Modifying the Solution for Heat Flow PDE Problem

[mrjeffy321]

I have got a heat flow partial differential equation problem that is giving me a little problem due to the direction the temperature is changing.

I have a bar (which lies along the X axis) which is initially at a uniform temperature which (for simplicity sake) we will call zero degrees.
At time, t, = 0 and onward one side of the bar is heated so that it is kept at a constant temperature of, say, 100 degrees.
I am asked to find the temperature of the bar as a function of the position, x, along the bar, and time t.

Setting up the heat flow equation, I have,
Del square u = 1/alpha * the first partial derivative of u with respect to time.
I can rearrange this and solve for a product solution of u(x, t) as X(x) * T(t). then separate out thr variables into two ordinary differential equations for X(x) and T(t) and get,
X’’ + k^2 * X = 0,
And,
T’ + (k*alpha)^2 * T = 0

The X equation is going to give me some sine and/or cosine solution, and the T equation should give me a exponential solution.

Solving for T,
T = A * e^-(K * alpha)^2 * t
Where A is just some constant.
I am ignoring the + exponent solution since this is non-physical, temperature cannot go to infinite as t goes to infinity.

But what worries me is that the equation does not satisfy the initial boundary conditions.

At t = 0, then entire rod should be at zero. And at t = infinite, the temperature of the rod should be some constant function of x (steady state solution when the two ends of the rod are held at different temperatures).
But my equation says that the temperature drops to zero as t goes to infinite and has some maximum value (A) when t = 0.

Clearly I need to modify my solution for T(t) so get it to show an increasing temperature which will approach some non-zero value.

 

[HallsofIvy]

At time, t, = 0 and onward one side of the bar is heated so that it is kept at a constant temperature of, say, 100 degrees.

Do you mean that one end, say x= 0, is kept at 0 degrees and the other at 100 degrees? And what is the length of the bar? I'll call it "L".
What happens is that your initial temperature distribution is not continuous. Since you'll be integrating to find the temperature function, that's not a problem.

Calling the temperature \phi(x,t), you are solving
\frac{\partial^2 \phi}{\partial x^2} = \frac{1}{\alpha^2}\frac{\partial \phi}{\partial t}
with boundary conditions \phi (0,t)= 0, \phi(L,t)= 100 and initial condition \phi(x,0)= 0 if x is not L, \phi(L,0)= 100.

This may not be the way you were taught to do this but I would recommend subtracting of the function 100x/L for all t, which satisfies the boundary conditions. Since
\frac{\partial^2 (100x/L)}{\partial x^2}= 0[/itex] <br /> u(x,t)= \phi(x,t)- 100x/L satisfies the same differential equation as \phi(x,t), u(0,t)= u(L,t)= 0, and u(x,0)= -100x/L for x not equal to L, u(L,0)= 0. Since in integrating, the value at a single point is unimportant, you can ignore that last value. Now, it should be easy to solve the equation for X, determine the correct values for k and use that in the T equation.

 

[mrjeffy321]

Do you mean that one end, say x= 0, is kept at 0 degrees and the other at 100 degrees? And what is the length of the bar? I'll call it "L".

Calling the temperature \phi(x,t), you are solving
\frac{\partial^2 \phi}{\partial x^2} = \frac{1}{\alpha^2}\frac{\partial \phi}{\partial t}
with boundary conditions \phi (0,t)= 0, \phi(L,t)= 100 and initial condition \phi(x,0)= 0 if x is not L, \phi(L,0)= 100.

Yes.

Since the x = 0 side of the bar will always be 0°, I could eliminate the possible solution to the X(x) function involving Cosine, since cos (0) = 1. I figured out that if I said that the length of the bar, L, was only 1/4 of the period of the Sine function solution, this would keep the x = L side from being zero.

I also found the solution for T(t) as being e^-(kα)^2 * t. But the problem I was having was that as time goes to infinite, my T(t) function goes to zero. Since T(t) is multiplied by X(x) to get the temperature function u(x, t), this would make the entire function zero.
What should happen is that some linear temperature distribution should develop in the bar to make u(x, infinity) = 100 * x / L.

This may not be the way you were taught to do this but I would recommend subtracting of the function 100x/L for all t, which satisfies the boundary conditions.

I don’t know about subtracting….seems like adding this value (100x/L) would work though. T(t) goes to zero as time goes to infinite, u(x, infinity) will just be = +100x/L, which does give the correct distribution. Although this would work at t = infinity, won't it interfere at other times….for example t = 0? When t = 0, T(t) = 1, so u(x, t) would be = sin (n * pi * x / (2L)) * 1 + 100x/L. Which is not true.