PDE by Laplace

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August 28th 2010, 07:08 AM
padawan

PDE by Laplace

Hi all,

I have been looking at the following problem and have hit a wall.

Q:

$\frac {\partial^2u} {\partial x^2}= \frac {\partial u} {\partial t}$ where $u(x,0) = 0$ , $\frac {\partial u}{\partial x} (0,t) = -1$ , $u(x,t) -> 0$ as $x -> \infty$

let $v(x,p) = \int_{0}^{\infty} e^{-pt}u(x,t)dt$

Prove $\frac {\partial^2 v}{\partial x^2} = pv$ and $\frac {\partial v}{\partial x} (0,p)= - \frac {1}{p}$ and $v(x,p) -> 0$ as $x -> \infty$

I can find the first proof, but am having problems figuring out the second part

$\frac {\partial v}{\partial x} (0,p)= - \frac {1}{p}$

.... I am not sure where to start- any one have any ideas on the method for this one?

Thanks for reading!

(original post subsequently corrected)
August 28th 2010, 09:38 AM
Ackbeet

Right. Showing that $v_{xx}=pv$ is a matter of plugging things in and doing the usual integration by parts for the Laplace Transform.

I'm very puzzled by the second part. I don't buy it. Are you sure you've typed it up correctly? If you have the second-order ODE

$\displaystyle{\frac{d^{2}v}{dx^{2}}=pv,}$ then the solutions are either exponentials or hyperbolic trig functions (they're equivalent).

For simplicity, I'll write

$v(x,p)=Ae^{\sqrt{p}\,x}+Be^{-\sqrt{p}\,x}.$

Taking the derivative of this expression with respect to $x$ definitely does not give you $-1/p.$

This is a very interesting problem, and very similar to the diffusion problem I did with my father for my senior seminar project at Grove City College. The fun part is doing the inverse Laplace Transform! You have to go to the line integral definition and use residue calculus: you get an infinite series solution.
August 28th 2010, 09:54 AM
padawan

apoligies... missed out a input:

$\frac{\partial v}{\partial x} (0,p) = - \frac{1}{p}$

I'll change original too...........
sorry ackbeet
August 28th 2010, 10:13 AM
Ackbeet

Danny reminded me in a PM that the coefficients are dependent on p. So I should write

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

Now you can get two limit conditions. One is straight from the original function, and the other is from the Final Value Theorem. What are those conditions?
August 28th 2010, 10:31 AM
padawan

these ones:

$u(x,0) = 0$ , $\frac {\partial u}{\partial x} (0,t) = -1$ , $u(x,t) -> 0$ as $x -> \infty$

or are you talking about other ones?
August 28th 2010, 10:54 AM
Ackbeet

Ah, but unless you have access to the original function (the solution to the pde!), I recommend you work in the Laplace domain. That is, try to find conditions on v. One of your original conditions will translate over directly to the transformed function, the other will be through the Final Value Theorem to which I linked.
August 28th 2010, 12:39 PM
padawan

Thanks Akbeet.

$\frac{\partial v}{\partial x}$ can't be $pv - v_0$ can it? surely that is $\frac{\partial v}{\partial t}$

I have been trying to find the Laplace of $\frac{\partial u}{\partial x}$ but can't seem to break down the integration:

$L(\frac{\partial v}{\partial x}) = \int_{0}^{\infty} e^{-pt}\frac{\partial v}{\partial x} dt$ integration by parts:

$u= e^{-pt}; \frac{\partial u}{\partial t} = -pe^{-pt}$

$v = \frac{\partial v}{\partial x}; v= \int \frac{\partial u}{\partial x} dt$

I can't see a way of integrating this part.....
August 28th 2010, 12:43 PM
Ackbeet

It's so difficult it's easy:

$L\{u_{x}(x,t)\}=v_{x}(x,p).$

That is, you can take the x-derivative outside the Laplace integral. Make sense?
August 28th 2010, 12:57 PM
padawan

I am not sure I undertand that:

$L(\frac{\partial v}{\partial x}) = \int_{0}^{\infty} e^{-pt}\frac{\partial v}{\partial x} dt$

$\frac{\partial v}{\partial x}$ can be a function of $t$ , so will it be possible to take this outside?

Given what you say, I guess you can, but I can't see how.
August 28th 2010, 01:01 PM
Ackbeet

I think you're confusing where $u$ and $v$ appear. Try this:

$\displaystyle{\int_{0}^{\infty}e^{-pt}u_{x}(x,t)\,dt=<br /> \int_{0}^{\infty}e^{-pt}\frac{\partial}{\partial x}\,u(x,t)\,dt=<br /> \frac{\partial}{\partial x}\int_{0}^{\infty}e^{-pt}u(x,t)\,dt=<br /> \frac{\partial}{\partial x}v(x,p)=<br /> v_{x}(x,p).}$

Therefore, $L\{u_{x}(x,t)\}=v_{x}(x,p).$
August 28th 2010, 01:35 PM
padawan

I have been trying to break this down with integration by parts, so guess I am getting the functions I am operating on mixed up with the intermediate u's and v's!

Thanks for spelling that out explicitly- think I need a coffee break and I'll come back with a fresh piece of paper and new, different variable letters.
August 28th 2010, 03:09 PM
padawan

ok... I see that now. Take the transform of -1 to equate that into the laplace domain given the proof you highlighted... perfect!

I'll take a crack at the last proof now using the final value theorem you linked.
August 28th 2010, 03:41 PM
padawan

I have had a read through the final value theorem, and am not sure its application here; that is..

$lim_{x\rightarrow \infty}f(\infty,t) = \lim_{p \rightarrow 0} v(x,p)$ ... in the case here, how would we assess without knowing the function $u(x,t)$

We have been given the condition that as $lim_{x\rightarrow \infty}$ is independant of 't'.... I am not sure how this tallies with $lim_{p\rightarrow 0} v(x,p)$ as suggested by the final value theorem.
August 28th 2010, 04:17 PM
Danny

Since

$\displaystyle \lim_{x \to \infty} u(x,t) \to 0$ then

$\displaystyle \lim_{x \to \infty} v(x,p) = \lim_{x \to \infty} \int_0^{\infty} e^{-pt} u(x,t)dt = \int_0^{\infty} e^{-pt} \lim_{x \to \infty} u(x,t) dt \to 0.$

Thus,

$\displaystyle \lim_{x \to \infty} v(x,p) \to 0$ .

This will take of one of your exponentials.
August 28th 2010, 04:21 PM
Ackbeet

You're also told in the original conditions that

$\displaystyle{u(x,0)=0.}$

What, then, does the Final Value Theorem tell you?

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