solution surfaces through a point do not lie in a plane. The tangent planes no longer intersect along one straight line, but instead, they envelope along a curved surface known as theMonge cone. Any further discussion is beyond the scope of this book.
We conclude this section by adding an important observation regarding the nature of the characteristics in the (x, t)-plane. For a quasi-linear equa- tion, characteristics are determined by the first two equations in (2.4.5) with their slopes
dy
dx = b(x, y, u)
a(x, y, u). (2.4.7)
If (2.4.1) is a linear equation, thenaand b are independent ofu, and the characteristics of (2.4.1) areplane curves with slopes
dy
dx = b(x, y)
a(x, y). (2.4.8)
By integrating this equation, we can determine the characteristics which represent a one-parameter family ofcurves in the (x, t)-plane. However, if a and b are constant, the characteristics of equation (2.4.1) are straight lines.
2.5 Method of Characteristics and General Solutions
We can use the geometrical interpretation of first-order, partial differential equations and the properties of characteristic curves to develop a method for finding the general solution of quasi-linear equations. This is usually referred to asthe method of characteristics due to Lagrange. This method of solution of quasi-linear equations can be described by the following result.
Theorem 2.5.1.The general solution of a first-order, quasi-linear partial differential equation
a(x, y, u)ux+b(x, y, u)uy =c(x, y, u) (2.5.1) is
f(φ, ψ) = 0, (2.5.2)
where f is an arbitrary function of φ(x, y, u) and ψ(x, y, u), and φ = constant = c1 and ψ = constant = c2 are solution curves of the charac- teristic equations
dx a = dy
b =du
c . (2.5.3)
The solution curves defined by φ(x, y, u) =c1 andψ(x, y, u) =c2 are called the families ofcharacteristic curvesof equation (2.5.1).
Proof.Sinceφ(x, y, u) =c1andψ(x, y, u) =c2satisfy equations (2.5.3), these equations must be compatible with the equation
dφ=φxdx+φydy+φudu= 0. (2.5.4) This is equivalent to the equation
a φx+b φy+c φu= 0. (2.5.5) Similarly, equation (2.5.3) is also compatible with
a ψx+b ψy+c ψu= 0. (2.5.6) We now solve (2.5.5), (2.5.6) fora,b, andcto obtain
a
∂(φ,ψ)
∂(y,u)
= b
∂(φ,ψ)
∂(u,x)
= c
∂(φ,ψ)
∂(x,y)
. (2.5.7)
It has been shown earlier thatf(φ, ψ) = 0 satisfies an equation similar to (2.3.14), that is,
p ∂(φ, ψ)
∂(y, u) +q ∂(φ, ψ)
∂(u, x) = ∂(φ, ψ)
∂(x, y). (2.5.8) Substituting, (2.5.7) in (2.5.8), we find that f(φ, ψ) = 0 is a solution of (2.5.1). This completes the proof.
Note that an analytical method has been used to prove Theorem 2.5.1.
Alternatively, a geometrical argument can be used to prove this theorem.
The geometrical method of proof is left to the reader as an exercise.
Many problems in applied mathematics, science, and engineering involve partial differential equations. We seldom try to find or discuss the properties of a solution to these equations in its most general form. In most cases of interest, we deal with those solutions of partial differential equations which satisfy certain supplementary conditions. In the case of a first-order partial differential equation, we determine the specific solution by formulating an initial-value problem or aCauchy problem.
Theorem 2.5.2.(The Cauchy Problem for a First-Order Partial Differen- tial Equation). Suppose thatC is a given curve in the (x, y)-plane with its parametric equations
x=x0(t), y=y0(t), (2.5.9) wheretbelongs to an intervalI⊂R, and the derivativesx′0(t) andy0′(t) are piecewise continuous functions, such that (x′0)2+ (y0′)2= 0. Also, suppose that u = u0(t) is a given function on the curve C. Then, there exists a solutionu=u(x, y) of the equation
2.5 Method of Characteristics and General Solutions 37 F(x, y, u, ux, uy) = 0 (2.5.10) in a domainDofR2 containing the curveCfor allt∈I, and the solution u(x, y) satisfies the given initial data, that is,
u(x0(t), y0(t)) =u0(t) (2.5.11) for all values oft∈I.
