PDES: GENERAL AND PARTICULAR SOLUTIONS
We now tackle the problem of solving some types of second-order PDE with
constant coefficients by seeking solutions that are arbitrary functions of particular
combinations of independent variables, just as we did for first-order equations.
Following the discussion of the previous section, we can hope to find such
solutions only if all the terms of the equation involve the same total number
of differentiations, i.e. all terms are of the same order, although the number
of differentiations with respect to the individual independent variables may be
different. This means that in (20.19) we require the constantsD,EandFto be
identically zero (we have, of course, already assumed thatR(x, y)iszero),sothat
we are now considering only equations of the form
A
∂^2 u
∂x^2
+B
∂^2 u
∂x∂y
+C
∂^2 u
∂y^2
=0, (20.20)
whereA,BandCare constants. We note that both the one-dimensional wave
equation,
∂^2 u
∂x^2
−
1
c^2
∂^2 u
∂t^2
=0,
and the two-dimensional Laplace equation,
∂^2 u
∂x^2
+
∂^2 u
∂y^2
=0,
are of this form, but that the diffusion equation,
κ
∂^2 u
∂x^2
−
∂u
∂t
=0,
is not, since it contains a first-order derivative.
Since all the terms in (20.20) involve two differentiations, by assuming a solution
of the formu(x, y)=f(p), wherepis some unknown function ofxandy(ort),
we may be able to obtain a common factord^2 f(p)/dp^2 as the only appearance of
fon the LHS. Then, because of the zero RHS, all reference to the form offcan
be cancelled out.
We can gain some guidance on suitable forms for the combinationp=p(x, y)
by considering∂u/∂xwhenuis given byu(x, y)=f(p), for then
∂u
∂x
=
df(p)
dp
∂p
∂x
.
Clearly differentiation of this equation with respect tox(ory) will not lead to a
single term on the RHS, containingfonly asd^2 f(p)/dp^2 , unless the factor∂p/∂x
is a constant so that∂^2 p/∂x^2 and∂^2 p/∂x∂yare necessarily zero. This shows that
pmust be a linear function ofx. In an exactly similar waypmust also be a linear
function ofy,i.e.p=ax+by.
If we assume a solution of (20.20) of the formu(x, y)=f(ax+by), and evaluate