# Solving PDE in semi-infinite domain using Laplace Transform

For a given PDE:

It is required to solve the PDE by method of Laplace Transform. I am able to convert the PDE into the Laplace domain to establish an equation in terms of second differential of u wrt x in terms of s. However, I am stuck with solving the particular and cumulative solution. Through one BC, it is clear that c1=0. However, I am confused as to how to proceed further. I am also slightly confused as to how to assign a particular solution to the same. The scope of the exercise does NOT permit the use of Method of Variation of Parameters.

Can anyone kindly shed some light on this? Also, I would greatly appreciate it if anyone had similar examples they'd be so kind as to share.

Best regards.

Let $$U(x,s)$$ be the Laplace transform of $$u(x,t)$$, then the Laplace transform of $$u_t$$ is given by

$$\mathcal \{u_t(x,t)\} = sU(x,s) - u(x,0) = sU - 1$$

Then we have

$$sU-1 = \alpha U_{xx}$$

Treating $$s$$ as constant, we can solve the above equation as a second-order ODE in $$x$$. The general solution is

$$U(x,s) = \frac{1}{s} + A(s) e^{\sqrt{s/\alpha}\ x} + B(s)e^{-\sqrt{s\alpha}\ x}$$

Since the function is required to be bounded at $$x\to \infty$$, we need $$A(s) = 0$$

The boundary condition $$U_x(0,s) = 0$$ forces $$B(s)=0$$ as well.

Therefore, the solution is just $$U(x,s) = \frac{1}{s}$$, or $$u(x,t) = 1$$

I assume the Laplace transform is applied in $$t$$. The Laplace transform is $$\mathscr{L}\{ u \} = \int_{0}^{\infty}e^{-st}u(x,t)dt$$

Assuming $$s > 0$$, and assuming $$u(x,\infty)$$ exists, \begin{align} \mathscr{L}\{u_t\}& = \int_{0}^{\infty}e^{-st}u_t(x,t)dt \\ &= e^{-st}u(x,t)|_{t=0}^{\infty}+s\int_{0}^{\infty}e^{-st}u(x,t)dt \\ &= -u(x,0)+s\int_{0}^{\infty}e^{-st}u(x,t)dt \\ &= -1+s\mathscr{L}\{u\}. \end{align} Therefore, $$-1+s\mathscr{L}\{u\} =\mathscr{L}\{u_t\}=\mathscr{L}\{\alpha u_{xx}\}=\alpha(\mathscr{L}\{u\})_{xx}$$ or $$\alpha\mathscr{L}\{u\}_{xx}-s\mathscr{L}\{u\}=-1$$

Therefore, $$\mathscr{L}\{u\}_{xx}-\frac{s}{\alpha}\mathscr{L}\{u\}=-\frac{1}{\alpha} \\ \therefore \;\mathscr{L}\{u\} = C(s)e^{-\sqrt{s/\alpha}\,x}+\frac{1}{s}$$ So $$u$$ is the inverse Laplace transform of the aboe.

• Thank you for clarifying. I had managed up to that step myself earlier. However, I am a bit confused as to how the second boundary condition du/dx is applied and how C(s) is obtained. How do I solve for u when C(s) remains unknown? Commented Nov 9, 2018 at 6:49
• @ShripathiRamakrishnan : $C(s)$ is determined by setting $\mathscr{L}\{u\}_x(x,s)=0$ at $x=0$. Commented Nov 9, 2018 at 6:52