Chapter 3. Second Order Linear PDEs

Transcription

1 Chapter 3. Second Order Linear PDEs 3.1 Introduction The general class of second order linear PDEs are of the form: ax, y)u xx + bx, y)u xy + cx, y)u yy + dx, y)u x + ex, y)u y + f x, y)u = gx, y). 3.1) The three PDEs that lie at the cornerstone of applied mathematics are: the heat equation, the wave equation and Laplace s equation,i.e. i) u t = u xx, the heat equation ii) u tt = u xx, the wave equation iii) u xx + u yy = 0, Laplace s equation or, using the same independent variables, x and y i) u xx u y = 0, the heat equation 3.3a) ii) u xx u yy = 0, the wave equation 3.3b) iii) u xx + u yy = 0. Laplace s equation 3.3c) Analogous to characterizing quadratic equations ax 2 + bxy + cy 2 + dx + ey + f = 0, as either hyperbolic, parabolic or elliptic determined by b 2 4ac > 0, b 2 4ac = 0, b 2 4ac < 0, hyperbolic, parabolic, elliptic,

2 2 Chapter 3. Linear Second Order Equations we do the same for PDEs. So, for the heat equation a = 1, b = 0, c = 0 so b 2 4ac = 0 and so the heat equation is parabolic. Similarly, the wave equation is hyperbolic and Laplace s equation is elliptic. This leads to a natural question. Is it possible to transform one PDE to another where the new PDE is simpler? Namely, under a change of variable r = rx, y), s = sx, y), can we transform to one of the following canonical forms: u rr u ss + l.o.t.s. = 0, hyperbolic, 3.5a) u ss + l.o.t.s. = 0, parabolic, 3.5b) u rr + u ss + l.o.t.s. = 0, elliptic, 3.5c) where the term l.o.t.s stands for lower order terms. For example, consider the PDE 2u xx 2u xy + 5u yy = ) This equation is elliptic since the elliptic b 2 4ac = 4 40 = 36 < 0. If we introduce new coordinates, r = 2x + y, s = x y, then by a change of variable using the chain rule u xx = u rr r 2 x + 2u rs r x s x + u ss s 2 x + u r r xx + u s s xx, u xy = u rr r x + u rs r x s y + s x ) + u ss s x s y + u r r xy + u s s xy, u yy = u rr ry 2 + 2u rs s y + u ss s 2 y + u r y + u s s yy, gives u xx = 4u rr + 4u rs + u ss, u xy = 2u rr u rs u ss, u yy = u rr 2u rs + u ss.

3 3.1. Introduction 3 Under 3.7), equation 3.6) becomes u rr + u ss = 0, which is Laplace s equation also elliptic). Before we consider transformations for PDEs in general, it is important to determine whether the equation type could change under transformation. Consider the general class of PDEs au xx + bu xy + cu yy = 0 3.7) where a, b, and c are functions of x and y and noting that we have suppressed the lower terms as they will not affect the type. Under a change of variable x, y) r, s) with the change of variable formulas 3.7) gives ) a u rr r 2 x + 2u rs r x s x + u ss s 2 x + u r r xx + u s s xx + b ) u rr r x + u rs r x s y + s x ) + u ss s x s y + u r r xy + u s s xy 3.8) ) + c u yy + u rr ry 2 + 2u rs s y + u ss s 2 y + u r y + u s s yy = 0 Rearranging 3.8), and again neglecting lower order terms, gives ar 2 x + br x + cry)u 2 rr + 2ar x s x + br x s y + s x ) + 2c s y )u rs 3.9) + as 2 x + bs x s y + cs 2 y)u ss = 0. Setting A = ar 2 x + br x + cr 2 y, B = 2ar x s x + br x s y + s x ) + 2c s y, 3.10) C = as 2 x + bs x s y + cs 2 y, gives again suppressing lower order terms) Au rr + Bu rs + Cu ss = 0, whose type is given by B 2 4AC = b 2 4ac) ) 2 r x s y s x,

