# Clarification of Solution of PDE Laplace Transform Problem

I have the following problem:

The transverse displacement $$u(x, t)$$ of a semi-infinite elastic string satisfies

$$\dfrac{\partial^2{u}}{\partial{t}^2} = c^2 \dfrac{\partial^2{u}}{\partial{x}^2}$$

for $$x > 0$$, $$t > 0$$,

with initial conditions

$$u(x, 0) = \dfrac{\partial{u(x, 0)}}{\partial{t}} = 0$$

for $$x > 0$$,

and boundary condition

$$-\beta \dfrac{\partial{u(0, t)}}{\partial{x}} = f(t)$$

for $$t > 0$$.

Show, using Laplace transforms, that the solution can be written

$$u(x, t) = \dfrac{c}{\beta} H \left( t - \dfrac{x}{c} \right) \int^{t - x/c}_0 f(u) \ du,$$

where $$H(x)$$ is the Heaviside function. Can you interpret this result physically?

And I have the following solution:

Taking the Laplace transform of the PDE yields

$$s^2 \mathcal{L} \{u\} - su(x, 0) - u_t(x, 0) = c^2 \dfrac{d^2}{dx^2} \mathcal{L} \{ u(x, t) \}$$.

Substituting the initial conditions gets us

\begin{align} &s^2 \mathcal{L} \{ u \} = c^2 \dfrac{d^2}{dx^2} \mathcal{L} \{ u(x, t) \} \\ &\Rightarrow \dfrac{d^2}{dx^2} \mathcal{L} \{ u(x, t) \} = \dfrac{s^2}{c^2} \mathcal{L} \{ u(x, t) \} \\ &\Rightarrow \dfrac{d^2}{dx^2} \mathcal{L} \{ u(x, t) \} - \dfrac{s^2}{c^2} \mathcal{L} \{u(x, t) \} = 0 \end{align}

We solve this homogeneous second-order linear DE:

\begin{align} &m^2 - \dfrac{s^2}{c^2} = 0 \\ &\Rightarrow m = \pm \dfrac{s}{c} \end{align}

Therefore, the general solution to this DE is

$$\mathcal{L} \{ u(x, t) \} = A(s) e^{x\frac{s}{c}} + B(s)e^{-x \frac{s}{c}},$$

where $$A(s)$$ and $$B(s)$$ are arbitrary functions.

To apply the boundary conditions, we differentiate the general solution with respect to $$x$$:

$$\dfrac{\partial}{\partial{x}} \mathcal{L} \{ u(x, t) \} = \dfrac{sA(s)}{c}e^{\frac{sx}{c}} - \dfrac{sB(s)}{c} e^{-\frac{sx}{c}}$$

And since we had the boundary condition

$$- \beta \dfrac{\partial}{\partial{x}} u(0, t) = f(t),$$

we have

$$\dfrac{\partial}{\partial{x}} \mathcal{L} \{ u(0, t) \} = \dfrac{sA(s)}{c} - \dfrac{sB(s)}{c}$$

Taking the Laplace transform of the boundary condition, we get

$$- \beta \dfrac{\partial}{\partial{x}} \mathcal{L} \{ u(0, t) \} = \mathcal{L} \{ f(t) \}$$

So we get

\begin{align} &-\dfrac{\mathcal{L} \{f(t) \}}{\beta} = \dfrac{sA(s)}{c} - \dfrac{sB(s)}{c} \\ & \Rightarrow -\dfrac{\mathcal{L} \{ f(t) \}}{\beta} = \dfrac{s}{c}(A(s) - B(s)) \\ & \Rightarrow \dfrac{c}{\beta} \dfrac{ \mathcal{L} \{ f(t) \}}{s} + A(s) = B(s) \end{align}

Using this to eliminate $$B(s)$$ from the general solution leaves

$$\dfrac{\partial}{\partial{x}} \mathcal{L} \{ u(x, t) \} = \dfrac{sA(s)}{c} e^{\frac{sx}{c}} - \left( \dfrac{c}{\beta} \dfrac{ \mathcal{L} \{ f(t) \}}{s} + A(s) \right) \dfrac{s}{c} e^{-\frac{sx}{c}}$$

We can now constrain $$A(s)$$ by using the initial value theorem with the initial conditions:

\begin{align} 0 &= u(x, 0) \\ &= \lim_{t \to 0} u(x, t) \\ &= \lim_{s \to \infty} s \mathcal{L} \{ u(x, t) \} \end{align}

for all $$x > 0$$.

Using the general solution, we see that

$$\lim_{s \to \infty} s \mathcal{L} \{ u(x, t) \} = \lim_{s \to \infty} A(s) e^{\frac{sx}{c}} s,$$

since $$\lim_{t \to 0} f(t) = 0$$ means that $$\lim_{s \to \infty} s \mathcal{L} \{ f(t) \} = 0$$.

Using the general solution, we see that

$$\lim_{s \to \infty} s \mathcal{L} \{ u(x, t) \} = \lim_{s \to \infty} A(s) e^{\frac{sx}{c}} s,$$

since $$\lim_{t \to 0} f(t) = 0$$ means that $$\lim_{s \to \infty} s \mathcal{L} \{ f(t) \} = 0$$.

