PDE Math Homework Help: Solving BVPs for Periodic Functions

[r.a.c.]

Homework Statement

We are given f \epsilon C(T) [set of continuous and 2pi periodic functions] and PS(T) [set of piecewise smooth and 2pi periodic functions]

SOlve the BVP

u_t(x,t) = u_xx(x,t) ; (x,t) belongs to R x (0,inf)
u(x,0) = f(x) ; x belongs to R


Find a solution of this PDE

Homework Equations

The only relevant equation I can think of is a Fourier Inversion which states that if is continuous and piecewise smooth then
f(x) = \sum f^h(n) e^i^n^x ; where f^h = 1/2\pi \int f(x)e^-^i^n^x dx

The Attempt at a Solution

I have tried solving the first equation ODE till I get by separation of varaibles
S''(x) - AS(x) = 0 and T'(t) - AT(t) = 0

A is real. Three cases: A>0 in which case the solution is S(x) = C(sin(Lx)) + B(cos(Lx))
where C,B are constants and L = (-L)^(1/2)
Then I tried using power series expansion of sin and cos to be able to relate it to the Fourier series of f (seeing as f(x) = S(x)) but to only get stuck.

When A = 0 S(x) = Cx + B
Then I can do this by finding the Fourier coefficients(f^h) and so on

When A<0 things become complicated because we get and exponential.

Anyways, first off, how do we know which A to use!? Then I can try to reach some sort of conclusion.

 

[kfdleb]

first, the best thing is to solve it using separation of variable i.e. as an ODE as u did

u get X''(x)-CX(x)=0 and T'(t)-CT=0

i think ur missing something in the given such as u(0,t)=u(2\pi,t)
if so the solution would be X''(x)+L^2X(x)=0
X(x)=Acos(Lx)+Bsin(Lx)
with L=-n^2*(pi)^2/(2pi)^2

the solution would be the sum of X_n(x)*T_n(t)

 

[r.a.c.]

No I'm not missing anything in the solution. That's the problem. If the constraints on the extremities of x were there it would be a piece of cake. And if we had them then there would be no need for f to be belong to C or PS. Even in the question it tells us to solve the PDE using Fourier series.

 

[r.a.c.]

Thanks but I found the answer to this. I think this is done.