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Second-order Partial Differential Equations 1 Classiﬁcation and Characteristics The general second-order partial differential equation in two independent variables is of the form: A(x, y)uxx +2B(x, y)uxy +C(x, y)uyy +D(x, y)ux +E(x, y)uy +F (x, y)u = f (x, y), which may conveniently be written as A(x, y)uxx + 2B(x, y)uxy + C(x, y)uyy = f (x, y, u, ux , uy ), Prepared by Swaroop Nandan Bora (1) where u(x, y) is a dependent variable and x, y are independent variables. The coefﬁ- cients are, in general, functions of x and y. For simplicity let us consider the possibility of obtaining solutions of the form u = g(x + λy), (2) where (x + λy) is the argument of g, for the homogeneous equation corresponding to equation (1) which is Auxx + 2Buxy + Cuyy = 0, (3) where A, B and C are assumed constants. If (2) is a solution of (3), then substituting, we obtain Ag (x + λy) + 2Bλg (x + λy) + Cλ2 g (x + λy) = 0, which will be an identity if and only if Cλ2 + 2Bλ + A = 0. (4) This is a quadratic equation in λ, and solving for λ, we have √ −B ± B 2 − CA λ= . (5) C Thus, (i) λ has two distinct roots if B 2 − CA > 0; (ii) λ has two equal roots if B 2 − CA = 0; and (iii) λ has two complex roots if B 2 − CA < 0. Hence, there are two, one or no real values of λ for which solutions of the form g(x + λy) exist, according to the sign of the discriminant B 2 − AC. Equation (1) is said to be 1 1. Hyperbolic if B 2 − AC > 0, 2. Parabolic if B 2 − AC = 0, and 3. Elliptic if B 2 − AC < 0. Now in each case, the loci x + λ1 y = C1 , (6) x + λ2 y = C2 where λ1 and λ2 are the roots of equation (4), are called the characteristic curves or simply the characteristics of equation (3). These two characteristics deﬁne two Prepared by Swaroop Nandan Bora families of parallel lines in the xy-plane. Please note that for the classiﬁcation of the PDEs, though we consider equation (3), i.e., only three terms of the equation (1), the classiﬁcation applies to the general equa- tion. The above classiﬁcation does not hold for PDEs in three independent variables. Consider the characteristic in its simplest form: x + λy = c. Then, dy 1 =− . dx λ Now eliminating λ from equation (4) by the slope dy/dx, we obtain the following differential equation 2 dy dy A − 2B + C = 0. (7) dx dx This differential equation must be satisﬁed by any characteristics of the form x+λy = c and thus characteristics are found by the solution of (7), and hence this equation is known as the characteristic differential equation. Even if A, B and C are not constants but functions of x and y, still the classiﬁcation as done above is valid and the characteristic equation in that case can be written as 2 dy dy A(x, y) − 2B(x, y) + C(x, y) = 0. (8) dx dx Solving (7) or (8) for dy/dx, we get √ dy B± B 2 − AC = . (9) dx A 2 Equation (9) represents a pair of ordinary differential equations the solutions of which exist in the form φ(x, y) = C1 = ξ . (10) ψ(x, y) = C2 = η These two are said to be the characteristics of the partial differential equation (1). The simplest, but very useful, examples of hyperbolic, parabolic and elliptic partial differential equation are, respectively, • The one-dimensional wave equation: utt = c2 uxx . • The one-dimensional heat conduction or diffusion equation: ut = αuxx . Prepared by Swaroop Nandan Bora • The two-dimensional Laplace’s equation: uxx + uyy = 0. Using the concepts of characteristics, a general second-order partial differential equa- tion can be reduced to one of the above forms. 