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Solve the initial and boundary value problem for the wave equation

u_{tt} = c^2 u_{xx}, 0 < x < L, t>0,
but with the following initial and boundary conditions:
1. L = 1, C^2 = 1, u (0, x) = x, u_t (0, x) = 1, u(t, 0) = u(t, 1) = 0.
2. L = 1, C^2 = 1, u (0, x) = 1, u_t (0, x) = x, u'(t, 0) = u'(t, 1) = 0.
3. L = 2, C^2 = 2, u (0, x) = 1, u_t (0, x) = 1, u'(t, 0) = u(t, 2) = 0.

1 Answer

6 votes

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(

User Jarrod Nettles
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