# The Heat Equation

• April 18th 2010, 09:28 PM
Aryth
The Heat Equation
Find the Steady State Solution and state the Transient Problem.

$u_t = ku_{xx}$

$0 < x < L$

$t > 0$

Initial Condition: $u(x,0) = f(x)$

a) Boundary Conditions:
$u_x(0,t) = 0$

$u_x(L,t) = -1$

b) Boundary Conditions:
$-u_x(0,t) + \frac{1}{2}u(0,t) = 0$

$-u_x(L,t) - \frac{1}{2}u(L,t) = 1$

I've handled conditions of the first kind, u(x,t) = c. But I have not dealt with either of these before. Any help would be appreciated.
• April 18th 2010, 10:12 PM
AllanCuz
Quote:

Originally Posted by Aryth
Find the Steady State Solution and state the Transient Problem.

$u_t = ku_{xx}$

$0 < x < L$

$t > 0$

Initial Condition: $u(x,0) = f(x)$

a) Boundary Conditions:
$u_x(0,t) = 0$

$u_x(L,t) = -1$

b) Boundary Conditions:
$-u_x(0,t) + \frac{1}{2}u(0,t) = 0$

$-u_x(L,t) - \frac{1}{2}u(L,t) = 1$

I've handled conditions of the first kind, u(x,t) = c. But I have not dealt with either of these before. Any help would be appreciated.

For case 1, this changes our X(x) equation a little bit.

We regularly solve by seperating and finding lambda using

$X(x) = acos ( \sqrt { \lambda} x) + b sin ( \sqrt { \lambda} x)$

And we use the boundary conditions to compute. Now, we have a new set of boundary conditions we must use

$\frac{du}{dx} (0,t)= 0$ and $\frac{du}{dx} (L,t)=-1$

So,

$X = -a \lambda sin ( \sqrt { \lambda} x) + b \lambda cos( \sqrt { \lambda} x)$

$X(0) = 0 = b$

Noting that -a is a constant, we would normally remove this as it will be included in our summation at the end. But in this case it serves a crucial purpose.

$X(L)=-1 = -a \lambda sin ( \sqrt { \lambda} L)$

This is true when

$\sqrt { \lambda } L = \frac {3 \pi }{2}$

and

$-a \lambda = 1$

Of course then we have

$\lambda = \frac { 9 \pi^2 }{4L^2}$

We can then find what a makes this equation work. Of course we then continue from here, noting that in addition to the regular summation we must add

$C_0 \frac{1}{2}$ to satisfy the initial conditions.

I will explain how to do the second condition tomorrow when im not so tired. Also, I would like to add I've never delt with this boundary condition. I have delt with

$\frac{du}{dx} (0,t)= 0$ and $\frac{du}{dx} (L,t)=0$

Which makes it a lot easiar to understand what's going on. In this particular case I kept in the a instead of discarding it. I don't know if this is absolutely correct or not...Anyways, it's late bed is calling.

Cheers
• April 19th 2010, 07:36 AM
Aryth
Are you sure you're solving my problem??

I'm supposed to be finding the steady state solution... Which, as far as I've seen, does not have the same form as $X(x)$, given that when you want to solve by splitting u into products you get:

$u(x,t) = T(t)X(x)$

And when you want a Steady State and Transient solution you get:

$u(x,t) = w(x,t) + v(x)$

I'm wanting to solve for v(x), and then produce a new differential equation with w(x,t) as opposed to u(x,t).
• April 19th 2010, 08:03 AM
AllanCuz
Quote:

Originally Posted by Aryth
Are you sure you're solving my problem??

I'm supposed to be finding the steady state solution... Which, as far as I've seen, does not have the same form as $X(x)$, given that when you want to solve by splitting u into products you get:

$u(x,t) = T(t)X(x)$

And when you want a Steady State and Transient solution you get:

$u(x,t) = w(x,t) + v(x)$

I'm wanting to solve for v(x), and then produce a new differential equation with w(x,t) as opposed to u(x,t).

ummm...

$u(x,t) = w(x,t) + v(x)$

Is for when we have initial temperature at both ends. This is where we find a linear function such that we can model the heat equation in a known form. From your initial post I thought we were solving,

$\frac{dy}{dt} = k \frac{d^2y}{dx^2}$

$\frac{dy}{dx} (0,t)= 0$

$\frac{dy}{dx} (L,t)=-1$

$
u(x,0) = f(x)
$

This is not the same as your latest post. What are we actually solving here, I am a little unclear?
• April 19th 2010, 09:25 AM
AllanCuz
Perhaps this is a new way of asking for a rod with constant temperature at the ends...I'll go through a general example, see if it is applicable here.

$\frac{dy}{dt} = k \frac{d^2 y}{dx^2}$

$u(0,t)=T1$ and $u(L,t)=T2$

$u(x,0)=f(x)$

Go through the normal seperation process

Now, let $u(x,t)=Q(x,t) + P(x)$

Getting this into the form of the heat equation we get

$\frac{dQ}{dt} = k \frac{d^2 Q}{dx^2} + p$

This is the regular heat equation if

$p` = 0$

Therefore,

$P(x) = mx+b$

Now fit this with the boundary conditions. So we have..

$P(0)=T1=b$

Also

$P(L)=T2=mL + T1$

$m = \frac {T2-T1}{L}$

$P(x) = \frac {T2-T1}{L} x + T1$

Now, let us note that

$u(x,0)=f(x)=Q(x,0) + P(x)$

This is true when

$Q(x,0)=f(x) - P(x)$

Therefore, our new heat equation becomes

$\frac{dQ}{dt} = k \frac{d^2 Q}{dx^2}$

$Q(0,t)=Q(L,t)=0$

$Q(x,0)=f(x) - (\frac {T2-T1}{L} x + T1)$

I'm not sure if this will help...but again, i'm confused at your original question. Perhaps it's a form I have never seen either but from what you described it doesn't seem to model what I have above (which is when we use what you suggested)
• April 19th 2010, 10:28 AM
Aryth
I've already solved a problem like that...

I'm stuck just like you are.... I don't need to solve the equation with a product solution and I don't need the eigenvalues of either the transient solution or the original solution.

I simply need a steady state equation and a statement of the new transient problem.

Since I've never dealt with this before, it could very well be impossible without initial conditions on u itself instead of all initial conditions relying on derivatives...

But I do not know.
• April 19th 2010, 01:42 PM
AllanCuz
Quote:

Originally Posted by Aryth
I've already solved a problem like that...

I'm stuck just like you are.... I don't need to solve the equation with a product solution and I don't need the eigenvalues of either the transient solution or the original solution.

I simply need a steady state equation and a statement of the new transient problem.

Since I've never dealt with this before, it could very well be impossible without initial conditions on u itself instead of all initial conditions relying on derivatives...

But I do not know.

Yeah, this isn't a standard condition. So....yeah, i'm fresh outta ideas. I'm not sure what else you can do here...