Answer to Question #194758 in Math for Arpan Chakraborty

Question

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

Answer 1

We apply the Laplace transform to the function u(x,t) in t, considering x as a parameter

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

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

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

Applying the Laplace transform to the equation, we have

$$U_{xx}(x,s)=sU(x,s)$$

This is a second order linear differential equation with a constant coefficient s>0: the solution can be written as

$$U(x,s)=C_1 e^{x \sqrt s}+C_2 e^{-x \sqrt s}$$

To ensure physical feasibility, we assume that

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

from the boundary condition

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

then

$$C_2 =\frac 1 s, \space s>0$$

So, the solution is

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

apply the inverse Laplace transform

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

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