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I was using Laplace to find solutions for $$\frac{\partial u}{\partial t}=\frac{\partial^2 u}{\partial x^2}$$

with boundary conditions $$u(0,t)=1 \\ u(1,t)=1 \\ u(x,0)=1+ \sin \pi x$$

I used $U(x,s)= \mathcal L^{-1}{u(x,t)}$ and got $$U(x,s)=Ae^{x\sqrt s} + Be^{-x\sqrt s} + \frac{1}{s} + \frac{\sin \pi x}{s+\pi ^2}$$

$$u(x,t)= A\mathcal L ^{-1} \{e^{x\sqrt s} \} + B\mathcal L ^{-1} \{e^{-x\sqrt s} \} + 1 + e^{-\pi ^2t}\sin \pi x$$

How to solve for the first two inverse Laplace terms?

S.Dan
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  • http://math.stackexchange.com/questions/347933/compute-the-inverse-laplace-transform-of-e-sqrtz/348021#348021 – Ron Gordon Feb 12 '16 at 08:30
  • $L^{-1}{e^{\sqrt{x} s}}$ or $L^{-1}{e^{a \sqrt{s}}}$ ? the former is easy but is a distribution, it is $f(t) = \delta(t+\sqrt{x})$ (so you made a mistake elsewhere as your initial condition is not a distribution) – reuns Feb 12 '16 at 08:37
  • @user1952009 sorry about it - just corrected it – S.Dan Feb 12 '16 at 09:26

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