Chapter 5 Higher Dimensions and Other Coordinates 297
Adding up forces in the vertical direction and equating the sum to the mass
times acceleration (in the vertical direction) we obtain
σ y(
sin(β)−sin(α))
+σ x(
sin(δ)−sin(γ ))
=ρ x y∂(^2) u
∂t^2
,
whereρis the surface density [m/L^2 ]. Because the anglesα, β, γ,andδare
small, the sine of each is approximately equal to its tangent:
sin(α)∼=tan(α)=∂u
∂x(x,y,t),and so forth. With these approximations used throughout, the preceding equa-
tion becomes
σ y(∂y
∂x(x+^ x,y,t)−∂u
∂x(x,y,t))
+σ x(∂u
∂y(x,y+ y,t)−∂u
∂y(x,y,t))
=ρ x y∂(^2) u
∂t^2
.
On dividing through by x y, we recognize two difference quotients in
the left-hand member. In the limit they become partial derivatives, yielding
the equation
σ(∂ (^2) u
∂x^2 +
∂^2 u
∂y^2
)
=ρ∂^2 u
∂t^2 ,
or
∂^2 u
∂x^2
+∂
(^2) u
∂y^2
=^1
c^2∂^2 u
∂t^2,
ifc^2 =σ/ρ. This is the two-dimensional wave equation.
If the membrane is fixed to the flat frame, the boundary condition would be
u(x,y,t)= 0 for (x,y)on the boundary.
Naturally, it is necessary to give initial conditions describing the displacement
and velocity of each point on the membrane att=0:
u(x,y, 0 )=f(x,y),
∂u
∂t(x,y, 0 )=g(x,y).EXERCISES
- Suppose that the frame is rectangular, bounded by segments of the lines
x=0,x=a,y=0,y=b. Write an initial value–boundary value problem,
complete with inequalities, for a membrane stretched over this frame.