Question

Solve the following IBVP using Laplace transform technique:
u_t = u_{xx} , 0<x<1, t>0
u(0,t) = 1, u(1,t) = 1, t>0
u(x,0) = 1 + sin \pi x, 0<x<1

15 Feb 2024
Answer :
Word Count : 764

To solve the given initial-boundary value problem (IBVP) using the Laplace transform technique, let's denote \( \mathcal{L} \{u(x, t)\} \) as the Laplace transform of \( u(x, t) \) with respect to \( t \), and \( \mathcal{L} \{u(x, t)\} = U(x, s) \). The Laplace transform of the partial derivatives \( u_t \) and \( u_{xx} \) are given by:

\[
\mathcal{L} \{u_t\} = sU(x, s) - u(x, 0)
\]

\[
\mathcal{L} \{u_{xx}\} = x^2U(x, s) - xu(x, 0) - u_x(0, s) + u_x(1, s)
\]

where \( u_x(0, s) \) and \( u_x(1, s) \) are the Laplace transforms of the spatial derivatives \( u_x(0, t) \) and \( u_x(1, t) \) with respect to \( t \), respectively.

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