Mathematical Methods for Physics and Engineering : A Comprehensive Guide

(Darren Dugan) #1

PDES: GENERAL AND PARTICULAR SOLUTIONS


were discussed. Comparing (20.41) with (20.12) we see that the characteristics are


merely those curves along whichpis constant.


Since the partial derivatives∂u/∂xand∂u/∂ymay be evaluated provided the

boundary curveCdoesnotlie along a characteristic, definingu(x, y)=φ(s)


alongCis sufficient to specify the solution to the original problem (equation


plus boundary conditions) near the curveC, in terms of a Taylor expansion


aboutC. Therefore the characteristics can be considered as the curves along


which information about the solutionu(x, y) ‘propagates’. This is best understood


by using an example.


Find the general solution of

x

∂u
∂x

− 2 y

∂u
∂y

= 0 (20.42)


that takes the value 2 y+1on the linex=1betweeny=0andy=1.

We solved this problem in subsection 20.3.1 for the case whereu(x, y) takes the value
2 y+1 along theentirelinex= 1. We found then that the general solution to the equation
(ignoring boundary conditions) is of the form


u(x, y)=f(p)=f(x^2 y),

for some arbitrary functionf. Hence the characteristics of (20.42) are given byx^2 y=c
wherecis a constant; some of these curves are plotted in figure 20.2 for various values of
c. Furthermore, we found that the particular solution for whichu(1,y)=2y+1for ally
was given by


u(x, y)=2x^2 y+1.

In the present case the value ofx^2 yis fixed by the boundary conditions only between
y=0andy= 1. However, since the characteristics are curves along whichx^2 y, and hence
f(x^2 y), remains constant, the solution is determined everywhere along any characteristic
that intersects the line segment denoting the boundary conditions. Thusu(x, y)=2x^2 y+1
is the particular solution that holds in the shaded region in figure 20.2 (corresponding to
0 ≤c≤1).
Outside this region, however, the solution is not precisely specified, and any function of
the form
u(x, y)=2x^2 y+1+g(x^2 y)


will satisfy both the equation and the boundary condition, providedg(p)=0for
0 ≤p≤1.


In the above example the boundary curve was not itself a characteristic and

furthermore it crossed each characteristiconce only. For a general boundary curve


Cthis may not be the case. Firstly, ifCis itself a characteristic (or is just a single


point) then information about the solution cannot ‘propagate’ away fromC,and


so the solution remains unspecified everywhere except onC.


The second possibility is thatC(although not a characteristic itself) crosses

some characteristics more than once, as in figure 20.3. In this case specifying the


value ofu(x, y) along the curvePQdetermines the solution along all the character-


istics that intersect it. Therefore, also specifyingu(x, y) alongQRcanoverdetermine


the problem solution and generally results in there being no solution.

Free download pdf