324 10 Partial Differential Equations
in our setup of the initial conditions.
10.4.1Closed-form Solution
We develop here the closed-form solution for the 2 + 1 dimensional wave equation with the
following boundary and initial conditions
c^2 (uxx+uyy) =utt x,y∈( 0 ,L),t> 0
u(x,y, 0 ) =f(x,y) x,y∈( 0 ,L)
u( 0 , 0 ,t) =u(L,L,t) = 0 t> 0
∂u/∂t|t= 0 =g(x,y) x,y∈( 0 ,L).
Our first step is to make the ansatz
u(x,y,t) =F(x,y)G(t),resulting in the equation
F Gtt=c^2 (FxxG+FyyG),
or
Gtt
c^2 G
=
1
F
(Fxx+Fyy) =−ν^2.The lhs and rhs are independent of each other and we obtain twodifferential equations
Fxx+Fyy+Fν^2 = 0 ,and
Gtt+Gc^2 ν^2 =Gtt+Gλ^2 = 0 ,
withλ=cν. We can in turn make the following ansatz for thexandydependent part
F(x,y) =H(x)Q(y),which results in
1
H
Hxx=−^1
Q
(Qyy+Qν^2 ) =−κ^2.
Since the lhs and rhs are again independent of each other, we can separate the latter equation
into two independent equations, one forxand one fory, namely
Hxx+κ^2 H= 0 ,and
Qyy+ρ^2 Q= 0 ,
withρ^2 =ν^2 −κ^2.
The second step is to solve these differential equations, which all have trigonometric func-
tions as solutions, viz.
H(x) =Acos(κx)+Bsin(κx),
and
Q(y) =Ccos(ρy)+Dsin(ρy).
The boundary conditions require thatF(x,y) =H(x)Q(y)are zero at the boundaries, meaning
thatH( 0 ) =H(L) =Q( 0 ) =Q(L) = 0. This yields the solutions