We apply the Laplace transform to the function u(x,t) in t, considering x as a parameter
L{u(x,t)}=U(x,s)
L{dtdu(x,t)}=sU(x,s)−u(x,0)=sU(x,s)
L{dx2d2u(x,t)}=Uxx(x,s) Applying the Laplace transform to the equation, we have
Uxx(x,s)=sU(x,s) This is a second order linear differential equation with a constant coefficient s>0: the solution can be written as
U(x,s)=C1exs+C2e−xs To ensure physical feasibility, we assume that
u(∞,t)=0 → U(∞,s)=0 → C1=0 from the boundary condition
u(0,t)=1=L−1{U(0,s)}=L−1{C2}, t>0 then
C2=s1, s>0 So, the solution is
U(x,s)=s1e−xs apply the inverse Laplace transform
u(x,t)=L−1{s1e−xs}=erfc(2tx) Answer: u(x,t)=erfc(2tx), t≥0, x≥0
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