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Chapter 4 (Second order PDE) Note: This module is prepared from Chapter 4 of the text book (T M.-U and L. Debnath, Linear Partial Differential Equations for Scientists and Engineers, Birkhauser, 4th ed., 2007) just to help the students. The study material is expected to be useful but not exhaustive. For detailed study, the students are advised to attend the lecture/tutorial classes regularly, and consult the text book.

Appeal: Please do not print this e-module unless it is really necessary.

Dr. Suresh Kumar, Department of Mathematics, BITS Pilani, Pilani Campus

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Contents Second order PDE . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

3

Canonical form . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

3

Exercise 4.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

8

Second order PDE with constant coefficients . . . . . . . . . . . . . . . . . . . . . . . . .

9

Exercise 4.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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Second order PDE

Dr. Suresh Kumar, BITS Pilani

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Second order PDE A second order and linear PDE in two independent variables x and y is of the form Auxx + Buxy + Cuyy + Dux + Euy + F u = G, where u is dependent variable; A, B, C, D, E and F are functions of x, y. This PDE is said to be hyperbolic, parabolic and elliptic according as B 2 − 4AC is positive, zero and negative, respectively. Further, it is homogeneous if G = 0 otherwise non-homogeneous.

Canonical form To transform the second order PDE to canonical form, we use the transformations α = α(x, y),

β = β(x, y)

assuming that α and β are twice continuously differentiable with the Jacobian αx βy − αy βx 6= 0 in the region of xy-plane under consideration. We have ux = uα α x + uβ β x uy = uα α y + uβ β y uxx = uααα2x + 2uαβ αx βx + uββ β 2x + uααxx + uβ βxx uxy = uαααx αy + uαβ (αx βy + αy βx ) + uββ βx βy + uααxy + uβ βxy uyy = uααα2y + 2uαβ αy βy + uββ β 2y + uααyy + uβ βyy Therefore the given PDE becomes A∗ uαα + B ∗ uαβ + C ∗ uββ + D∗ uα + E ∗ uβ + F ∗ u = G∗ , where A∗ = Aα2x + Bαx αy + Cαy2 B ∗ = 2Aαx βx + B(αx βy + αy βx ) + 2Cαy βy C ∗ = Aβ 2x + Bβx βy + Cβy2 D∗ = Aαxx + Bαxy + Cαyy + Dαx + Eα y E ∗ = Aβxx + Bβxy + Cβyy + Dβx + Eβy F∗ = F G∗ = G

Second order PDE

Dr. Suresh Kumar, BITS Pilani

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It can be verified that B ∗2 − 4A∗ C ∗ = J 2 (B 2 − 4AC). This shows that under the transformations α = α(x, y) and β = β(x, y), the nature of the PDE does not change since J 2 > 0. Next we assume that A, B, C are non-zero, and α and β are such that A∗ = 0 and C ∗ = 0. It follows that 

γx A γy

2

+B



γx γy



+ C = 0,

where γ is α or β. Let γ(x, y) = const. be a solution. Then dy γx =− γy dx So the previous equation becomes    2 dy dy −B A + C = 0. dx dx √ dy B + B 2 − 4AC =⇒ = or dx 2A

B− dy = dx

√ B 2 − 4AC . 2A

These are known as the characteristic equations of the PDE. Let the solutions of these equations be φ(x, y) = c1 ,

ψ(x, y) = c2 ,

known as the characteristic curves. Hence the transformations α = φ(x, y) and β = ψ(x, y) transform the second order PDE to the canonical form B ∗ uαβ + D∗ uα + E ∗ uβ + F ∗ u = G∗ , provided B 2 − 4AC > 0, the hyperbolic case. In the parabolic case, we have B 2 − 4AC = 0. Also, A∗ = 0 assuming α = const. and β arbitrary (we take β = y). So the relation B ∗2 − 4A∗ C ∗ = J 2 (B 2 − 4AC) gives B ∗ = 0. Therefore the canonical form in the parabolic case is C ∗ uββ + D∗ uα + E ∗ uβ + F ∗ u = G∗ . In case, we assume β = const. and α arbitrary (we choose α = x), then C ∗ = 0 and the canonical form reads as A∗ uαα + D∗ uα + E ∗ uβ + F ∗ u = G∗ Finally, in the elliptic case, we have B 2 −4AC < 0. Consequently, φ and ψ are complex, say φ = α+iβ and ψ = α − iβ so that α=

