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I need to solve the following PDE, given the boundary conditions and initial condition, using Laplace Transforms:

$$ \begin{aligned} \displaystyle K_1\frac{\partial^2u}{\partial x^2}+K_2u&=\frac{\partial u}{\partial t} \\ u(0,t)&=u_0 \\ u(\infty,t)&=0 \\ u(x,0)&=0 \end{aligned} $$

So far, I've taken the Laplace transform of the PDE to get the following result:

$$K_1\frac{\partial ^2U}{\partial x^2}+K_2U=sU$$

Solving the above ODE results in the following:

$$U(x,s)=C_1\exp\left(\sqrt{\frac{k_2+s}{k_1}}x\right)+C_2\exp\left(-\sqrt{\frac{k_2+s}{k_1}}x\right)$$

By taking the Laplace transform of the two boundary conditions, I get the following:

$$U(0,s)=\frac{u_0}{s}$$

$$U(\infty,s)=0$$

Using the second boundary condition, I can calculate that $C_2=0$, and that the PDE in terms of $x$ and $s$ is:

$$U(x,s)=\frac{u_0}{s}\exp\left(\sqrt{\frac{K_2+s}{K_1}}x\right)$$

Now, I have to take the inverse Laplace transform, but I'm having problems at this step. I have the result which is:

$$ u(x,t)=\frac{2u_0}{\sqrt{\pi}}\int_{\frac{x}{2\sqrt{K_1t}}}^{\infty} \exp\left(-\lambda^2-\frac{K_2x^2}{4K_1\lambda^2}\right) \ d\lambda $$

It's suggested that I use one of the shifting theorems to obtain this solution, but I'm not sure how. Thanks!

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  • $\begingroup$ Are you allowed to use complex analysis? $\endgroup$
    – Dylan
    Commented Nov 28, 2017 at 2:34
  • $\begingroup$ We haven't been taught that in the course...but how would you do it using complex analysis? $\endgroup$ Commented Nov 28, 2017 at 4:24
  • $\begingroup$ Generally the inverse transform is defined as an contour integral parallel to the imaginary line. $\endgroup$
    – Dylan
    Commented Nov 28, 2017 at 4:38

1 Answer 1

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There is some simplification if you rewrite the PDE as $$ K_1\frac{\partial^2u}{\partial x^2}=\frac{\partial u}{\partial t}-K_2u \\ K_1 \frac{\partial^2 (e^{-K_2t}u)}{\partial^2 x}=\frac{\partial (e^{-K_2t}u)}{\partial t} $$ Let $v(x,t)=e^{-K_2t}u(x,t)$. Then the new PDE becomes $$ v_t = K_1 v_{xx} \\ v(0,t) = u_0e^{-K_2t} \\ v(\infty,t) = 0 \\ v(x,0) = 0. $$ Now when you transform in $t$: \begin{align} \mathscr{L}\{v_t\} & =\int_{0}^{\infty}v_t(x,t)e^{-st}dt \\ & =v(x,t)e^{-st}|_{t=0}^{\infty}-\int_{0}^{\infty}v(x,t)e^{-st}(-s)dt \\ & = s\mathscr{L}\{v\}. \end{align} So the transformed equations for $\mathscr{L}\{v\}(s,x)$ are \begin{align} s\mathscr{L}\{v\}(s,x) &=K_1\mathscr{L}\{v\}_{xx}(s,x), \\ \mathscr{L}\{v\}(s,0)&= \mathscr{L}\{u_0e^{-K_2t}\}=\frac{u_0}{s+K_2}. \end{align} The solution in $s,x$ is $$ \mathscr{L}\{v\}(s,x) = \frac{u_0}{s+K_2}\exp\left(-\sqrt{\frac{s}{K_1}}x\right). $$ The inverse Laplace transform is a convolution in $t$ of inverse transforms of the two terms multiplying each other on the right above: \begin{align} v(t,x) & = \left( u_0 e^{-K_2 t}\right)*\left( \frac{1}{\pi}\int_{0}^{\infty}e^{-st}\sin\left(\sqrt{\frac{s}{K_1}}x\right)ds\right) \\ & = \frac{u_0}{\pi}\int_{0}^{\infty}\int_{0}^{t}e^{-K_2t'}e^{-s(t-t')}dt'\sin\left(\sqrt{\frac{s}{K_1}}x\right)ds \end{align} The integral in $t'$ is easily simplified and the solution $u(x,t)=e^{K_2t}v(x,t)$ of the original equation recovered, but I'll leave it there, before I compound my errors. :)

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