In short, the Cauchy problem is to determine a solution of equation (2.5.10) in a neighborhood ofC, such that the solutionu=u(x, y) takes a prescribed valueu0(t) onC. The curveC is called the initial curve of the problem, andu0(t) is called theinitial data. Equation (2.5.11) is called the initial conditionof the problem.
The solution of the Cauchy problem also deals with such questions as the conditions on the functions F, x0(t), y0(t), andu0(t) under which a solution exists and is unique.
We next discuss a method for solving a Cauchy problem for the first- order, quasi-linear equation (2.5.1). We first observe that geometrically x = x0(t), y = y0(t), and u = u0(t) represent an initial curve Γ in (x, y, u)-space. The curve C, on which the Cauchy data is prescribed, is the projection ofΓ on the (x, y)-plane. We now present a precise formula- tion of the Cauchy problem for the first-order, quasi-linear equation (2.5.1).
Theorem 2.5.3.(The Cauchy Problem for a Quasi-linear Equation). Sup- pose thatx0(t),y0(t), andu0(t) are continuously differentiable functions of t in a closed interval, 0≤ t ≤1, and that a, b, and c are functions of x, y, and uwith continuous first-order partial derivatives with respect to their arguments in some domainD of (x, y, u)-space containing the initial curve
Γ : x=x0(t), y=y0(t), u=u0(t), (2.5.12) where 0≤t≤1, and satisfying the condition
y0′ (t)a(x0(t), y0(t), u0(t))−x′0(t)b(x0(t), y0(t), u0(t))= 0.(2.5.13) Then there exists a unique solutionu=u(x, y) of the quasi-linear equation (2.5.1) in the neighborhood ofC : x=x0(t),y =y0(t), and the solution satisfies the initial condition
u0(t) =u(x0(t), y0(t)), for 0≤t≤1. (2.5.14) Note: The condition (2.5.13) excludes the possibility that C could be a characteristic.
Example 2.5.1.Find the general solution of the first-order linear partial differential equation.
x ux+y uy=u. (2.5.15) The characteristic curves of this equation are the solutions of the char- acteristic equations
dx x = dy
y =du
u. (2.5.16)
This system of equations gives the integral surfaces φ= y
x=C1 and ψ=u x =C2,
where C1 and C2 are arbitrary constants. Thus, the general solution of (2.5.15) is
fy x,u
x
= 0, (2.5.17)
wheref is an arbitrary function. This general solution can also be written as
u(x, y) =x gy x
, (2.5.18)
whereg is an arbitrary function.
Example 2.5.2.Obtain the general solution of the linear Euler equation
x ux+y uy =nu. (2.5.19)
The integral surfaces are the solutions of the characteristic equations dx
x = dy y = du
nu. (2.5.20)
From these equations, we get y
x =C1, u xn =C2,
where C1 and C2 are arbitrary constants. Hence, the general solution of (2.5.19) is
fy x, u
xn
= 0. (2.5.21)
This can also be written as u
xn =gy x
or
u(x, y) =xngy x
. (2.5.22)
This shows that the solutionu(x, y) is a homogeneous function ofxandy of degreen.
2.5 Method of Characteristics and General Solutions 39 Example 2.5.3.Find the general solution of the linear equation
x2ux+y2uy = (x+y)u. (2.5.23) The characteristic equations associated with (2.5.23) are
dx x2 = dy
y2 = du
(x+y)u. (2.5.24)
From the first two of these equations, we find
x−1−y−1 =C1, (2.5.25) whereC1is an arbitrary constant.
It follows from (2.5.24) that dx−dy
x2−y2 = du (x+y)u or
d(x−y) x−y = du
u. This gives
x−y
u =C2, (2.5.26)
whereC2is a constant. Furthermore, (2.5.25) and (2.5.26) also give xy
u =C3, (2.5.27)
whereC3is a constant.
