MHB How Does Separation of Variables Solve the Heat Equation in a Metal Rod?

[jmorgan]

The temperature distribution u(x, t), at time t > 0, along a homogeneous
metal rod can be obtained by solving the 1d heat equation;

ut = kuxx (1)

where k = 2 is a constant. The length of the rod is 1m and the temperature
at either end of the rod is zero for all time, so that the boundary conditions
are

u(0, t) = u(1, t)=0

and the initial temperature distribution is:

u(x, 0) = ⇢ 0, 0 < x < 0.5
1, 0.5 < x < 1

QUESTION :

Let u(x, t) = F(x)G(t) and using the separation of variables method,
show that the solution of the the 1d heat equation (1) requires the
solution of the following two ODEs:
G'(t) = kµG and F''(x) = µF, where µ is a constant.

 

[Prove It]

Re: Seperation of Variables/ODE

The temperature distribution u(x, t), at time t > 0, along a homogeneous
metal rod can be obtained by solving the 1d heat equation;

ut = kuxx (1)

where k = 2 is a constant. The length of the rod is 1m and the temperature
at either end of the rod is zero for all time, so that the boundary conditions
are

u(0, t) = u(1, t)=0

and the initial temperature distribution is:

u(x, 0) = ⇢ 0, 0 < x < 0.5
1, 0.5 < x < 1

QUESTION :

Let u(x, t) = F(x)G(t) and using the separation of variables method,
show that the solution of the the 1d heat equation (1) requires the
solution of the following two ODEs:
G'(t) = kµG and F''(x) = µF, where µ is a constant.

Well first of all, this isn't an ODE, it's a PDE.

What have you tried so far?

 

[jmorgan]

Re: Seperation of Variables/ODE

I have began by :

XT' = kX''T

rearrange: T'/T=kX''/X=\alpha

trying to solve: T'/T=\alpha and kX''/X=\alpha

and that is the furthest I can go

 

[jmorgan]

Re: Seperation of Variables/ODE

Ignore my previous reply, I now have:

XT' = kX''T

divide both by kXT to get: T'/kT = X''/X

so T'/kT = µ and X''/x = µ (this is where G'(t)=k µG and F''(x)= µF is required)

now I am stuck on how to solve these.

 

[Prove It]

Re: Seperation of Variables/ODE

Ignore my previous reply, I now have:

XT' = kX''T

divide both by kXT to get: T'/kT = X''/X

so T'/kT = µ and X''/x = µ (this is where G'(t)=k µG and F''(x)= µF is required)

now I am stuck on how to solve these.

The first is a separable first order DE, and the second can be rearranged into a linear homogeneous second order DE with constant coefficients. Both have very standard methods of solution.