φ+ψ , 2

β=

φ−ψ . 2i

Second order PDE

Dr. Suresh Kumar, BITS Pilani

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Then we find a canonical form of the type A∗ uαα + A∗ uββ + D∗ uα + E ∗ uβ + F ∗ u = G∗ . Ex. Find the canonical form of y2 uxx − x2 uyy = 0. Sol. Comparing the given PDE with Auxx + Buxy + Cuyy + Dux + Euy + F u = G we find A = y2 , B = 0, C = −x2 , D = 0, E = 0, F = 0, G = 0. So B 2 − 4AC = 4x2 y2 > 0 except x = 0 or y = 0. So the given PDE is hyperbolic everywhere in the xy-plane except on the coordinate axes. The characteristic equations are √ √ B + B 2 − 4AC B − B 2 − 4AC 2xy −2xy x dy −x dy = = = 2 = , = = . dx 2A 2A 2y 2y2 y dx y Integrating the two equations, we get the characteristic curves y 2 x2 y 2 x2 = c2 . − = c1 , + 2 2 2 2 Consider the transformations α=

y 2 x2 − , 2 2

β=

y2 x2 + . 2 2

Then x2 = β − α, y2 = β + α, and αx = −x, αy = y, αxx = −1, αxy = 0, αyy = 1, βx = x, βy = y, βxx = 1, βxy = 0, βyy = 1. It follows that A∗ = Aα2x + Bαx αy + Cαy2 = y2 x2 − x2 y2 = 0 B ∗ = 2Aαx βx + B(αx βy + αy βx ) + 2Cαy βy = −2y2 x2 − 2x2 y2 = −4x2 y2 C ∗ = Aβ 2x + Bβx βy + Cβy2 = y2 x2 − x2 y2 = 0 D∗ = Aαxx + Bαxy + Cαyy + Dαx + Eαy = −y2 − x2 E ∗ = Aβxx + Bβxy + Cβyy + Dβx + Eβy = y2 − x2 F∗ = F = 0 G∗ = G = 0

Second order PDE

Dr. Suresh Kumar, BITS Pilani

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Therefore the canonical form of the given PDE is A∗ uαα + B ∗ uαβ + C ∗ uββ + D∗ uα + E ∗ uβ + F ∗ u = G∗ , =⇒

−4x2 y2 uαβ + (−y2 − x2 )uα + (y2 − x2 )uβ = 0.

=⇒

uαβ =

β α x2 − y 2 x2 + y 2 u + uβ = uα − uβ . α 2 2 2 2 2 2 2 −4x y 2(α − β ) −4x y 2(α − β 2 )

Ex. Find the canonical form of x2 uxx + 2xyuxy + y2 uyy = 0, and hence find its general solution. Sol. Comparing the given PDE with Auxx + Buxy + Cuyy + Dux + Euy + F u = G we find A = x2 , B = 2xy, C = y2 , D = 0, E = 0, F = 0, G = 0. So B 2 − 4AC = 4x2 y2 − 4x2 y2 = 0, and therefore the given PDE is parabolic everywhere in the xy-plane. Here, the characteristic equation is dy B 2xy y = = 2 = . x dx 2A 2x Integrating, we get the characteristic curve y = c1 . x Consider the transformations α=

y , x

β = y.

Then we have αx = −y/x2 , αy = 1/x, αxx = 2y/x3 , αxy = −1/x2 , αyy = 0, βx = 0, βy = 1, βxx = 0, βxy = 0, βyy = 0. It follows that A∗ = Aα2x + Bαx αy + Cαy2 = 0 B ∗ = 2Aαx βx + B(αx βy + αy βx ) + 2Cαy βy = 0 C ∗ = Aβ 2x + Bβx βy + Cβy2 = y2 D∗ = Aαxx + Bαxy + Cαyy + Dαx + Eαy = 0 E ∗ = Aβxx + Bβxy + Cβyy + Dβx + Eβy = 0 F∗ = F = 0 G∗ = G = 0

Second order PDE

Dr. Suresh Kumar, BITS Pilani

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Therefore the canonical form of the given PDE is A∗ uαα + B ∗ uαβ + C ∗ uββ + D∗ uα + E ∗ uβ + F ∗ u = G∗ , =⇒

y2 uββ = 0

or

uββ = 0

for

y 6= 0.

