1.1 Partial Differential Equations-Some Basics
9
Since b 2 - ac > 0, these roots are real and distinct. Denoting the roots by -VI
and -V2, one may choose transformed coordinates such that dy jdx = VI along
lines of constant and dy fdx = V2 along lines of constant 11. This choice of
coordinates satisfies (1.17) and (1.18) because
and
Those curves along which either
for the hyperbolic equation (1.11) .
or 11 is constant are the characteristic curves
After zeroing A and C, the canonical form (1.12) is obtained by dividing (1.15)
by
11), which must be nonzero by (1.16) because
-
=
- VI) i= O.
In the case a = 0, a similar expression for the transformed coordinates can be
obtained by dividing the relations
11) = 0 and
11) = 0 by c instead of
a. If both a and c are zero, the partial differential equation is placed in canonical
form simply by dividing by b (which is nonzero because b 2 - ac > 0).
If i. is zero, the canonical hyperbolic equation (1.12) has solutions of the form
and h(I1). One circumstance in which i is zero occurs when a, b, and c are
constant and L = 0 in (1.11) . Then the characteristics are the straight lines
= y - VIX
and
11 = Y - V2X ,
and there exist solutions of the form g(y - VIX) and h(y - V2X) . When (1.11)
serves as a mathematical model for wave-propagation problems, it usually includes a second-order derivative with respect to time . Suppose, therefore, that
a i= 0 and that x represents the time coordinate. Then the speed of signal propagation along the characteristics is given by their slope in the y -x plane, which is
VI for the constant-s characteristics and V2 for the constant-n characteristics.
In the parabolic case with a i= 0, the quadratic equation (1.17) has the double
root -b/a, and there is a single characteristic defined such that
= -i:
(1.19)
Let I1(X, y) be any simple function such that
-
i= O.
These choices for and 11 imply that A = 0 and B
2 - AC = 0, which in turn
implies that B = O. The canonical parabolic form (1.13) is obtained by dividing
(1.15) by C, which must be nonzero, or else neither (1.11) nor (1.15) will be
a second-order differential equation. If, on the other hand, a = 0 in (1.11), a
similar transformation can be performed after dividing through by c, which must
be nonzero if a second-order partial derivative is present in (1.11) because b 2 =
b 2 - ac = O.
9
Since b 2 - ac > 0, these roots are real and distinct. Denoting the roots by -VI
and -V2, one may choose transformed coordinates such that dy jdx = VI along
lines of constant and dy fdx = V2 along lines of constant 11. This choice of
coordinates satisfies (1.17) and (1.18) because
and
Those curves along which either
for the hyperbolic equation (1.11) .
or 11 is constant are the characteristic curves
After zeroing A and C, the canonical form (1.12) is obtained by dividing (1.15)
by
11), which must be nonzero by (1.16) because
-
=
- VI) i= O.
In the case a = 0, a similar expression for the transformed coordinates can be
obtained by dividing the relations
11) = 0 and
11) = 0 by c instead of
a. If both a and c are zero, the partial differential equation is placed in canonical
form simply by dividing by b (which is nonzero because b 2 - ac > 0).
If i. is zero, the canonical hyperbolic equation (1.12) has solutions of the form
and h(I1). One circumstance in which i is zero occurs when a, b, and c are
constant and L = 0 in (1.11) . Then the characteristics are the straight lines
= y - VIX
and
11 = Y - V2X ,
and there exist solutions of the form g(y - VIX) and h(y - V2X) . When (1.11)
serves as a mathematical model for wave-propagation problems, it usually includes a second-order derivative with respect to time . Suppose, therefore, that
a i= 0 and that x represents the time coordinate. Then the speed of signal propagation along the characteristics is given by their slope in the y -x plane, which is
VI for the constant-s characteristics and V2 for the constant-n characteristics.
In the parabolic case with a i= 0, the quadratic equation (1.17) has the double
root -b/a, and there is a single characteristic defined such that
= -i:
(1.19)
Let I1(X, y) be any simple function such that
-
i= O.
These choices for and 11 imply that A = 0 and B
2 - AC = 0, which in turn
implies that B = O. The canonical parabolic form (1.13) is obtained by dividing
(1.15) by C, which must be nonzero, or else neither (1.11) nor (1.15) will be
a second-order differential equation. If, on the other hand, a = 0 in (1.11), a
similar transformation can be performed after dividing through by c, which must
be nonzero if a second-order partial derivative is present in (1.11) because b 2 =
b 2 - ac = O.