4 4 Chapter 3. Linear Second Order Equations from which we deduce that b 2 4ac > 0, B 2 4AC > 0, b 2 4ac = 0, B 2 4AC = 0, b 2 4ac < 0, B 2 4AC < 0, giving that the equation type is unchanged under transformation. We now consider transformations to canonical form. As there are three types of canonical forms, hyperbolic, parabolic and elliptic, we will deal with each type separately. 3.2 Canonical Forms If we introduce the change of coordinates r = rx, y), s = sx, y), 3.11) the derivatives change according to: First Order u x = u r r x + u s s x, u y = u r + u s s y, 3.12) Second Order u xx = u rr r 2 x + 2u rs r x s x + u ss s 2 x + u r r xx + u s s xx, u xy = u rr r x + u rs r x s y + s x ) + u ss s x s y + u r r xy + u s s xy, 3.13) u yy = u rr ry 2 + 2u rs s y + u ss s 2 y + u r y + u s s yy, If we substitute 3.12) and 3.13) into the general linear equation 3.1) and re-arrange we obtain ar 2 x + br x + cry)u 2 rr + ) 2ar x s x + br x s y + s x ) + 2c s y urs + as 2 x + bs x s y + cs 2 y)u ss + l.o.t.s. = ) Our goal now is to target a given canonical form and solve a set of equations for the new coordinates r and s.

5 3.2.1 Parabolic Canonical Form 3.2. Canonical Forms 5 Comparing 3.14) with the parabolic canonical form 3.5b) leads to choosing ar 2 x + br x + cr 2 y = 0, 3.15a) 2ar x s x + br x s y + s x ) + 2c s y = 0, 3.15b) Since in the parabolic case b 2 4ac = 0, then substituting c = 4a b2 both equations of 3.15) are satisfied if we find 2ar x + b = ) with the choice of sx, y) arbitrary. The following examples demonstrate. Example 1. Consider u xx + 6u xy + 9u yy = ) Here, a = 1, b = 6 and c = 9 showing that b 2 4ac = 0, so the PDE is parabolic. Solving r x + 3 = 0, gives r = f 3x y). As we wish to find new coordinates as to transform the original equation to canonical form, we choose r = 3x y, s = y. Calculating second derivatives u xx = 9u rr, u xy = 3u rr + 3u rs, u yy = u rr 2u rs + u ss. 3.18) Substituting 3.18) into 3.17) gives u ss = 0! Not to be confused with factorial!).

6 6 Chapter 3. Linear Second Order Equations Solving gives u = f r)s + gr). where f and g are arbitrary functions. In terms of the original variables, we obtain the solution u = y f 3x y) + g3x y). Example 2. Consider x 2 u xx 4xyu xy + 4y 2 u yy + xu x = ) Here, a = x 2, b = 4xy and c = 4y 2 showing that b 2 4ac = 0, so the PDE is parabolic. Solving x 2 r x 2xy = 0, or gives xr x 2y = 0, r = f x 2 y). As we wish to find new coordinates, i.e. r and s, we choose simple r = x 2 y, s = y. Calculating first derivatives gives u x = 2xyu r. 3.20) Calculating second derivatives u xx = 4x 2 y 2 u rr + 2yu r, u xy = 2x 3 yu rr + 2xyu rs + 2xu r, u yy = x 4 u rr + 2x 2 u rs + u ss. 3.21a) 3.21b) 3.21c) Substituting 3.20) and 3.21) into 3.19) gives 4y 2 u ss 4x 2 yu r = 0.

7 3.2. Canonical Forms 7 or, in terms of the new variables, r and s, u ss r s 2 u r = ) An interesting question is whether different choices of the arbitrary function f and the variable s would lead to a different canonical forms. For example, suppose we chose r = 2 ln x + ln y, s = ln y, we would obtain u ss u r u s = 0, 3.23) a constant coefficient parabolic equation, whereas, choosing r = 2 ln x + ln y, s = 2 ln x, we would obtain the heat equation. u ss u r = 0, 3.24) Hyperbolic Canonical Form In order to obtain the canonical form for the hyperbolic type, i.e. u rr u ss + l.o.t.s. = 0, 3.25) it is necessary to choose ) ar 2 x + br x + cry 2 = as 2 x + bs x s y + cs 2 y, 2ar x s x + br x s y + s x ) + 2c s y = ) The problem is that this system is still a very hard problem to solve both PDEs are nonlinear and coupled!). Therefore, we introduce a modified hyperbolic form that is much easier to work with.