Integrating two times with respect to β, we get the general solution u = βf (α) + g(α) = yf (y/x) + g(y/x), where f and g are arbitrary functions appeared due to integration. Ex. Find the canonical form of uxx + x2 uyy = 0. Sol. Comparing the given PDE with Auxx + Buxy + Cuyy + Dux + Euy + F u = G we find A = 1, B = 0, C = x2 , D = 0, E = 0, F = 0, G = 0. So B 2 − 4AC = −4x2 < 0 except x = 0. So the given PDE is elliptic except everywhere in the xy-plane except on y-axis. The characteristic equations are √ √ dy dy B + B 2 − 4AC B − B 2 − 4AC = ix, = −ix. = = 2A 2A dx dx Integrating the two equations, we get the characteristic curves 2y − ix2 = c1 , 2y + ix2 = c2 . Letting φ = 2y − ix2 and ψ = 2y + ix2 , consider α = (φ + ψ)/2 = 2y,

β = (φ − ψ)/(2i) = −x2 .

Then we have αx = 0, αy = 2, αxx = 0, αxy = 0, αyy = 0, βx = −2x, βy = 0, βxx = −2, βxy = 0, βyy = 0. It follows that A∗ = Aα2x + Bαx αy + Cαy2 = 4x2 B ∗ = 2Aαx βx + B(αx βy + αy βx ) + 2Cαy βy = 0 C ∗ = Aβ 2x + Bβx βy + Cβy2 = 4x2 D∗ = Aαxx + Bαxy + Cαyy + Dαx + Eα y = 0 E ∗ = Aβxx + Bβxy + Cβyy + Dβx + Eβy = −2 F∗ = F = 0 G∗ = G = 0

Second order PDE

Dr. Suresh Kumar, BITS Pilani

Therefore the canonical form of the given PDE is A∗ uαα + B ∗ uαβ + C ∗ uββ + D∗ uα + E ∗ uβ + F ∗ u = G∗ , =⇒

4x2 uαα + 4x2 uββ − 2uβ = 0.

=⇒

uαα + uββ =

=⇒

uαα + uββ

1 uβ . 2x2 1 = − uβ . 2β

Note: It is not always possible to solve/integrate the canonical form of a PDE.

Exercise 4.1 1. Find the canonical forms of the following PDE: (a) uxx + 2uxy + 3uyy + 4ux + 5uy + u = ex . (b) 2uxx − 4uxy + 2uyy + 3u = 0. (c) uxx + 5uxy + 4uyy + 7uy = sin x. (d) uxx + yuyy + 4yux + 12 uy = 0. (e) uxx + xuyy = 0. (f) uxx − 2 cos x uxy + (1 + cos2 x)uyy + u = 0. 2. Determine general solution of the following equations: (a) uxx + uyy = 0. (b) uxx + 10uxy + 9uyy = y. 3. Obtain the solution of the Cauchy problem uxx + uyy = 0, u(x, 0) = f (x), uy (x, 0) = g(x). 4. Transform uxx + yuyy + sin(x + y) = 0 to canonical form and hence find its general solution.

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Second order PDE

Dr. Suresh Kumar, BITS Pilani

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Second order PDE with constant coefficients A second order and linear PDE with constant coefficients in two independent variables x and y is of the form auxx + buxy + cuyy + dux + euy + fu = Q, where a, b, c, d, e, f are constants, and Q is function of x, y. If Q = 0, it is homogeneous otherwise non-homogeneous. Letting D =