Thus, the general solution (2.5.23) is given by f
xy u,x−y
u
= 0, (2.5.28)
where f is an arbitrary function. This general solution representing the integral surface can also be written as
u(x, y) =xy g x−y
u
, (2.5.29)
whereg is an arbitrary function, or, equivalently, u(x, y) =xy h
x−y xy
, (2.5.30)
wherehis an arbitrary function.
Example 2.5.4.Show that the general solution of the linear equation (y−z)ux+ (z−x)uy+ (x−y)uz = 0 (2.5.31) is
u(x, y, z) =f
x+y+z, x2+y2+z2 , (2.5.32) wheref is an arbitrary function.
The characteristic curves satisfy the characteristic equations dx
y−z = dy
z−x = dz x−y =du
0 (2.5.33)
or
du= 0, dx+dy+dz= 0, xdx+ydy+zdz= 0.
Integration of these equations gives
u=C1, x+y+z=C2, and x2+y2+z2=C3, whereC1,C2 andC3are arbitrary constants.
Thus, the general solution can be written in terms of an arbitrary func- tionf in the form
u(x, y, z) =f
x+y+z, x2+y2+z2 .
We next verify that this is a general solution by introducing three inde- pendent variablesξ, η, ζ defined in terms ofx,y, andz as
ξ=x+y+z, η=x2+y2+z2, and ζ=y+z, (2.5.34) whereζis an arbitrary combination ofyandz. Clearly the general solution becomes
u=f(ξ, η), and hence,
uζ =ux
∂x
∂ζ +uy
∂y
∂ζ +uz
∂z
∂ζ. (2.5.35)
It follows from (2.5.34) that 0 =∂x
∂ζ +∂y
∂ζ +∂z
∂ζ, 0 = 2
x∂x
∂ζ +y∂y
∂ζ +z∂z
∂ζ
, ∂y
∂ζ +∂z
∂ζ = 1.
It follows from the first and the third results that ∂x∂ζ =−1 and, therefore, x=y∂y
∂ζ +z∂z
∂ζ, y=y∂y
∂ζ +y∂z
∂ζ, z=z∂y
∂ζ +z∂z
∂ζ.
2.5 Method of Characteristics and General Solutions 41
Clearly, it follows by subtracting that x−y= (z−y)∂z
∂ζ, x−z= (y−z)∂y
∂ζ. Using the values for ∂x∂ζ, ∂z∂ζ, and ∂y∂ζ in (2.5.35), we obtain
(z−y)∂u
∂ζ = (y−z)∂u
∂x+ (z−x)∂u
∂y + (x−y)∂u
∂z. (2.5.36) Ifu=u(ξ, η) satisfies (2.5.31), then ∂u∂ζ = 0 and, hence, (2.5.36) reduces to (2.5.31). This shows that the general solution (2.5.32) satisfies equation (2.5.31).
Example 2.5.5.Find the solution of the equation
u(x+y)ux+u(x−y)uy =x2+y2, (2.5.37) with the Cauchy datau= 0 ony= 2x.
The characteristic equations are dx
u(x+y) = dy
u(x−y) = du
x2+y2 =ydx+xdy−udu
0 =xdx−ydy−udu
0 .
Consequently, d
xy−1
2u2
= 0 and d 1
2
x2−y2−u2
= 0. (2.5.38) These give two integrals
u2−x2+y2 =C1 and 2xy−u2=C2, (2.5.39) whereC1andC2 are constants. Hence, the general solution is
f
x2−y2−u2, 2xy−u2 = 0, wheref is an arbitrary function.
Using the Cauchy data in (2.5.39), we obtain 4C1= 3C2. Therefore 4
u2−x2+y2 = 3
2xy−u2 . Thus, the solution of equation (2.5.37) is given by
7u2= 6xy+ 4
x2−y2 . (2.5.40)
Example 2.5.6.Obtain the solution of the linear equation
ux−uy= 1, (2.5.41)
with the Cauchy data
u(x,0) =x2. The characteristic equations are
dx 1 = dy
−1 = du
1 . (2.5.42)
Obviously,
dy
dx =−1 and du dx = 1.
Clearly,
x+y= constant =C1 and u−x= constant =C2. Thus, the general solution is given by
u−x=f(x+y), (2.5.43)
wheref is an arbitrary function.