8 8 Chapter 3. Linear Second Order Equations Modified Hyperbolic Form The modified hyperbolic canonical form is defined as u rs + l.o.t.s. = 0, 3.27) noting that a = 0, b = 1 and c = 0 and that b 2 4ac > 0 still! In order to target the modified hyperbolic form, it is now necessary to choose ar 2 x + br x + cr 2 y = 0, as 2 x + bs x s y + cs 2 y = a) 3.28b) If we re-write 3.28a) and 3.28b) as follows a a rx sx s y ) 2 + 2b r x + c = 0, ) 2 + 2b s x + c = 0, s y 3.29a) 3.29b) then we can solve equations 3.29a) and 3.29b) separately for r x and s x s y. This leads to two first order linear PDEs for r and s. The solutions of these then gives rise to the correct canonical variables. The following examples demonstrate. Example 3. Consider u xx 5u xy + 6u yy = ) Here, a = 1, b = 5 and c = 6 showing that b 2 4ac = 1 > 0, so the PDE is hyperbolic. Thus, 3.57a) becomes r 2 x 5r x + 6ry 2 = 0, s 2 x 5s x + 6s 2 y = 0, and factoring gives ) ) ) ) rx 2 rx 3 = 0, sx 2s y sx 3s y = 0,

9 3.2. Canonical Forms 9 from which we choose giving rise to solutions r x 2 = 0, s x 3s y = 0, r = f 2x + y), s = g3x + y). As we wish to find new co-ordinates as to transform the original equation to canonical form, we choose r = 2x + y, s = 3x + y. Calculating second derivatives u xx = 4u rr + 12u rs + 9u ss, u xy = 2u rr + 5u rs + 3u ss, 3.31) u yy = u rr + 2u rs + u ss. Substituting 3.31) into 3.30) gives u rs = 0. Solving gives u = f r) + gs). where f and g are arbitrary functions. In terms of the original variables, we obtain the solution u = f 2x + y) + g3x + y). Example 4. Consider xu xx x + y)u xy + yu yy = ) Here, a = x, b = x + y) and c = y showing that b 2 4ac = x y) 2 > 0, so the PDE is hyperbolic. Solving xr 2 x x + y)r x + yr 2 y = 0,

10 10 Chapter 3. Linear Second Order Equations or, upon factoring xrx y ) rx ) = 0. As s satisfies the same equation, we choose the first factor for r and the second for s xr x y = 0, s x s y = ) Upon solving 3.33), we obtain r = f xy), s = gx + y). As we wish to find new co-ordinates, i.e. r and s, we choose simple r = xy, s = x + y. Calculating first derivatives gives u x = yu r + u s, u y = xu r + u s. 3.34) Calculating second derivatives u xx = y 2 u rr + 2yu rs + u ss, u xy = xyu rr + x + y)u rs + u ss + u r, u yy = x 2 u rr + 2xu rs + u ss. 3.35a) 3.35b) 3.35c) Substituting 3.34) and 3.35) into 3.32) gives 4xy x + y) 2) u rs x + y)u r = 0, or, in terms of the new variables, r and s, u rs + s s 2 4r u r = Regular Hyperbolic Form We now wish to transform a given hyperbolic PDE to its regular canonical form u rr u ss + l.o.t.s. = )

11 3.2. Canonical Forms 11 First, let us consider the following example. x 2 u xx y 2 u yy = ) If we were to transform to modified canonical form, we would solve xr x y = 0, xs x + ys y = 0, which gives r = f xy), s = gx/y). As we wish to find new co-ordinates, i.e. r and s, we choose simple r = xy, s = x/y. In doing so, the original PDE then becomes However, if we choose then the original PDE becomes u rs 1 2r u s = ) r = ln x + ln y, s = ln x ln y, u rs u s = 0, 3.39) which is clearly an easier PDE. However, if we introduce new coordinates α and β such that noting that derivatives transform α = r + s 2, β = r s 2, u r = 1 2 u α u β, u s = 1 2 u α 1 2 u β u rs = 1 4 u αα 1 4 u ββ, 3.40) and the PDE 3.39) becomes u αα u ββ 2u α + 2u β = 0,