∂ ∂x

and D′ =

∂ , ∂y

we can write the PDE as

(aD2 + bDD′ + cD′2 + dD + eD′ + f )u = Q. or F (D, D′ )u = Q, where F (D, D′ ) = aD2 + bDD′ + cD′2 + dD + eD′ + f . The following results facilitate to find the general solution of the PDE F (D, D′ )u = Q. Theorem If u1 and u2 are two independent solutions of F (D, D′ )u = 0, then its general solution is uh = u1 + u2 . Proof. By the given F (D, D′ )u1 = 0 and F (D, D′ )u2 = 0. Adding the two, we get F (D, D′ )(u1 +u2 ) = 0. So u1 + u2 is a solution of F (D, D′ )u = 0. Also, given that u1 and u2 are independent. It follows that uh = u1 + u2 is general solution of F (D, D′ )u = 0. Theorem The general solution of this PDE F (D, D′ )u = Q is given by u = uh + up , where uh is general solution of the corresponding homogeneous PDE F (D, D′ )u = 0, and up is a particular solution of F (D, D′ )u = Q. Proof. By the given F (D, D′ )uh = 0 and F (D, D′ )up = Q. Adding the two, we get F (D, D′ )(uh + up ) = Q. It follows that u = uh + up is general solution of F (D, D′ )u = Q. Method to find uh Consider the homogeneous PDE F (D, D′ )u = 0, where F (D, D′ ) = aD2 + bDD′ + cD′2 + dD + eD′ + f . Suppose F (D, D′ ) is reducible to linear factors in D and D′ , that is, F (D, D′ ) = (a1 D + b1 D′ + c1 )(a1 D + b1 D′ + c1 ) for some real constants a1 , a2 , b1 , b2 , c1 , c2 . Then we have (a1 D + b1 D′ + c1 )(a2 D + b2 D′ + c2 )u = 0. We shall obtain two independent solutions u1 and u2 by solving the equations (a1 D + b1 D′ + c1 )u = 0 and (a2 D + b2 D′ + c2 )u = 0. Let us first find the solution of (a1 D + b1 D′ + c1 )u = 0. It is first order linear PDE with auxiliary equations dx dy du = = . −c1 u b1 a1

Second order PDE

Dr. Suresh Kumar, BITS Pilani

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From the first two fractions, we get b1 x − a1 y = k1 , where k1 is an integration constant. From the first and third fractions, we find u = k2 e

−c1 a1

x

provided a1 6= 0 otherwise we could choose second

and third fractions. So the general solution is c1

k2 = φ(k1 ) ∴u=e

−c1 a1

x

or ue a1 = φ1 (b1 x − a1 y). x

φ1 (b1 x − a1 y).

Here φ1 is an arbitrary function. So u1 = e

−c1 x a1

φ1 (b1 x − a1 y).

Similarly, the general solution of (a2 D + b2 D′ + c2 )u = 0 reads as u=e

−c2 x a2

φ2 (b2 x − a2 y),

where φ2 is an arbitrary function. So u2 = e

−c2 x a2

φ2 (b2 x − a2 y).

Thus, the general solution of (a1 D + b1 D′ + c1 )(a2 D + b2 D′ + c2 )u = 0 is given by uh = u1 + u2 = e

−c1 a1

x

−c2 x

φ1 (b1 x − a1 y) + e a2 φ2 (b2 x − a2 y).

In case, a1 = a2 , b1 = b2 and c1 = c2 , then we have (a1 D + b1 D′ + c1 )2 u = 0. Its general solution can be proved to be uh = e

−c1 a1

x

φ1 (b1 x − a1 y) + xe

−c1 x a1

φ1 (b1 x − a1 y).

Notice the multiplication by x in the second term. Method to find up Let up be a particular solution of the PDE f (D, D′ )u = Q so that f (D, D′ )up = Q. Applying the inverse operator up =

1 F (D,D ′ )

on both sides, we get

1 Q. F (D, D′ )

Thus up is obtained by applying the operator

1 F (D,D ′ )

on Q.

In the following, we describe the methods to obtain up for some particular functional forms of Q. (1) If Q = eax+by , then up =

1 eax+by F (D,D ′ )

=

1 eax+by F (a,b)

provided F (a, b) 6= 0, that is, replace D by a

1 F (D,D ′ )

on both sides and dividing by f (a, b), we get

and D′ by b. For F (D, D′ )eax+by = F (a, b)eax+by . Operating 1 eax+by F (D,D ′ )

=

1 eax+by . F (a,b)

Second order PDE

Dr. Suresh Kumar, BITS Pilani

(2) If Q = sin(ax + by), then up =

1 F (D,D ′ )

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sin(ax + by) is solved by replacing D2 by −a2 , DD′ by −ab

and D′2 by −b2 . Similar is the case for Q = cos(ax + by).

