Partial Diff. Egn.

I have the following question:

Consider

du/dt = D * d^2(u)/dx^2 + sin ((pi*x) / L)
{where my derivatives are partial derivatives}

where 0<=x<=L and D>0 is a constant.

Initial conditions given: u(0,t) = 0, u(L,t) = 1, u(x,0) = x/L

You may assume that the solution can be written in the form
u(x,t) = z(x) + v(x,t)

a) Show that the steady state part z(x) satisfies the ODE
d^2(z)/dx^2 = -1/D * sin(pi*x/L) {These are ordinary derivatives}
Also obtain the boundary conditions for z(x) and solve for z(x)

b) Show that v(x,t) satisfies the diffusion equation

dv/dt = D * d^2(v) / dx^2 {These are partial derivatives}

Find the boundary and initial conditions for v(x,t)

My workings:

For part a)

To do this I substitute u(x,t) = z(x) + v(x,t) in the original PDE and I obtain dv/dt = D * d^2(v) / dx^2 {partial derivative} + d^2(z) / dx^2 {ordinary derivative} + sin( Pi*x/L)

Then since lim v(x,t) as t -> inf = 0 we say that d^2(v) / dx^2 and dv / dt {these are partial derivatives} are equal to zero as t-> inf so the steady part satisfies the ODE d^2(z)/dx^2 = -1/D * sin (Pi*x/L)

Then by integration twice I obtain that z(x) = L^2/D*Pi^2 * sin(Pi*x/L) + c1*x + c2

Then I don't know how do I find boundary conditions for z(x) so I can deduce what c1 and what c2 is!

For part b) I know that if I substitute z(x) = L^2/D*Pi^2 * sin(Pi*x/L) + c1*x + c2 would give me that v(x,t) satisfies the diffusion equation

dv/dt = D * d^2(v) / dx^2 {These are partial derivatives}

So basically my questions are:

1) How do I obtain the initial conditions for z(x)? (so I can deduce constants c1 and c2)

2) How do I obtain the initial conditions for v(x,t)?

Basically, the question, is an one - dimensional diffusion problem with a source term, I would appreciate any help!

Thank you very much, I will be glad of any help! :smile:

Reply 1

ghostwalker

Original post by Darkprince

1) How do I obtain the initial conditions for z(x)? (so I can deduce constants c1 and c2)

Not commenting on the working, and in lieu of anyone else replying:

For 1) when x=0, and when x=L, consider

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

This equals

\displaystyle \lim_{t \to \infty} \left(z(x)+v(x,t)\right)

That will give you a couple of conditions.

PS: Your working would be considerably more readable in LaTex.

(edited 12 years ago)

Reply 2

Darkprince

You just gave me a (your) brilliant idea, thank you!

Since we know that u(0,t) = 0 and
u(L,t) = 1

Also u(x,t) = z(x) + v(x,t), but v(x,t) -> o as t -> infinity (the exercise told us we are able to assume that), so we end up with z(L) = 1 and z(0)=0

so I find c2 = 0 and c1 = 1/L

Is this correct?

Am I right?

Reply 3

ghostwalker

Original post by Darkprince

I've never studied partial derivatives, and differential equations are not my favrourite topic, hence I wasn't wishing to comment on your original working.

I can't say whether they are correct or incorrect with regard to the whole question , but they are correct for the function z(x) that you're using.

(edited 12 years ago)

Reply 4

Darkprince

It guides me where I want, the only thing I only need to know know is the initial conditions for v(x,t) and then I am done :smile:

Thanks though!

Reply 5

ghostwalker

Original post by Darkprince

It guides me where I want, the only thing I only need to know know is the initial conditions for v(x,t) and then I am done :smile:

Thanks though!

Since you have the initial conditions for u(x,t), and z(x), then the initial conditions for v(x,t) follow by using u(x,t) = z(x) + v(x,t) and just substituting in - I think.

(edited 12 years ago)

Reply 6

Darkprince

Yes, I tried something similar. I am not too sure.

So I have u(0,t)=0, u(L,t)=1 and u(x,0) = x/L

Also z(L) = 1 and z(0)=0

and u(x,t) = z(x) + v(x,t)

How you would approach it?

Reply 7

ghostwalker

Original post by Darkprince

Yes, I tried something similar. I am not too sure.

So I have u(0,t)=0, u(L,t)=1 and u(x,0) = x/L

Also z(L) = 1 and z(0)=0

and u(x,t) = z(x) + v(x,t)

How you would approach it?

Straight forward substitution:

Consider the points (0,t), sub in, and what do you get?

Similarly (L,t).

Both give simple values.

(x,0), will be the trickier one, if you need it (bare in mind my previous disclaimers!).