# Thread: PDE by Laplace

1. Thanks a lot Danny. You and Ackbeet are always helping me undertand this vast subject of DE I am just scratching the surface of!

Is it the initial value theorom about the initial condition of $u(x,0) = 0$ ?

This to me would suggest, from having a read through some control engineering notes (that I have dusted down from my attic from many years ago!) that the behaviour of the system as the complex laplace variable, $p$ in our case, hits the limit of infinity the system output will tend to zero- this interprets as high frequency response in control systems.... is this what you were thinking ackbeet?

2. I have been working through the solution, and as suggested by Ackbeet at the beginning, the PDE drops to a ODE in the complex $p$ plane.... and with the initial condition... the general form drops to:

$v(x,p) = Ae^{-\sqrt p x}$ where 'A' is the arbitary constant.

Following from the discussion of the limit theorems, would it be possible to say that the value of:

$u(0,t)$ can be found from using the limit of $\lim_{p \to \infty} v(x,p)$ ? from our expression, will this leave $v = 0$... will this imply 0 in the time domain (u) then?

3. Whoops. I've been invoking the wrong theorem. You're going to want the initial value theorem. In our case, that theorem states that

$\displaystyle{\lim_{t\to 0^{+}}u(x,t)=\lim_{p\to\infty}p\,v(x,p)=0.}$

This is assuming continuity of the solution near the origin, since you're not technically given the limit but the function value at the origin.

The coefficients depend on $p$. Thus, while you've eliminated the correct term, you haven't written the remaining term down correctly. You should have, after using the condition Danny mentioned in Post # 14, that

$v(x,p)=B(p)\,e^{-\sqrt{p}\,x}.$

Now, plug this expression into the limit I've just given you. What do you get?

4. I have just seen something which makes me laugh. You can prove your elusive second result by simply looking at this equation:

$\displaystyle{\frac {\partial u}{\partial x} (0,t) = -1}$ in the right way.

5. Sorry for the late reply Ackbeet... had a few days off!

I can see the general expression, and in the limit of 'p' to infinity (as t =0+), does the exponential not drop to zero, leaving $v(x,t) = 0$?

6. Uh, which part of the problem does this refer to?

7. should be clearer!

$\displaystyle{\lim_{t\to 0^{+}}u(x,t)=\lim_{p\to\infty}p\,v(x,p)=0.}$

8. Right. That's one of the limit theorems for Laplace Transforms. Let me go back to the original problem: what parts have you solved, and what parts haven't you solved?

9. I think I am actually ok with all the parts now thanks Ackbeet... the help you and Danny showed kicked me in the right direction! thanks again in any case.

10. Oh, good. You're very welcome. Have a good one!

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