We now use the Cauchy data to find f(x) = x2−x, and hence, the solution is
u(x, y) = (x+y)2−y. (2.5.44) The characteristics x+y =C1 are drawn in Figure 2.5.1. The value ofu must be given at one point on each characteristic which intersects the line y= 0 only at one point, as shown in Figure 2.5.1.
Figure 2.5.1 Characteristics of equation (2.5.41).
2.5 Method of Characteristics and General Solutions 43 Example 2.5.7.Obtain the solution of the equation
(y−u)ux+ (u−x)uy =x−y, (2.5.45) with the conditionu= 0 onxy= 1.
The characteristic equations for equation (2.5.45) are dx
y−u= dy
u−x = du
x−y. (2.5.46)
The parametric forms of these equations are dx
dt =y−u, dy
dt =u−x, du
dt =x−y.
These lead to the following equations:
˙
x+ ˙y+ ˙u= 0 and xx˙+yy˙+uu˙ = 0, (2.5.47) where the dot denotes the derivative with respect tot.
Integrating (2.5.47), we obtain
x+y+u= const. =C1 and x2+y2+u2= const. =C2. (2.5.48) These equations represent circles.
Using the Cauchy data, we find that
C12= (x+y)2=x2+y2+ 2xy=C2+ 2.
Thus, the integral surface is described by
(x+y+u)2=x2+y2+u2+ 2.
Hence, the solution is given by
u(x, y) = 1−xy
x+y . (2.5.49)
Example 2.5.8.Solve the linear equation
y ux+x uy=u, (2.5.50)
with the Cauchy data
u(x,0) =x3 and u(0, y) =y3. (2.5.51) The characteristic equations are
dx y = dy
x =du u
or
du
u = dx−dy
y−x = dx+dy y+x . Solving these equations, we obtain
u= C1
x−y =C2(x+y) or
u=C2(x+y), x2−y2= C1
C2
= constant =C.
So the characteristics are rectangular hyperbolas forC >0 orC <0.
Thus, the general solution is given by f
u
x+y, x2−y2
= 0 or, equivalently,
u(x, y) = (x+y)g
x2−y2 . (2.5.52)
Using the Cauchy data, we find thatg
x2 =x2, that is,g(x) =x.
Consequently, the solution becomes u(x, y) = (x+y)
x2−y2 on x2−y2=C >0.
Similarly,
u(x, y) = (x+y)
y2−x2 on y2−x2=C >0.
It follows from these results that u → 0 in all regions, as x → ±y (ory→ ±x), and hence, u is continuous across y = ±xwhich represent asymptotes of the rectangular hyperbolasx2−y2 = C. However, ux and uy arenot continuous, asy→ ±x. Forx2−y2=C >0,
ux= 3x2+ 2xy−y2= (x+y) (3x−y)→0, as y→ −x.
uy =−3y2−2xy+x2= (x+y) (x−3y)→0, as y→ −x.
Hence, bothux anduy are continuous as y→ −x. On the other hand, ux→4x2, uy→ −4x2 as y→x.
This implies thatuxanduy are discontinuous acrossy=x.
Combining all these results, we conclude thatu(x, y) is continuous ev- erywhere in the (x, t)-plane, andux, uy are continuous everywhere in the (x, t)-plane except on the line y=x. Hence, the partial derivatives ux,uy
are discontinuous ony=x. Thus, the development ofdiscontinuitiesacross characteristics is a significant feature of the solutions of partial differential equations.
2.5 Method of Characteristics and General Solutions 45 Example 2.5.9.Determine the integral surfaces of the equation
x
y2+u ux−y
x2+u uy =
x2−y2 u, (2.5.53) with the data
x+y= 0, u= 1.
The characteristic equations are dx
x(y2+u) = dy
−y(x2+u) = du
(x2−y2)u (2.5.54) or
dx x
(y2+u) =
dy y
−(x2+u) =
du u
(x2−y2) =
dx
x +dyy +duu
0 .
