Problem 1.3.1

1.3 #1

Heat flows longitudinally through a laterally insulated metal bar of length 10 centimeters, and the temperature $u = u(x,t)$ satisfies the diffusion equation $u_t = ku_{xx}$, where $k = 0.02$ square centimeters per second. Suppose the temperatures at some fixed time T at $x = 4, 6, 8$ cm are 58, 64, and 72 degrees, respectively. Estimate $u_{xx}(6,T)$ using a difference approximation. Will the temperature at $x = 6$ increase or decrease in the next instant of time? Estimate the temperature at $x = 6$ at $T + 0.5$ seconds. Hint: Recall from calculus that $$f''(a) \approx \frac{f(a-h)-2f(a)+f(a+h)}{h^2}$$ where h is a small increment. This approximation is also derived in Chapter 6.

So approximating $u_{xx}(6,T)$ we have $$f''(a)\approx\frac{f(6-2)-2f(6)+f(6+2)}{2^2}\approx\frac{f(4)-2f(6)+f(8)}{2^2}\approx\frac{58-128+72}{4}\approx\frac{1}{2}$$ Because $u_t=ku_{xx}$ and we just found $u_{xx}$ to be greater than 0, we can expect the temperature to increase in the next instant of time. Now $u_t$ can be approximated using $$u_t(6,T)\approx\frac{u(6,T+.5)-u(6,T)}{.5}$$ and we know $u_t(6,T)\approx.01$ and $u(6,T)=64$, so we can solve for $u(6,T+.5)\approx64.005$.

$u_t(6,T)=ku_{xx}(6,T)$ as stated in the problem, with $k=.02$. I showed that $u_{xx}(6,T)\approx\frac{1}{2}$, so $u_t(6,T)=(.02)(\frac{1}{2})=.01$. I hope that clears it up!

I as well have a question, i can follow the math but i was wondering if there was a reason to why you chose $h=2$?

The temperatures the problem gave us were at x values of 4, 6, and 8, so it made sense to use a step value of 2.

Okay thank you! i see now, "step value" makes sense and helped me understand the big equation better