12 12 Chapter 3. Linear Second Order Equations a PDE in regular hyperbolic form. Thus, combining the variables r and s and α and β gives directly In fact, one can show that if α = ln x, β = ln y. α = r + s 2, β = r s 2, where r and s satisfies 3.28a) and 3.28b) then α and β satisfies ) aα 2 x + bα x α y + cα 2 y = aβ 2 x + bβ x β y + cβ 2 y, 3.41a) 2aα x β x + bα x β y + α y β x ) + 2cα y β y = b) which is 3.58a) with r and s replaces with α and β. This give a convenient way to go directly to the coordinates that lead to the regular hyperbolic form. We note that so we can essentially consider α, β = r ± s 2, 3.42) ar 2 x + br x + cr 2 y = 0, as 2 x + bs x s y + cs 2 y = a) 3.43b) but instead of factoring, treat each as a quadratic equation in r x / or s x /s y and solve according. We demonstrate with an example. Example 5. Consider 8u xx 6u xy + u yy = ) The corresponding equations for r and s are 8r 2 x 6r x + r 2 y = 0, 8s 2 x 6s x s y + s 2 y = 0, 3.45a) 3.45b)

13 3.2. Canonical Forms 13 but as they are identical it suffices to only consider one. Dividing 3.57a) by r 2 y gives 8 rx Solving by the quadratic formula gives or The method of characteristics gives which gives which we choose which leads to the chose ) 2 6 r x + 1 = 0. r x = 6 ± 2 16, 8r x 3 ± 1) = 0. dx 8 = dy ; dr = 0. 3 ± 1 r = f 3 ± 1)x + 8y), r = 3x + 8y ± x, r = 3x + 8y, s = x, Under this transformation, the original equation 3.44) becomes the desired canonical form. u rr u ss = 0, Example 6. Consider xy 3 u xx x 2 y 2 u xy 2x 3 yu yy y 2 u x + 2x 2 u y = ) The corresponding equations for r and s are xy 3 r 2 x x 2 y 2 r x 2x 3 yr 2 y = 0, xy 3 s 2 x x 2 y 2 s x 2x 3 ys 2 y = 0, 3.47a) 3.47b)

14 14 Chapter 3. Linear Second Order Equations and choosing the first gives y 2 rx Solving by the quadratic formula gives or ) 2 xy r x 2x 2 = 0. r x = 1 ± 3)x, 2y 2yr x 1 ± 3) x = 0. Solving gives r = f x 2 + 2y 2 ± 3x 2). If we choose f to be simple and split according to the ± gives r = x 2 + 2y 2, s = 3x 2, Under this transformation, the original equation 3.46) becomes u rr u ss = 0, the desired canonical form Elliptic Canonical Form In order to obtain the canonical form for the elliptic type, i.e. u rr + u ss + l.o.t.s. = 0, it is necessary to choose ar 2 x + br x + cr 2 y = ) as 2 x + bs x s y + cs 2 y, 2ar x s x + br x s y + s x ) + 2c s y = ) The problem, like the regular hyperbolic type, is still difficult to solve. However, we find that if we let r = α + β, s = α β 2 2i Please note the switch in the variables r and s and α and β. 3.49)

15 3.2. Canonical Forms 15 where α and β satisfy aα 2 x + bα x α y + cα 2 y = 0, aβ 2 x + bβ x β y + cβ 2 y = 0, 3.50a) 3.50b) then 3.48) is satisfied. This is much like the connection between modified and regular hyperbolic canonical form. As solving 3.50a) gives rise to complex roots, the formulas 3.49) will take real and complex parts of the solved αandβ equations as new variables. The next few examples will illustrate. Example 7. Consider u xx 4u xy + 5u yy = ) The corresponding equations for r and s are r 2 x 4r x + 5r 2 y = 0, s 2 x 4s x s y + 5s 2 y = 0, 3.52a) 3.52b) but as they are identical it suffices to only consider one. Dividing 3.52a) by r 2 y gives rx Solving by the quadratic formula gives or The method of characteristics gives which gives ) 2 4 r x + 5 = 0. r x = 2 ± i, r x 2 ± i) = 0. dx 1 = dy, dr = 0. 2 ± i r = f 2x + y ± ix),

