I have a boundary-value problem that need to be solved using laplace transform

Question

∂^2u/∂x^2 = ∂^2u/∂t^2 + 2∂u/∂t + u , x>0 , t>0

u(0,t) = sin 3t, t>0

u(x,0)=0, ∂u/∂t |t=0 =0, x>0

The answer is: u(x,t) = e^-x sin 3(t-x)u(t-x)

Answer 1

Taking Laplace transform of both sides with respect to t, and denoting the Laplace transform of u(x,t) by U(x,s), we get:

L{∂^2u/∂t^2}(x,s) + 2L{∂u/∂t}(x,s) + L{u}(x,s) = s^2U(x,s) - su(x,0) - u'(x,0) + 2sU(x,s) + U(x,s) = (s^2 + 2s + 1)U(x,s)

Taking Laplace transform of the initial and boundary conditions, we get:

u(0,t) = sin 3t => U(0,s) = L{sin 3t}(s) = 3/(s^2 + 9) u(x,0) = 0 => U(x,0) = 0 ∂u/∂t |t=0 =0 => sU(x,0) - u(x,0) = 0

Using these Laplace transforms, the original partial differential equation and initial/boundary conditions can be transformed into an algebraic equation in the Laplace domain. Solving for U(x,s), we get:

U(x,s) = 3s/(s^2 + 9) * (1/(s^2 + 2s + 1)) = 3s/(s+1)^2 * 1/(s^2 + 9)

Now we need to find the inverse Laplace transform of U(x,s) to get the solution u(x,t). We can use partial fraction decomposition to simplify U(x,s) into the following form:

U(x,s) = A/(s+1) + B/(s+1)^2 + C/(s^2 + 9)

where A, B, and C are constants to be determined. Multiplying both sides by the common denominator and solving for A, B, and C, we get:

A = 3/8, B = 3/4, C = -3/8

Therefore, we have:

U(x,s) = 3/8/(s+1) + 3/4/(s+1)^2 - 3/8/(s^2 + 9)

Taking the inverse Laplace transform of each term using standard Laplace transform tables, we obtain:

u(x,t) = 3/8e^(-t) + 3/4te^(-t) - 3/8sin(3t)*e^(-3x)

Therefore, the solution to the given partial differential equation is:

u(x,t) = 3/8e^(-t) + 3/4te^(-t) - 3/8sin(3t)*e^(-3x), x>0, t>0.