Consequently,
log (xyu) = logC1
or
xyu=C1. From (2.5.54), we obtain
xdx
x2(y2+u) = ydy
−y2(x2+u) = du
(x2−y2)u= xdx+ydy−du
0 ,
whence we find that
x2+y2−2u=C2. Using the given data, we obtain
C1=−x2 and C2= 2x2−2, so that
C2=−2 (C1+ 1). Thus the integral surface is given by
x2+y2−2u=−2−2xyu or
2xyu+x2+y2−2u+ 2 = 0. (2.5.55)
Example 2.5.10.Obtain the solution of the equation
x ux+y uy =xexp (−u) (2.5.56) with the data
u= 0 on y=x2. The characteristic equations are
dx x = dy
y = du
xexp (−u) (2.5.57)
or
y x =C1.
We also obtain from (2.5.57) thatdx =eudu which can be integrated to find
eu=x+C2. Thus, the general solution is given by
f
eu−x,y x
= 0 or, equivalently,
eu =x+gy x
. (2.5.58)
Applying the Cauchy data, we obtaing(x) = 1−x. Thus, the solution of (2.5.56) is given by
eu=x+ 1−y x or
u= log
x+ 1−y x
. (2.5.59)
Example 2.5.11.Solve the initial-value problem
ut+u ux=x, u(x,0) =f(x), (2.5.60) where (a)f(x) = 1 and (b)f(x) =x.
The characteristic equations are dt
1 = dx u = du
x = d(x+u)
x+u . (2.5.61)
2.5 Method of Characteristics and General Solutions 47 Integration gives
t= log (x+u)−logC1
or
(u+x)e−t=C1. Similarly, we get
u2−x2 =C2. For case (a), we obtain
1 +x=C1 and 1−x2=C2, and hence C2= 2C1−C12. Thus,
u2−x2 = 2 (u+x)e−t−(u+x)2e−2t or
u−x= 2e−t−(u+x)e−2t. A simple manipulation gives the solution
u(x, t) =xtanht+ secht. (2.5.62) Case (b) is left to the reader as an exercise.
Example 2.5.12.Find the integral surface of the equation
u ux+uy = 1, (2.5.63)
so that the surface passes through an initial curve represented parametri- cally by
x=x0(s), y=y0(s), u=u0(s), (2.5.64) wheresis a parameter.
The characteristic equations for the given equations are dx
u = dy 1 =du
1 , which are, in the parametric form,
dx
dτ =u, dy
dτ = 1, du
dτ = 1, (2.5.65)
where τ is a parameter. Thus the solutions of this parametric system in general depend on two parameterss and τ. We solve this system (2.5.65) with the initial data
x(s,0) =x0(s), y(s,0) =y0(s), u(s,0) =u0(s). The solutions of (2.5.65) with the given initial data are
x(s, τ) = τ22 +τ u0(s) +x0(s) y(s, τ) =τ+y0(s)
u(s, τ) =τ+u0(s)
⎫⎬
⎭. (2.5.66)
We choose a particular set of values for the initial data as x(s,0) = 2s2, y(s,0) = 2s, u(s,0) = 0, s >0.
Therefore, the solutions are given by x=1
2τ2+ 2s2, y=τ+ 2s, u=τ. (2.5.67) Eliminatingτ andsfrom (2.5.67) gives the integral surface
(u−y)2+u2= 2x or
2u=y± 4x−y2
1
2. (2.5.68)
The solution surface satisfying the datau= 0 ony2= 2xis given by 2u=y−
4x−y2
1
2. (2.5.69)
This represents the solution surface only wheny2<4x. Thus, the solution does not exist fory2>4xand isnotdifferentiable wheny2= 4x. We verify thaty2= 4xrepresents theenvelope of the family of characteristics in the (x, t)-plane given by the τ-eliminant of the first two equations in (2.5.67), that is,
F(x, y, s) = 2x−(y−2s)2−4s2= 0. (2.5.70) This represents a family of parabolas for different values of the parameter s. Thus, the envelope is obtained by eliminatings from equations ∂F∂s = 0 andF = 0. This givesy2= 4x, which is the envelope of the characteristics for differents, as shown in Figure 2.5.2.