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

I'm confused about this last part:

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.

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

I hope this answer is helpful to you.

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  • $\begingroup$ Thanks for the answer. Shouldn't it be $$\dfrac{\partial}{\partial{x}} \mathcal{L} \{ u(x, t) \} = \dfrac{sA(s)}{c} e^{\frac{sx}{c}} - \left( \dfrac{c}{\beta} \dfrac{\mathcal{L}\{f(t)\}}{s} + A(s) \right) \dfrac{s}{c} e^{-\frac{sx}{c}} \Rightarrow \dfrac{\partial}{\partial{x}} \mathcal{L} \{ u(x, t) \} = \dfrac{sA(s)}{c} e^{\frac{sx}{c}} - \left( \dfrac{1}{\beta} \mathcal{L}\{f(t)\} + \dfrac{s}{c} A(s) \right) e^{-\frac{sx}{c}}$$ $\endgroup$ – The Pointer Sep 21 '18 at 3:06
  • $\begingroup$ Multiplying both sides by $s$, we get $$s\mathcal{L} \{ u(x, t) \} = \dfrac{s^2}{c} A(s) \left( e^{\frac{sx}{c}} - e^{\frac{-sx}{c}} \right) - \dfrac{s}{\beta} \mathcal{L} \{ f(t) \} e^{-\frac{sx}{c}}$$ $\endgroup$ – The Pointer Sep 21 '18 at 3:34
  • $\begingroup$ 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.) $\endgroup$ – Ernie060 Sep 21 '18 at 7:48
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    $\begingroup$ @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. $\endgroup$ – Ernie060 Sep 29 '18 at 11:46
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    $\begingroup$ Ahh, ok, now it makes sense to me. Thank you for the clarification! :) $\endgroup$ – The Pointer Sep 29 '18 at 11:49

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