Answer:
Step-by-step explanation:
Case 1:
L = 1, c^2 = 1, u(0, x) = x, u_t(0, x) = 1, u(t, 0) = u(t, 1) = 0.
We assume the solution can be expressed as u(t, x) = T(t)X(x). Substituting this into the wave equation, we get:
T''(t)X(x) = X''(x)T(t) / c^2
Dividing both sides by T(t)X(x) and setting them equal to a separation constant -λ^2, we have:
T''(t) / T(t) = X''(x) / (c^2X(x)) = -λ^2
Solving the time part, we have T''(t) + λ^2T(t) = 0, which has solutions of the form:
T(t) = Aλsin(λt) + Bλcos(λt)
Next, we solve the spatial part, X''(x) + λ^2c^2X(x) = 0, with the given boundary conditions:
X(x) = sin(nπx), where n is a positive integer.
Since the boundary condition u(0, x) = x implies X(0) = 0, we can ignore the cosine term in the spatial solution. So, our general solution becomes:
u(t, x) = Σ(Aλsin(λt))sin(nπx)
Applying the initial condition u(0, x) = x, we can determine the coefficients Aλ using Fourier sine series:
x = Σ(Aλsin(0))sin(nπx)
x = Σ(0)sin(nπx) (since sin(0) = 0)
Therefore, the coefficient Aλ is 0 for all λ. The solution for case 1 is:
u(t, x) = 0
Case 2:
L = 1, c^2 = 1, u(0, x) = 1, u_t(0, x) = x, u'(t, 0) = u'(t, 1) = 0.
Following the same separation of variables, we have:
T''(t) / T(t) = X''(x) / (c^2X(x)) = -λ^2
The time part is T(t) = Aλsin(λt) + Bλcos(λt), same as before.
For the spatial part, we have X''(x) + λ^2X(x) = 0. The general solution for this is:
X(x) = Asin(λx) + Bcos(λx)
Applying the boundary condition u'(t, 0) = 0, we get:
u'(t, 0) = ∑(Aλλcos(λt))sin(λx) = 0
This implies Aλλ = 0, except for λ = 0.
Therefore, the spatial part becomes X(x) = Bcos(λx).
The complete solution for case 2 is:
u(t, x) = Bcos(λx)(Aλsin(λt) + Bλcos(λt))
Applying the initial condition u(0, x) = 1, we find:
1 = Bcos(λx)(Aλsin(