Heat Equation With Seperable Variables

[Rweir]

Homework Statement

du/dt = d²u/dx² + u
Bc: u'(0) = u'(1) = 0
Ic: u(x,0) = 1

Homework Equations

Using sturm-liouville to solve for eigenvalues.

The Attempt at a Solution

After first separating variables in the equation
we get G^'/G - 1 = F^'' = λ
after using Sturm-Liouville we find that
F(x) = Acos(n*Pi*x)
G(t) = Ae^(-n2pi2-1)t

So after multiplying them together and then taking the initial condition of u(x,0) = 1
we get A*cos(n*pi*x) = 1 and thus the problem arises after using Fourier expansion we get A = 0 which makes everything 0. Any suggestions as to why it is coming out like this?

 

[lanedance]

first the u makes it a little different, its not your standard heat conduction equation, so you have
u_t = u_xx + u

so i get to
X'' + (λ-1)X = 0
T' + λT = 0

so you won't be able to satisfy the BCs for (λ-1)<0

now how about when λ = 1?

 

[lanedance]

have a think about the normal heat equation
u_t = u_xx
the rate of change with time of temp is proportional to the spatial curvature of the temp, ie everything gets smoothed out...

u_t = u_xx + u
now the rate of change also has a component proportional to temp as well...

and note the IC is initially uniform everywhere

now think about your BCs
u'(0) = u'(1) = 0
which in the normal heat equation are equivalent to insulated ends

 

[lanedance]

but also stepping back a bit, as the basis functions are orthogonal, shouldn't the coefficient in the sum be proportional to
a_n = \int dx.cos(n \pi x ) .u(x,0) = \int dx.cos(n \pi x ) .u(x,0)

which is different from every basis function having to satisfy the IC, only their sum has to