16 16 Chapter 3. Linear Second Order Equations which we choose r = 2x + y ± x, which leads to the chose r = 2x + y, s = x, Under this transformation, the original equation 3.51) becomes u rr + u ss = 0, the desired canonical form. Example 8. Consider x 2) 2 uxx x 2) 1 + y 2) u xy y 2) 2 uyy + 4x 1 + x 2) u x = ) The corresponding equations for r and s are x 2) 2 r 2 x x 2) 1 + y 2) r x y 2) 2 r 2 y = 0, 3.54a) x 2) 2 s 2 x x 2) 1 + y 2) s x s y y 2) 2 s 2 y = 0, 3.54b) but as they are identical it suffices to only consider one. Solving by the quadratic formula gives or r x = 2 ± i) 1 + y 2 ) 1 + x 2, x 2) r x 1 ± i) 1 + y 2) = 0. The method of characteristics gives the solution as ) r = f tan 1 x + 2 tan 1 y ± tan 1 x, which we choose r = tan 1 x + 2 tan 1 y ± tan 1 x,

17 3.2. Canonical Forms 17 which leads to the choice r = tan 1 x + 2 tan 1 y, s = tan 1 x, Under this transformation, the original equation 3.53) becomes u rr + u ss 2yu r = 0, and upon using the original transformation gives the desired canonical form. u rr + u ss 2 tan r s 2 u r = 0, Exercises 1. Determine the type of the following second order PDEs i) x 2 u xx y 2 u yy = u x + u y ii) u xx + 2u xy + u yy = 0 iii) y 2 u xx + 2yu xy u yy = 0 iv) u xy + u = u x + u y 2. Transform the following parabolic PDEs to canonical form. Find the general solution if possible. i) u xx + 2u xy + u yy = 0 ii) y 2 u xx 2xyu xy + x 2 u yy = 0. iii) y 2 u xx + 2xyu xy + x 2 u yy 2xu x = Reduce the following second order PDEs to modified hyperbolic canonical form i) 2u xx 3u xy + u yy = u x + u y, ii) x 2 u xx 3xyu xy + 2y 2 u yy = 0.

18 18 Chapter 3. Linear Second Order Equations 4. Reduce the following second order PDEs to canonical form i) 4u xx 8u xy + 3u yy = 0, ii) 4u xx + 4u xy + 5u yy = 1, iii) x 2 u xx + y 2 u yy = 1, iv) u xx 1 + y 2 ) 2 u yy = If a linear second order PDE ax, y)u xx + 2bx, y)u xy + cx, y)u yy + lots = 0. is hyperbolic, then it is possible to transform to a modified hyperbolic canonical form u rs + l.o.t.s. = 0, by choosing r and s such that they satisfy ar 2 x + 2br x + cr 2 y = 0, as 2 x + 2bs x s y + cs 2 y = 0. 5i) Show that by introducing new variables α and β such that α = r + s, β = r s, then the following equations are satisfied ) aα 2 x + 2bα x α y + cα 2 y = aβ 2 x + 2bβ x β y + cβ 2 y, aα x β x + bα x β y + β x ) + cα y β y = 0. This then leads to the hyperbolic canonical form u αα u ββ + l.o.t.s. = 0. 5ii). If, instead of the α and β introduced in 5i), we introduce α and β such that α = c 11 r + c 12 s, β = c 21 r + c 22 s,

19 3.2. Canonical Forms 19 where c 11, c 12, c 21 and c 22 are constant and r and s satisfy the above first order PDEs, namely ar 2 x + 2br x + cr 2 y = 0, as 2 x + 2bs x s y + cs 2 y = 0, find conditions on c 11, c 12, c 21 and c 22 such that ) aα 2 x + 2bα x α y + cα 2 y = aβ 2 x + 2bβ x β y + cβ 2 y, aα x β x + bα x β y + β x ) + cα y β y = 0. are still satisfied.