Question

Find the separable solutions to the differential equation (the heat equation)

\[u_t=u_{xx}\]

Given conditions:
\[u_x(0,t)=0\]\[u_x(\pi,t)=0\]

My professor didn't finish any example during class...

Answer 1

how about initial condition?

Answer 2

here is a complete solution of heat equation with L instead pi and with different boundary conditions
i hope this may help u

Answer 3

I'll read through this, thats all the info we're given

Answer 4

Looks like parametric questions with t representing time. What do you mean by heat equation? I've quite familiar with heat ;-)

Definition: Heat is the flow of thermal kinetic energy (temperature) from an region of high concentration to lower concentration as per natural entropy flow. The 3 modes of heat transfer are conduction, convection (which is really just advection plus conduction trough the no-slip zone), and radiation (transfer by electromagnetic particles)

I think the document @mukushla gave you is probably your best bet.

Answer 5

through* the no-slip zone

Answer 6

Agent - It's a partial differential equation called the heat equation

Answer 7

Context?

FYI:
heat \(\neq\) temperature In contrast to what most people say when they talk.

Answer 8

I'm taking that this is from Differential Equation mathematics, a subject I've heard of but alas am not familiar with myself. My experience is from the experimental and application sides of heat. I guess what I was asking was if this "initial value problem" is in anyway related to a mode of heat transfer. I'll see if I can find some people who might be able to help with the mathematics explanation. :-)

Answer 9

"The heat equation is an important partial differential equation which describes the distribution of heat (or variation in temperature) in a given region over time."
- http://en.wikipedia.org/wiki/Heat_equation

I can separate it to:

\[\lambda = 0, X(x)=Ax+b\]
\[\lambda <0, X(x)=Acosh(\sqrt{\mu}x)+Bsinh(\sqrt{\mu}x)\]
\[\lambda >0,X(x)=Acos(\sqrt{\lambda}x)+Bsin(\sqrt{\lambda}x)\]

Note sure exactly how to proceed, i'll read through the pdf.

Answer 10

It's from Differential equations 2 class (partial differential equations, unlike diffeq 1 - ordinary differential equations.

Answer 11

partial differential equation

Answer 12

I'd have to review a lot of stuff to give you a full answer, so I'm just going to refer you to something similar (but perhaps a bit simpler) that what
@mukushla gave you

http://tutorial.math.lamar.edu/Classes/DE/SolvingHeatEquation.aspx

Answer 13

again it has L instead of \(\pi\) but that shouldn't be a big problem to fix...

Answer 14

Cool, i'll take a look through it

Answer 15

Looks like i'm going to have to figure out eigenfunctions/values for systems of ode's, we skipped that chapter in ODE class...

Answer 16

bummer... well all the info on how to do that is in those notes, but learning all that would probably take at least a day or two...

Answer 17

In the case of
\[\lambda =0\]\[X(x)=Ax+B\]\[X'(x)=A\]\[X'(0)=0=A \rightarrow A=0\]
Thus,\[X(x)=B\]is a non-trivial solution


In case of
\[\lambda <0\]\[Let -\lambda = \mu\]\[X(x)=Acosh(\sqrt{\mu}x)+Bsinh(\sqrt{\mu}x)\]\[X'(x)=A \sqrt{\mu}\sinh(\sqrt{\mu}x)+B \sqrt{\mu}\cosh({\sqrt{\mu}x)}\]\[X'(0)=0=B \sqrt{\mu} \rightarrow B=0\]B must equal zero since lambda must be positive.
\[X'(\pi)=0=A \sqrt{\mu}\sinh(\sqrt{\mu}\pi)\rightarrow A=0\]A must equal zero since mu <0 and pi is positive.
No non-trivial solutions for lambda <0


In case of
\[\lambda >0\]\[X(x)=Acos(\sqrt{\lambda}x)+Bsin(\sqrt{\lambda}x)\]\[X'(x)=-A \sqrt{\lambda}\sin(\sqrt{\lambda}x)+B \sqrt{\lambda}\cos(\sqrt{\lambda}x)\]\[X'(0)=0=B \sqrt{\lambda}\rightarrow B=0\]\[X'(\pi)=0=-A \sqrt{\lambda}\sin(\sqrt{\lambda}\pi)\]Here we have non trivial solutions when sqrt(lambda) is an integer, \[n=\sqrt{\lambda}\]Thus,
\[X(x)=Acos(n \pi)\]Is a non-trivial solution when n=1,2,3,4...

Continued on next post!

Answer 18

Missed a posst, this one goes before the last one...

From the last post we got the solutions for X(x)
\[X_0(x)=B\]\[X_n(x)=Acos(nx)\]n = 1,2,3,4....

Now we need to find the corresponding solutions for T(t)
\[T'(t)+\lambda T(t)=0\]\[T(t)=Ce^{-\lambda t}\]

The corresponding solutions are:
\[T_0(t)=C\]\[T_n(t)=Ce^{-n^2t}\]Note:\[\sqrt{\lambda}=n \rightarrow \lambda =n^2\]

Since we assumed a solution of the form\[u(x,t)=X(x)T(t)\]
\[u_0(x,t)=BC=E\]\[u_n(x,t)=De^{-n^2t}\cos(nx)\]

These are the separable solutions, we must take into consideration the solutions that are a linear combination of these, next post!

Answer 19

it deleted one of my posts...

We were also given the initial condition:\[u(x,0)=f(x)\]

We will put the solutions of the last post together in this way:
\[u(x,t)=E+\sum_{n=1}^{\infty }Dcos(nx)\]
I'll rewrite this in a convenient way that resembles a fourier series:
\[u(x,t)=\frac{a_0}{2}+\sum_{n=1}^{\infty }a_n cos(nx)\]
So we can now find a_n
\[f(x)=u(x,0)=\frac{a_0}{2}+\sum_{n=1}^{\infty}a_ncos(nx)\]
\[f(x)\cos(kx)=\frac{a_0}{2}\cos(kx)+\sum_{n=1}^{\infty}a_ncos(nx)\cos(kx)\]
\[\int\limits_{0}^{\pi} f(x)\cos(kx)dx=\int\limits_{0}^{\pi} \frac{a_0}{2}\cos(kx)dx+\int\limits_{0}^{\pi} a_ncos(nx)\cos(kx)dx\]
\[\int\limits_{0}^{\pi}f(x)\cos(nx)dx=a_n \int\limits_{0}^{\pi}\cos^2(nx)dx\]
\[\int\limits_{0}^{\pi}f(x)\cos(nx)dx=\frac{a_n}{2}\int\limits_{0}^{\pi}[1+\cos(2nx)]dx\]
\[\int\limits_{0}^{\pi}f(x)\cos(nx)dx=\frac{a_n \pi}{2}\]
\[\frac{2}{\pi}\int\limits_{0}^{\pi}f(x)\cos(nx)dx=a_n\]

Finally:
\[u(x,t)=\frac{a_0}{2}+\sum_{n=1}^{\infty}\cos(nx)\int\limits_{0}^{\pi}\frac{2}{\pi}f(x)\cos(nx)dx\]

Nasssty....

Answer 20

thats the right order...