yt = c(uτ − uy ); utt = c(uτ − uy )t == .... = c2 (uτ τ − 2uτ y + uyy ). In new variables the equation (2) becomes (we substitute the expressions for utt and uxx found into (2)): uτ y = 0 (3) which is much easier to solve. In fact, the equation (3) can be solved by simple integration in variables τ and y (in any order) . We obtain that u(τ, y) = G(τ ) + H(y) or u(t, x) = G(x + ct) + H(x − ct) (4) where G and H are arbitrary differentiable functions. ◦ The function u given in (4) describes the general solution of the wave equation (2). ◦ To find the unique solution of a particular wave equation, we need to use the additional boundary/initial conditions. ◦ We assume here that those conditions are specified as in the previous lecture and, for simplicity, we assume that g = 0. 3 Since u(0, x) = f (x) we obtain that f (x) = G(x) + H(x). (5) On another hand, differentiating the function u(t, x) in t, we obtain that ut (t, x) = cG0 (x + ct) − cH 0 (x − ct) so that the condition ut (0, x) = g(x) = 0 yields G0 (x) − H 0 (x) = 0. Formally, integrating the last equation in x gives us that G(x) − H(x) = K = const. Together with the equation (5) this leads to 1 1 G(x) = f (x) + K 2 2 and 1 1 H(x) = f (x) − K. 2 2 Formally, using instead of x the expressions x + ct and x − ct we have that 1 1 G(x + ct) = f (x + ct) + K 2 2 and 1 1 H(x − ct) = f (x − ct) − K. 2 2 Adding G(x + ct) and H(x − ct), we obtain that 1 u(t, x) = [f (x + ct) + f (x − ct)]. 2 (6) However, the last expression is merely a formal form of a function satisfying the wave equation. We haven’t fully used the conditions on u that should be satisfied: we used the initial conditions so far but not the boundary conditions. ◦ Using the boundary conditions, will lead to the function f that should be: - an odd function; 4 - must have the period 2L; Indeed: The conditions u(t, 0) = 0 yields that 1 [f (ct) + f (−ct)] = 0 2 so that f (−ct) = −f (ct) meaning that f must be an odd function as a function defined on IR. The conditions u(t, L) = 0 yields that 1 [f (ct + L) + f (−ct + L)] = 0 2 so that f (ct + L) = −f (−ct + L) = f (ct − L) since f is an odd function. But the relation f (ct + L) = f (ct − L) means that f must be a periodic function with period 2L defined as a function on IR. In summary, we established that the solution of the wave equation with the specified boundary/initial conditions is given in the form 1 u(t, x) = [f ? (x + ct) + f ? (x − ct)] 2 (7) wheer f ? is the odd periodic extension of the function f on IR. This solution coinsides exactly with the solution u(t, x) we obtained earlier using the method of separation of variables combined with the Fourier series techniques. Remark 1. The most important aspect in the change of variables idea we have demonstrated was to choose the change of variables in a way that the new PDE becomes a more easier one that can be solved using simple integration techniques. For a particular class of 2nd order PDEs, there is a procedure that suggests how that suitable change of variables can be found. This idea is due originally to D’Alembert so that the corresponding solution method in called D’Alembert’s method. D’Alembert’s method 5 Assume that one has to solve a second order PDE of the form Autt + 2Butx + Cuxx = F (t, x, u, ut , ux ) (8) where A, B, and C are the coefficients (they could be some real numbers but also some functions depending on (t, x)). Step 1: Consider what is called the characteristic equation of (8) given by A(x0 )2 − 2B(x0 ) + C = 0 (9) where x = x(t) is seen as the function depending on t so that x0 is the derivative of x in t. Step 2: Solve the equation (9) implicitely as an ODE. Since the eqution is quadratic in x0 , there will be two solutions and three possible cases: two different real solutions, or one real solution only (of the multiplicity 2), or two complex-conjugate solutions. Step 3: Since any implicit solution of the equation (9) can be written in a form Φ(t, x) = const, the corresponding change of variables to be made will be the following: Case I (two real roots called also the hyperbolic case): τ = Φ(t, x) and y = Ψ(t, x). In this case the change of variables reduces the equation (8) to the equation uτ y = F1 ; Case II( one real root calle also the parabolic case): τ = t and y = Φ(t, x); In this case the change of variables reduces the equation (8) to the equation uτ τ = F2 ; Case III (two complex-conjugate roots called also elliptic case): 1 1 τ = [Φ(t, x) + Ψ(t, x)] and y = [Φ(t, x) − Ψ(t, x)]. 2 2i In this case the change of variables reduces the equation (8) to the equation uτ τ + uyy = F3 ; 6 The functions Fi , i = 1, 2, 3 are some functions of the similar form as the function F . Remark 2. Change of variables in case III does not really simplify the original equation as the transformed PDE is of the same kind as the given PDE. In this case one has to use other means in trying to find a better change of variables as suggested by the D’Alembert’s method. Remark 3. The hyperbolic case is equivalent to the condition: AC −B 2 &lt; 0; The parabolic case is equivalent to the condition: AC − B 2 = 0; and The elliptic case is equivalent to the condition: AC − B 2 &gt; 0. Example 2. Find the general solution of the PDE utt − 16uxx = 0 using the D’Alembert’s method. Example 3. Do the same as in Example 2 for the PDE tutx − xuxx = 0. 7...
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• Fall '09
• Partial differential equation, wave equation, uτ τ

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