Find an equilibrium solution of heat equation Please help

[yaya10]

Hello

I need help with this heat equation

I need to find steady state solution please help me

I ve tried but I could not get it.

Homework Statement

http://www.alm5zn.com/upfiles/m0w00932.jpg

Homework Equations

The Attempt at a Solution

I ve got C= C^x/2 +ax+b

but i could not apply the boundry conditions?

any help would be apprecited

Thank you

 

[yaya10]

this is my working so far


Uploaded with ImageShack.us


the link

http://img52.imageshack.us/i/28244265.gif/

 

[LCKurtz]

Your equations are simple enough that you could just post them here instead of making everyone open a gif on imageshack.

Anyway, there are two obvious things wrong with what you did:

1. What happened to the D and k that should be in your steady state equation?
2. C is the independent variable and you are treating it like it is a constant. It depends on x even in the steady state so its antiderivative isn't Cx.

So correct your steady-state equation and solve it as a second order constant coefficient equation.

 

[HallsofIvy]

A "steady state" solution is one that does NOT depend on t. That means that the partial derivative with respect to t is 0 and the partial derivative with respect to x can be treated like an ordinary derivative:
d\frac{d^2C}{dx^2}- kC= 0
That's a "linear second order differential equation with constant coefficients", probably the simplest kind of differential equation. Do you know how to solve such equations? What is the "characteristic equation"?

 

[yaya10]

hello LCKurtz,

sorry about the image but I thould it would be much describable this way.

thank for your help

=-

HallsofIvy


thanks for being here

I ve got this

first I divide them by D to get K/C then I called them a=K/c

characteristic equation:

(m^2)-a=0

m^2=a---->>> m=-sqrt(a), sqrt(a)

--->C(x)= c1*exp(-sqrt(a)), and c2*exp(sqrt(a))

=-=-=-

(m^2)-a=0, m^2=a \mapsto m=-\sqrt{a},\sqrt{a} --->C(x)= c1*exp(\sqrt{a})+ c2*exp(-\sqrt{a}).

 

[olivermsun]

(m^2)-a=0, m^2=a \mapsto m=-\sqrt{a},\sqrt{a} --->C(x)= c1*exp(\sqrt{a})+ c2*exp(-\sqrt{a}).

Your characteristic equation comes from assuming solutions of form exp(mx), so your equilibrium solution needs to look like:

C(x)= c_1\exp(\sqrt{a}x)+ c_2\exp(-\sqrt{a}x).

But now consider the boundary conditions (which really dominate the equilibrium solution for this problem anyway).

 

[yaya10]

thanx for your help

Now I substitue the boundry conditions I got

C(infinty,t)= c1exp(infinity)+c2 exp(infinty)=0

-->> c1+c2=0-->>>>> c1=-c2

is that true ?!

I am sorry for bothering you guys

i need to learn

thanks again

 

[LCKurtz]

thanx for your help

Now I substitue the boundry conditions I got

C(infinty,t)= c1exp(infinity)+c2 exp(infinty)=0

That isn't how you write limits. Think about this. If you have a function of the form

f(x) = Ce^kx+De^-kx what happens to each term as x → ∞? That should determine one of the constants. And you also have x → -∞ to think about.

 

[yaya10]

the lim when x → ∞ = ∞

so from the equation you gave

D wil be zero and C is infinity so we have got D=0

and then

when → - ∞ = 0

we have got

C will approach zero and D is infinity.

So C=D+0

I hope this is correct.

Limits are hard for me so I reall hope I have got it right.


Thank you

 

[LCKurtz]

That isn't how you write limits. Think about this. If you have a function of the form

f(x) = Ce^kx+De^-kx what happens to each term as x → ∞? That should determine one of the constants. And you also have x → -∞ to think about.

the lim when x → ∞ = ∞

so from the equation you gave

D wil be zero and C is infinity so we have got D=0

and then

when → - ∞ = 0

we have got

C will approach zero and D is infinity.

So C=D+0

I hope this is correct.

Limits are hard for me so I reall hope I have got it right.


Thank you

The point is that you are given in your problem that as x → ±∞ the limit is zero. So you should be able to argue that the only way that could happen is if both C and D are zero.