1.1 Hyperbolic Class Equation (1) can be classiﬁed as hyperbolic type if B 2 − CA > 0. The characteristic will be given by the solution of the following pair of ordinary differential equations √ dy B ± B 2 − AC = , dx A which will give distinct and real roots. Integrating the pair we will get φ(x, y) = C1 . ψ(x, y) = C2 Thus the characteristics are deﬁned as ξ = φ(x, y) . (11) η = ψ(x, y) Under this transformation, the original equation (1) can be reduced to the standard form uξη = h(ξ, η, u, uξ , uη ). (12) 3 1.2 Parabolic Class Equation (1) can be classiﬁed as parabolic type if B 2 − CA = 0. As the roots will be real and equal, the characteristics will be given by the solution of the ODE dy B = . dx A Integrating we get φ(x, y) = c1 . Note that we will get only one characteristic from the solution of the above equation, say, η = φ(x, y). The other characteristic may be deﬁned as Prepared by Swaroop Nandan Bora ξ = x. The selection of ξ, i.e., the second characteristic, is based on the fact that it should not be parallel to the other characteristic η. In other words, we choose ξ such that the Jacobian of the transformation ∂(ξ, η) J= = 0. ∂(x, y) Therefore under the transformation ξ =x , (13) η = φ(x, y) equation (1) can be written as uξξ = h(ξ, η, u, uξ , uη ). (14) 1.3 Elliptic Class Equation (1) can be classiﬁed as elliptic type if B 2 − CA < 0. The characteristic will be given by the solution of the following pair of ordinary differential equations √ dy B ± i CA − B 2 = . dx A Taking positive sign and integrating, we get φ(x, y) = c, where φ(x, y) is a complex function. The characteristics in this case are given by ξ = Re{φ(x, y)} . (15) η = Im{φ(x, y)} 4 Under this transformation, equation (1) reduces to the standard form uξξ + uηη = h(ξ, η, u, uξ , uη ). (16) Alternatively, considering both positive and negative signs and then solving the differ- ential equations, we obtain φ(x, y) = c1 , ψ(x, y) = c2 . These two functions are complex conjugates. The characteristics are then given by ξ = φ(x, y) . (17) ¯ Prepared by Swaroop Nandan Bora η = ψ(x, y) = φ(x, y) Under this transformation, equation (1) reduces to the standard form uξη = h(ξ, η, u, uξ , uη ). (18) Example: Find the characteristics of the following equation and reduce it to the ap- propriate standard form and then obtain the general solution: 3uxx + 10uxy + 3uyy = 0. Solution: Here we have B 2 − CA = 16 > 0 and hence the equation is of hyperbolic type. The characteristics are given by the solutions of 2 dy dy A − 2B + C = 0, dx dx i.e., 2 dy dy 3 − 10 + 3 = 0, dx dx giving dy 5±4 1 = = 3, . dx 3 3 Integrating these, we get 3x − y = c1 , x − 3y = c2 . Hence the characteristics are deﬁned by ξ = 3x − y, η = x − 3y. 5 Here we have ξx = 3, ηx = 1, ξy = −1, ηy = −3. Therefore, under this transformation, we have ux = uξ ξx + uη ηx = 3uξ + uη , uy = uξ ξy + uη ηy = −uξ − 3uη . Now the expressions for the second order derivatives are uxx = 9uξξ + 6uξη + uηη , uxy = −3uξξ − 10uξη − 3uηη , uyy = uξξ + 6uξη + 9uηη . Prepared by Swaroop Nandan Bora Substituting these values in the given equation we get uξη = 0. Now integrating with respect to ξ partially uη = F (η). Again integrating with respect to η partially, we obtain u = F (η) + G(ξ). Hence the solution of the original equation is u(x, y) = F (x − 3y) + G(3x − y), where F and G are arbitrary functions. Example: Find the characteristics of the following equation and reduce it to the ap- propriate standard form and then obtain the general solution: uxx + 4uxy + 4uyy = 0. Solution: Here we have B 2 − CA = 0 and hence the equation is of parabolic type. The characteristics are given by the solutions of 2 dy dy A − 2B + C = 0, dx dx i.e., 2 dy dy −4 + 4 = 0, dx dx 6 giving dy = 2, dx which, on integration, gives y − 2x = c1 . Therefore the characteristics are given by. ξ = x, η = y − 2x. Here we have ξx = 1, ηx = −2, ξy = 0, ηy = 1. Therefore, under this transformation, Prepared by Swaroop Nandan Bora we have ux = uξ ξx + uη ηx = uξ − 2uη , uy = uξ ξy + uη ηy = uη . Now the expressions for the second order derivatives are uxx = uξξ − 4uξη + 4uηη , uxy = uξη − 2uηη , uyy = uηη . Substituting these values in the given equation we get uξξ = 0. Now integrating with respect to ξ partially uξ = f (η). Again integrating with respect to ξ partially, we obtain u = ξf (η) + g(η). Hence the solution of the original equation is u(x, y) = xf (y − 2x) + g(y − 2x), where f and g are arbitrary functions. Example: Find the characteristics of the following equation and reduce it to the ap- propriate standard form and then obtain the general solution: uxx + 4uxy + 5uyy = 0. 7 Solution: Here we have B 2 − CA = −1 < 0 and hence the equation is of elliptic type. The characteristics are given by the solutions of 2 dy dy A − 2B + C = 0, dx dx i.e., 2 dy dy −4 + 5 = 0, dx dx giving dy = 2 ± i. dx Prepared by Swaroop Nandan Bora Taking the positive sign and integrating, y − (2 + i)x = c. which can be rewritten as (2 − i)y − 5x = K1 . or, 2y − 5x − iy = K1 . The characteristics in this case are ξ = 2y − 5x − iy, η = 2y − 5x + iy. Therefore, under this transformation, we will ultimately have the canonical form of the given equation as uξη = 0. Integrating this with respect to η partially, uξ = φ (ξ). Again integrating this with respect to ξ partially u = φ(ξ) + ψ(η). 8 Now let us consider the solution for a nonhomogeneous equation. Example: Find the characteristics of the following equation and reduce it to the ap- propriate standard form and then obtain the general solution: uxx − 4uxy + 4uyy = cos(2x + y). Solution: It can readily be seen that the given equation is a parabolic one and the characteristics are given by the solution of 2 dy +2 = 0. dx Prepared by Swaroop Nandan Bora giving us y + 2x = c1 . Hence the characteristics are given by ξ = x, η = y + 2x. The canonical form will be uξξ = cos η. Integrating partially with respect to ξ, we get uξ = ξ cos η + f (η). Again integrating w.r.t. ξ, we get ξ2 u(ξ, η) = cos η + ξf (η) + g(η). 2 Writing back in the variables x and y, we have the solutions as: x2 u(x, y) = xf (y + 2x) + g(y + 2x) + cos(y + 2x), 2 where f and g are arbitrary functions. 1.4 The D’Alembert solution of the wave equation In applied sciences and engineering, method of characteristics is not of much use for parabolic and elliptic equations. On the other hand, the two families of characteristics of hyperbolic equations, being real and distinct, are of considerable practical value. 9 Especially, in one-dimensional progressive wave propagation, consideration of char- acteristics can give us a good deal of information about the propagation of the wave fronts. This solution of one-dimensional wave equation, known as D’Alembert’s solu- tion, was discovered by a French mathematician named Jean Le Rond D’Alembert. Consider the one-dimensional wave equation utt = c2 uxx . (19) Here B 2 − AC = c2 > 0. The characteristics are given by the solution of the ODE dx = ±c. (20) Prepared by Swaroop Nandan Bora dt Integrating we get x − ct = c1 . (21) x + ct = c2 Hence the characteristics are given by ξ = x − ct, η = x + ct. Under this transformation ux = uξ ξx + uη ηx = uξ + uη , ut = uξ ξt + uη ηt = −cuξ + cuη , uxx = (uξξ ξx + uξη ηx ) + (uηξ ξx + uηη ηx ) = uξξ + 2uξη + uηη , utt = −c(uξξ ξt + uξη ηt ) + c(uηξ ξt + uηη ηt ), = c2 (uξξ − 2uξη + uηη ). Substituting these into the given equation, we obtain, uξη = 0. (22) Integrating partially with respect to η, uξ = F (ξ). Integrating partially w.r.t. ξ, we obtain, u = F (ξ) + G(η). 