For, sin(ax + by) is the imaginary part of ei(ax+by) . So by the exponential rule (1), D is to be replaced by ia, and hence D2 by −a2 . Likewise, DD′ by −ab and D′2 by −b2 . (3) If Q = pn (x, y), a polynomial of degree n in x and y, then up =

1 p (x, y) F (D,D ′ ) n

is solved by expressing

F (D, D′ ) in ascending powers of D and D′ . (4) If Q = eax+by g(x, y), then up =

1 eax+by g(x, y) F (D,D ′ )

1 = eax+by F (D+a,D ′ +b) g(x, y). It is known as the

exponential shift rule. Ex. Find the general solution of 2uxx − uxy − uyy + ux − uy = e2x+3y . Sol. The given PDE is (2D2 − DD′ − D′2 + D − D′ )u = e2x+3y . To find uh , we solve the homogeneous PDE (2D2 − DD′ − D′2 + D − D′ )u = 0. It can be rewritten as (2D + D′ + 1)(D − D′ )u = 0. From the factor 2D + D′ + 1, we have a1 = 2, b1 = 1, c1 = 1. From the factor D − D′ , we have a2 = 1, b2 = −1, c2 = 0. ∴ uh = e

−c1 x a1

φ1 (b1 x − a1 y) + e

−c2 a2

x

φ1 (b2 x − a2 y) = e

−1 x 2

φ1 (x − 2y) + φ2 (−x − y).

Next, the particular solution of the given PDE is given by up =

(2D

+ D′

1 1 1 e2x+3y = e2x+3y = − e2x+3y . ′ 8 + 1)(D − D ) (2.2 + 3 + 1)(2 − 3)

Thus, the general solution of the given PDE is u = uh + up = e

−1 x 2

φ2 (x − 2y) + φ2 (−x − y) −

1 2x+3y e . 8

Ex. Find the general solution of 2uxx − 3uxy + uyy = sin(x − 2y). Sol. The given PDE is (2D2 − 3DD′ + D′2 )u = sin(x − 2y). To find uh , we solve the homogeneous PDE (2D2 − 3DD′ + D′2 )u = 0. It can be rewritten as (2D − D′ )(D − D′ )u = 0. From the factor 2D − D′ , we have a1 = 2, b1 = −1, c1 = 0. From the factor D − D′ , we have a2 = 1, b2 = −1, c2 = 0. ∴ uh = e

−c1 x a1

φ1 (b1 x − a1 y) + e

−c2 a2

x

φ1 (b2 x − a2 y) = φ1 (−x − 2y) + φ2 (−x − y).

Second order PDE

Dr. Suresh Kumar, BITS Pilani

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Next, the particular solution of the given PDE is given by up =

2D2

1 1 1 sin(x − 2y) = sin(x − 2y) = − sin(x − 2y). ′ ′2 2 2 12 − 3DD + D 2(−1 ) − 3(1.2) − 2

Thus, the general solution of the given PDE is u = uh + up = φ1 (−x − 2y) + φ2 (−x − y) −

1 sin(x − 2y). 12

Ex. Find a particular solution of (D2 − D′2 )u = x2 + y2 . Sol. We have 1 (x2 + y2 ) D2 − D′2 = D−2 (1 − D−2 D′2 )−1 (x2 + y2 )

up =

= D−2 (1 + D−2 D′2 + ....)(x2 + y2 ) = D−2 (x2 + y2 + D−2 D′2 (x2 + y2 )) = D−2 (x2 + y2 + D−2 (2)) = D−2 (x2 + y2 + x2 ) = D−2 (2x2 + y2 ) x4 x2 y 2 = + 2 6 Note that 1/D or D−1 stands for integration with respect to x. Likewise 1/D′ or D′−1 stands for integration with respect to y. Further, D−2 = D−1 D−1 etc. Ex. Find a particular solution of (D2 − 2DD′ + D′2 )u = x cos y. Sol. We have 1 (x cos y) − 2DD′ + D′2 = (D′ − D)−2 x cos y

up =

D2

= D′−2 (1 − D′−1 D)−2 (x cos y) = D′−2 (1 + 2D′−1 D + 3D′−2 D2 + .....)(x cos y) = D′−2 (x cos y + 2D′−1 cos y + 0 + ...) = D′−2 (x cos y + 2 sin y) = −x cos y − 2 sin y Ex. Find a particular solution of (D2 − 2DD′ + D′2 )u = ex+y cos(2x + y).

