Question #194758

Using Laplace transformers solve du/dt=d²u/dx², x>0 ,t>0 with the condition U(0,t)=1,u(x,0)=0


1
Expert's answer
2021-05-18T18:13:46-0400

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\{ u(x,t) \}=U(x,s)

L{du(x,t)dt}=sU(x,s)u(x,0)=sU(x,s)L\Big\{ \frac {d u(x,t)}{dt} \Big\}=sU(x,s)-u(x,0)=sU(x,s)

L{d2u(x,t)dx2}=Uxx(x,s)L\Big\{ \frac {d^2 u(x,t)}{dx^2} \Big\}=U_{xx}(x,s)

Applying the Laplace transform to the equation, we have


Uxx(x,s)=sU(x,s)U_{xx}(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+C2exsU(x,s)=C_1 e^{x \sqrt s}+C_2 e^{-x \sqrt s}

To ensure physical feasibility, we assume that


u(,t)=0  U(,s)=0  C1=0u(\infin ,t)=0 \space \rightarrow \space U(\infin,s)=0 \space \rightarrow \space C_1=0

from the boundary condition


u(0,t)=1=L1{U(0,s)}=L1{C2}, t>0u(0,t)=1=L^{-1}\{U(0,s) \}=L^{-1}\{C_2 \}, \space t>0

then


C2=1s, s>0C_2 =\frac 1 s, \space s>0

So, the solution is


U(x,s)=1sexsU(x,s)=\frac 1 s e^{-x \sqrt s}

apply the inverse Laplace transform


u(x,t)=L1{1sexs}=erfc(x2t)u(x,t)=L^{-1} \Big \{\frac 1 s e^{-x \sqrt s} \Big \}=erfc \Big (\frac{x}{2 \sqrt t} \Big)

Answer:  u(x,t)=erfc(x2t), t0, x0\space u(x,t)=erfc \Big (\frac{x}{2 \sqrt t} \Big), \space t \ge 0, \space x \ge 0


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