10 Hence the solution in physical variables is u(x, t) = F (x − ct) + G(x + ct) (23) where F and G are arbitrary functions. The physical interpretation of these functions is quite interesting. The functions F and G represent two progressive waves travelling in opposite directions with the speed c. To see this let us ﬁrst consider the solution u = F (x − ct). At t = 0, it deﬁnes the curve u = F (x), and after time t = t1 , it deﬁnes the curve u = F (x − ct1 ). But these curves are identical except that the latter is translated to the right a distance equal to ct1 . Prepared by Swaroop Nandan Bora Thus the entire conﬁguration moves along the positive direction of x-axis a distance of ct1 in time t1 . The velocity with which the wave is propagated is, therefore, ct1 v= =c t1 Similarly the function G(x+ct) deﬁnes a wave progressing in the negative direction of x-axis with constant velocity c. The total solution is, therefore, the algebraic sum of these two travelling waves. Solution (23) is very convenient representation for progressive waves which travel large distances through a uniform medium. Let us consider the following two initial conditions for a uniform medium −∞ < x < ∞. Displacement: u(x, 0) = φ(x), (24) Velocity: ut (x, 0) = ψ(x), (25) that is, we consider the vibration of a thin string of inﬁnite length. Then from the solution (23) we ﬁnd that F (x) + G(x) = φ(x), (26) −cF (x) + c G (x) = ψ(x), (27) for all values of x. Integrating the second equation with respect to x, we get x 1 −F (x) + G(x) = ψ(s)ds. (28) c x0 From (26) and (28), we obtain x 1 1 F (x) = φ(x) − ψ(s)ds , (29) 2 c x0 x 1 1 G(x) = φ(x) + ψ(s)ds , (30) 2 c x0 11 where A is an integration constant and s is a dummy variable. Substituting these expressions into (23), we obtain x+ct 1 1 u= [φ(x − ct) + φ(x + ct)] + ψ(s)ds. (31) 2 2c x−ct This is D’Alembert’s solution of wave equation. Thus for a given initial displacement and velocity in the vertical direction, the wave equation is completely solved and this solution is usually called the progressive wave solution. It is to be noted that the use of string problem to demonstrate the solution of the wave problem is a matter of conve- nience. However, any variables satisfying the wave equation possess the mathematical Prepared by Swaroop Nandan Bora properties developed for the string. It is clear that the wave equation can be handled very easily by introducing the characteristic variables (ξ, η). The relationship between the physical plane and the characteristic plane for this particular example can be demonstrated graphically. Equation (23) represents the solution as the sum of two progressive waves;one going to the right and the other to the left. The wave celerity is c. For each of the two progressive waves, we can also follow the wave motion by observing that in the xt- 1 1 plane 2 F (x−ct) is constant along each line x−ct = constant and similarly 2 G(x+ct) is constant along each line x + ct = constant. Thus there are two families of parallel lines called the characteristics along which the waves are propagated. Furthermore, along x-axis the values of u(x, 0) and ut (x, 0) are given as initial conditions of displacement and velocity and they just sufﬁce to determine the constant values of F and G along the individual characteristic. The characteristics, therefore, represent the paths in xt-plane along which disturbances in the medium propagated. Finally, since the solution of wave equation is u = F (x − ct) + G(x + ct) the value of u at any point in the xt-plane is the sum of the values of F and G on the respective characteristics which pass through that point. Suppose we need to ﬁnd the wave function at P (x, t), then draw two character- istics passing through that point. Measure the x-intercepts subtended by these two characteristics. Then using the formula (31), we obtain β 1 1 u(x, t) = [φ(α) + φ(β)] + ψ(s)ds, 2 2c α where α = x − ct and β = x + ct, which is as required. 12