Second order PDE

Dr. Suresh Kumar, BITS Pilani

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Sol. We have up = = = = =

1 [ex+y cos(2x + y)] − 2DD′ + D′2 1 ex+y cos(2x + y) (D + 1)2 − 2(D + 1)(D′ + 1) + (D′ + 1)2 1 ex+y 2 cos(2x + y) D − 2DD′ + D′2 1 ex+y cos(2x + y) −4 − 2(−2) + (−1) −ex+y cos(2x + y) D2

(by exponential shift rule)

General method for particular solution: If F (D, D′ ) is reducible to linear factors in D and D′ , but Q is any function of x and y (may be different from the ones considered earlier), then a particular solution of the PDE F (D, D′ )u = Q can be obtained by the procedure illustrated in the following example. Ex. Find a particular solution of (D2 + DD′ − 2D′2 )u = 8 ln(x + 5y). Sol. The given PDE can be rewritten as (D + 2D′ )(D − D′ )u = 8 ln(x + 5y). The particular solution is given by up = (D + 2D′ )−1 (D − D′ )−1 [8 ln(x + 5y)] Let v = (D − D′ )−1 [8 ln(x + 5y)] so that up = (D + 2D′ )−1 v. Now first we find v, and then up . The equation v = (D − D′ )−1 [8 ln(x + 5y)] is same as (D − D′ )v = 8 ln(x + 5y) with the auxiliary equations dx dy dv . = = 8 ln(x + 5y) 1 −1 From the first two fractions, we get y = c − x, where c is a constant of integration. Then considering first and third fractions, we have dv dx = . 1 8 ln(5c − 4x) Integrating and replacing back c = x + y, we get v = −2(x + 5y) ln(x + 5y) − 8x. Thus, up = (D + 2D′ )−1 v = (D + 2D′ )−1 [−2(x + 5y) ln(x + 5y) − 8x]. Again, it can be rewritten as the PDE (D + 2D′ )up = −2(x + 5y) ln(x + 5y) − 8x, with the auxiliary equations, dy dup dx = = . 1 2 −2(x + 5y) ln(x + 5y) − 8x

Second order PDE

Dr. Suresh Kumar, BITS Pilani

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From the first two fractions, we get y = 2x − d, where d is a constant of integration. Then considering first and third fractions, we have dup dx = . 1 −[8x + 2(11x − 5d) ln(11x − 5d)] Integrating and replacing back d = 2x − y, we get   1 up = − 4x2 + (x + 5y)2 {2 ln(x + 5y) − 1} . 22 Note. The rules discussed for finding the solutions of the second order PDE with constant coefficients can be generalized for the solutions of the third and higher order PDE with constant coefficients.

Exercise 4.2 1. Find the solutions of the following homogeneous PDE: (a) (2D2 + 3DD′ + 2D − 2D′2 − D′ )u = 0. [Ans. u = φ1 (x + 2y) + e−x φ2 (2x − y)] (b) (4D2 + 4D + 1)u = 0. [Ans. u = e−x/2 [φ1 (2y) + xφ2 (2y)]] (c) (4D′2 + 12D′ + 9)u = 0. [Ans. u = e−3y/2 [φ1 (2x) + yφ2 (2x)]] (d) (D3 − 3D2 D′ + 3DD′2 + D′3 )u = 0. [Ans. u = φ1 (x + y) + xφ2 (x + y) + x2 φ3 (x + y)] (e) (D3 + D2 D′ − DD′2 − D′3 )u = 0. [Ans. u = φ1 (x + y) + φ2 (x − y) + xφ3 (x − y)]

(f) (4D3 − 8D2 D′ + 4D2 − 8DD′ + D − 2D′ )u = 0. [Ans. u = φ1 (2x + y) + e−x/2 [ψ1 (2y) + xψ2 (2y)]]

(g) (4D4 − 4D3 D′ − 3D2 D′2 + 2DD′3 + D′4 )u = 0.