Using Laplace Transforms to solve a PDE

Question

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!

We haven't been taught that in the course...but how would you do it using complex analysis? — user2998682, Nov 28 '17 at 04:24
Generally the inverse transform is defined as an contour integral parallel to the imaginary line. — Dylan, Nov 28 '17 at 04:38

score 3 · Answer 1 · answered Nov 28 '17 at 16:23

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. :)

Using Laplace Transforms to solve a PDE

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