For a given PDE:

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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.


2 Answers 2


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.

  • $\begingroup$ 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? $\endgroup$ Commented Nov 9, 2018 at 6:49
  • $\begingroup$ @ShripathiRamakrishnan : $C(s)$ is determined by setting $\mathscr{L}\{u\}_x(x,s)=0$ at $x=0$. $\endgroup$ Commented Nov 9, 2018 at 6:52

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