I don't see how they got

$$\lim_{s \to \infty} s \mathcal{L} \{ u(x, t) \} = \lim_{s \to \infty} A(s) e^{\frac{sx}{c}} s$$

?

I would greatly appreciate it if someone could please take the time to clarify this.

First I want to mention that you forgot the $$\mathcal{L}$$ symbol at some places:

So we get

\begin{align} &-\dfrac{\mathcal{L}\{f(t)\}}{\beta} = \dfrac{sA(s)}{c} - \dfrac{sB(s)}{c} \\ & \Rightarrow -\dfrac{\mathcal{L}\{f(t)\}}{\beta} = \dfrac{s}{c}(A(s) - B(s)) \\ & \Rightarrow \dfrac{c}{\beta} \dfrac{\mathcal{L}\{f(t)\}}{s} + A(s) = B(s) \end{align}

Using this to eliminate $$B(s)$$ from the general solution leaves

$$\dfrac{\partial}{\partial{x}} \mathcal{L} \{ u(x, t) \} = \dfrac{sA(s)}{c} e^{\frac{sx}{c}} - \left( \dfrac{c}{\beta} \dfrac{\mathcal{L}\{f(t)\}}{s} + A(s) \right) \dfrac{s}{c} e^{-\frac{sx}{c}}$$

Now we can show the equation. Substituting the expression for $$B(s)$$ in $$\mathcal{L}\{u(x,t)\}$$ and multiplying with $$s$$ gives $$s\mathcal{L} \{ u(x, t) \} = s A(s) \left(e^{\frac{sx}{c}}+e^{-\frac{sx}{c}}\right) + \dfrac{c}{\beta} \dfrac{s\mathcal{L}\{f(t)\}}{s} e^{-\frac{sx}{c}}.$$

From the boundary condition $$f(t) =-\beta \frac{\partial u(0,t)}{\partial x}$$ for all $$t>0$$, and the initial condition $$u(x,0)=0$$ for all $$x>0$$, it follows that $$\lim_{t\to 0}f(t)=0$$. Then the initial value theorem implies $$\lim_{s\to\infty}s\mathcal{L}\{f(t)\}=0$$. This and the fact that $$x$$, $$s$$ and $$c$$ are positive implies that. $$\lim_{s\to\infty} \frac{c}{\beta} \frac{s\mathcal{L}\{f(t)\}}{s} e^{-\frac{sx}{c}}=0.$$

Edit: As was pointed out by The Pointer in the comments (pun not intended), it suffices to note that $$\lim_{s\to \infty} \mathcal{L}\{f(t)\}=0$$ (asymptotic property of Laplace transforms) and that the constants are positive.

Since $$\lim_{s\to\infty} \mathcal{L}\{u(x,t)\} = 0$$ (asymptotic property of Laplace transforms), also $$\lim_{s\to\infty} A(s)=0$$, and thus $$\lim_{s\to \infty} A(s) s e^{-\frac{sx}{c}}=0$$.

We conclude that $$\lim_{s\to \infty} s \mathcal{L}\{u(x,t)\} = \lim_{s\to\infty }s A(s) e^{\frac{sx}{c}}.$$

• Thanks for the answer. Shouldn't it be $$\dfrac{\partial}{\partial{x}} \mathcal{L} \{ u(x, t) \} = \dfrac{sA(s)}{c} e^{\frac{sx}{c}} - \left( \dfrac{c}{\beta} \dfrac{\mathcal{L}\{f(t)\}}{s} + A(s) \right) \dfrac{s}{c} e^{-\frac{sx}{c}} \Rightarrow \dfrac{\partial}{\partial{x}} \mathcal{L} \{ u(x, t) \} = \dfrac{sA(s)}{c} e^{\frac{sx}{c}} - \left( \dfrac{1}{\beta} \mathcal{L}\{f(t)\} + \dfrac{s}{c} A(s) \right) e^{-\frac{sx}{c}}$$ – The Pointer Sep 21 '18 at 3:06
• Multiplying both sides by $s$, we get $$s\mathcal{L} \{ u(x, t) \} = \dfrac{s^2}{c} A(s) \left( e^{\frac{sx}{c}} - e^{\frac{-sx}{c}} \right) - \dfrac{s}{\beta} \mathcal{L} \{ f(t) \} e^{-\frac{sx}{c}}$$ – The Pointer Sep 21 '18 at 3:34
• I think the expression for $s\mathcal{L}\{u(x,t)\}$ is correct. You forgot to take the factors $\frac{s}{c}$ in account when integrating with respect to $x$. (But please, do correct me if I am wrong, I didn't write down the calculations on paper, so a typo or missing constant quickly gets in.) – Ernie060 Sep 21 '18 at 7:48
• @The Pointer. Indeed, in hindsight I see that this argument is way too long. About your second remark. I forgot the $\lim$ in the right hand side. – Ernie060 Sep 29 '18 at 11:46
• Ahh, ok, now it makes sense to me. Thank you for the clarification! :) – The Pointer Sep 29 '18 at 11:49