7. (a) Solve the wave equation in three dimensions for t > 0
with the
initial conditions φ(x) = A for |x| < ρ, φ(x) = 0 for |x| >
ρ, and
ψ|x| ≡ 0, where A is a constant. (This is somewhat like the
plucked
string.) (Hint: Differentiate the solution in Exercise 6(b).)
((b) Solve the wave equation in three dimensions for t > 0
with the
initial conditions φ(x) ≡ 0,ψ(x) = A for |x| < ρ, and...
Consider the 1D wave equation d^2u/dt^2 = c^2( d^2u/dx^2) with
the following boundary conditions: u(0, t) = ux (L, t) = 0 . (a)
Use separation of variables technique to calculate the eigenvalues,
eigenfunctions and general solution. (b) Now, assume L = π and c =
1. With initial conditions u(x, 0) = 0 and ut(x, 0) = 1, calculate
the solution for u(x, t). (c) With initial conditions u(x, 0) =
sin(x/2) and ut(x, 0) = 2 sin(x/2) −...
A particle is described by the wave function ψ(x) =
b(a2 - x2) for -a ≤ x ≤ a and ψ(x)=0 for x ≤
-a and x ≥ a, where a and b are positive real
constants.
(a) Using the normalization condition, find b in terms of
a.
(b) What is the probability to find the particle at x =
0.21a in a small interval of width 0.01a?
(c) What is the probability for the particle to be found between x...
A particle is described by the wave function ψ(x) = b(a2 - x2)
for -a ≤ x ≤ a and ψ(x)=0 for x ≤ -a and x ≥ a , where a and b are
positive real constants.
(a) Using the normalization condition, find b in terms of a.
(b) What is the probability to find the particle at x = 0.33a in
a small interval of width 0.01a?
(c) What is the probability for the particle to be found...
Find the solution to the heat equation on 0 < x < l,
with u(0, t) = 0, ux(l, t) = 0, and u(x, 0) =
phi(x).
This is sometimes called a "mixed" boundary condition.
Solve Laplace's equation inside a rectangle 0 ≤ x ≤ L, 0 ≤ y ≤
H, with the following boundary conditions [Hint: Separate
variables. If there are two homogeneous boundary conditions in y,
let u(x,y) = h(x)∅(y), and if there are two homogeneous boundary
conditions in x, let u(x,y) = ∅(x)h(y).]:
∂u/∂x(0,y) = 0
∂u/∂x(L,y) = 0
u(x,0) = 0
u(x,H) = f(x)
Take the Laplace transform of the following initial value and
solve for X(s)=L{x(t)}X(s)=L{x(t)}:
x′′+16x={sin(πt),0}
0≤t<1
1≤t
x(0)=0
x′(0)=0.
a) X(s)=
Now find the inverse transform to find
b) x(t)=
Use u(t−a) for the Heaviside function shifted a units
horizontaly.