Question
Solve the following IBVP using the Laplace transform technique:
Answer :
Word Count : 511
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Skip in 10
We are asked to solve the initial-boundary value problem (IBVP) [ u_t = 2 u_{xx}, \quad 0 < x < 1, \ t>0, ] [ u(0,t) = 1, \quad u(1,t) = 1, \quad t>0, ] [ u(x,0) = 1 + \sin(\pi x), \quad 0 < x < 1, ] using the Laplace transform method. --- Step 1: Laplace transform in (t) Let (\bar{u}(x,s) = \mathcal{L}{u(x,t)} = \int_0^\infty e^{-st} u(x,t), dt). The Laplace transform of (u_t) gives [ \mathcal{L}{u_t} = s \bar{u}(x,s) - u(x,0) = s \bar{u}(x,s) - (1 + \sin(\pi x)). ] The PDE transforms to [ s \bar{u} - (1 + \sin \pi x) = 2 \bar{u}*{xx} \quad \implies \quad 2 \bar{u}*{xx} - s \bar{u} = -(1 + \sin \pi x), ] with transformed boundary conditions [ \bar{u}(0,s) = \frac{1}{s}, \quad \bar{u}(1,s) = \frac{1}{s}. ] --- Step 2: Solve the transformed ODE The ODE is [ \bar{u}_{xx} - \frac{s}{2} \bar{u} = -\frac{1 + \sin \pi x}{2}. ] Step 2a: Homogeneous solution Solve [ \bar{u}_{xx} - \frac{s}{2} \bar{u} = 0 \quad \implies \quad \bar{u}_h = A e^{\sqrt{s/2} , x}
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