# Math Help - Modelling with PDEs - Boundary Condition

1. ## Modelling with PDEs - Boundary Condition

I am working on this question for my Partial Differential Equation seminar and would like some help in understanding how to create boundary conditions.

An endothermic chemical reaction takes place within the region x [0,L], consuming
heat energy at a rate R = x(L − x). The ends of the region are perfectly insulated
and the region is initially heated to A degrees.
(a) Write down a partial differential equation, together with boundary and initial
conditions governing the temperature u(x, t) in the region.
[Hints: The rate of heat leaving the region at x = 0,L is assumed to be propor-
tional to du/dx (partial, curly d) . Your equation should contain an inhomogeneity corresponding to the heat sink, R, which is independent of u.]

So I worked out that the initial condition will be u(x,0)=A and from there on I am a little stuck. I believe that because of the hint that the boundary conditions will be Neumann conditions (ie u'(0,t)=? and u'(L,t)=? ) but I have looked at all kinds of examples and cannot work out how to find what they are equal to. I think either they would equal 0 but my instinct tells me they should maybe be a function of t since surely the temperature would depend on what time it is?

For the equation i have u_t = Du_xx -x(L-x) but i am not sure whether i need the D at the front or not.

Any help would be greatly appreciated, i would really like to understand what I am doing

Thanks

Do you understand the u' makes no sense here? You are not thinking in terms of more than one variable. u is a function of two variables and you must state whether you are differentiating with respect to x or t. Saying that the ends are insulated means there is no heat flow over them. "Heat flow", in one direction or another, is the change of temperature with respect to x. Your condition should be $u_x= 0$ at the ends of the interval.

For the equation i have u_t = Du_xx -x(L-x) but i am not sure whether i need the D at the front or not.
What does "D" represent? I suspect it is the "derivative operator" but, again, that makes no sense in a problem with two variables. You already have the derivatives with respect to x and t.

Perfectly insulated means

$u_x(0,t) = 0, u_x(L,t) = 0.$

oh woops little notation error, i knew it was wrt x, thanks for reminding me. The D is meant to be the proportionality constant, often set to one to give the heat equation in 1D only in this situation there is also a sink and i can't tell from the question whether the D is meant to be one or whether i should just leave it general as D. Thanks for your help.

so u_x(0,t)=0 and u_x(L,t)=0 because the ends of the region are insulated?

the thing i can't get my head round is that if the boundary conditions describe the temperature at any time at x=0,L why is there no t in the conditions, because surely the temperature would vary depending on what t is? or is this why we talk about u_x and not u. sorry if i'm talking nonsense i'm just struggling to get my head round some of these bits.

We usually set $D = 1$ because we can scale t. BTW - The solution will depend on t! That's what you are required to find - the solution of the PDE!

so now i proceed with separation of variables but because of the source term this can't be done directly right? i do the u(x,t)=v(x)+w(x,t) ? or is there a way of doing it with normal SOV?

Without the source term, a separation of variables would lead to

$u = \sum_{n=0}^\infty a_n e^{-k^2t} \cos \frac{ n \pi}{L} x$ where $k = \frac{D n \pi}{L}$.

So, with the source term we try a solution of the form

$u = \sum_{n=0}^\infty T_n(t) \cos \frac{n \pi}{L} x$.

If we expand the source term itself, i.e.

$Q = \sum_{n=0}^\infty q_n \cos \frac{n \pi}{L} x$

for appropriate $q_n$, then substitute and compare like terms of $\cos \frac{n \pi}{L} x$. This gives us a series of ODEs for $T_n$.

ok i feel totally lost now
Using u=T(t)X(x) on the equation without the source term i get X(x)=cos(npix/L) but i really have no idea what i should do after this. I can get to u=(cos(npix/L))e^((-n^2pi^2/L^2)*t) but i've no idea where to go from there. how do i involve my source term again as well as use my initial condition?

ok i actually think i may be onto something now.
i put u(x,t)=sum(1,inf) of u_n(t).cos(n*pi*x/L)
then subbed this into my original equation, then will take fourier series of source and then compare like cos terms
have i got it now? sorry its taken so long if i have!

right so this is what i have to show:

By seeking separable solutions, u(x, t) = T(t)X(x), show that the temperature
is given by
u(x, t) = u0(t)/2+sum(1,inf) u_n(t)cos((n*pi*x)/L) (1)

so far i have got to:
Using u=T(t)X(x) on the equation without the source term i get X(x)=cos(npix/L) but i don't know how to show it goes to (1). Is it to do with the linearity of solutions? andi can see it looks like a fourier series and cos is an even function which would explain u_0 being there and why there is a cos term but i feel that this isnt really relevent to what i'm trying to show.

Any Ideas?