Question

Solve the heat-flow problem by speration of variables:
\[∂u/∂t=∂^2u/∂x^2\]\[0 < x < 1\]\[0 < t < \infty\]
Boundary Conditions:
\[\left(\begin{matrix}u(0,t)=0 \\ u_x(1,t)=0\end{matrix}\right)0 < t < \infty\]
Initial Conditions:\[u(x,0)=x\]\[0 \le x \le 1\]

Answer 1

you assume \[u(t,x) = T(t)X(x) \] where T(t) is a function strictly of t and X is a function of x. Note theses are BIG X and T. Then substitute into the heat equation and separate.

Answer 2

\[ \frac{\partial{u} }{\partial{t}}=\frac{dT}{dt} (X(x))\]
and \[ \frac{\partial^2{u} }{\partial{x^2}}=T(t)\frac{d^2X(x)}{dx^2} \]

Answer 3

Now \[ \frac{\partial{u}}{\partial t}=\frac{\partial^2 u}{\partial x^2} \] becomes

\[ X(x) T'(t) = T(t)X''(x) \] get all of the "x" related terms on the right and the "t" terms on the left. That is the separation.

Answer 4

Since I think you are not there, I will just continue with the solution.
\[ \frac{T'}{T} =\alpha\frac{X''}{X}\]. Usually there is a thermal coefficient \[\alpha\] included as part of the heat equation. I will assume \[\alpha = 1\] anyway back to the problem. KEY CONCEPT.the only way for the equality \[ \frac{T'}{T} =\frac{X''}{X}\] to be true is if \[ \frac{T'}{T} =\frac{X''}{X}=constant\].

Answer 5

The constant may be less than 0, equal to zero, or greater than 0. You must look at each case. Often the constant is represented by \[\lambda^2\]. You will see why this is used at the end of this part of the solution process.
FIRST CASE: \[constant>0\] so \[ \lambda^2>0\] is used giving the ODE, \[\frac{X''}{X} = \lambda^2\] or \[ X''=\lambda^2 X\] or \[ X''- \lambda^2 X=0\].
WOW. This is a DE that you should be able to solve.

Answer 6

The general solution to the ODE \[ X'' - \lambda^2 X=0 \] is \[ X=c_1 e^{\lambda x}+c_2 e^{-\lambda x } \]

Answer 7

You need to end conditions to find the constants \[c_1,c_2\]. The end conditions are found by using \[u(0,t)=0=X(0)T(t)\]. This will be true for any T(t) if X(0)=0. So there is the first end/boundary condition. The second is \[u_x(1,t)=0=X'(1)T(t) \] this will require \[X'(1)=0\] since again we what to keep T(t) as general as possible.

Answer 8

Using the two BCs \[X(0)=0 \] temperature is 0 at this end of a bar, and \[X'(1)=0 \] you can try to solve the boundary value problem. I say try because there may be no solution of this type. Continuing, \[X(0)=0=c_1e^{\lambda 0} +c_2 e^{-\lambda 0) }\]

Answer 9

So \[c_1+c_2=0 \] or \[c_2 = -c_1 \]. Using the second BC \[X'(1)=0=\lambda c_1e^{\lambda 1} -\lambda c_2e^{-\lambda 1 }\]. \[ \lambda \neq 0 \] so divide both sides by \[ \lambda \] to get \[0=c_1e^{\lambda 1} -c_1e^{-\lambda 1}\] but wait this says that \[c_1e^{\lambda}=c_1e^{-\lambda} \] , this is only true if \[c_1=0\], and so \[c_2 =0 \] also.

Answer 10

Need to stop.