This is a two part problem that uses the Maximum/Minimum Principle and the Continuous Dependence. I already got the answer for the Maximum/Minimum Principle, but now I have to apply the Continuous Dependence Theorem to prove that there is a difference of $\frac{\pi^6}{640}$ which is kind of small because it's 1.497.
- Use the Maximum/Minimum Principles to deduce that the solution u of the problem
D.E. $u_t=ku_{xx} 0$
$0 \leq x \leq \pi $
$t \geq 0$
B.C. $u_x(0,t) =0, u_x(\pi,t)=0$
I.C $u(x,0)=sin(x) + \frac{1}{2} sin(2x)$
satisfies $0 \leq u(x,t) \leq \frac{3 \sqrt{3}}{4} $ for all $ 0 \leq x \leq \pi, t \geq 0$
By Proposition 1
$b_ne^{-(\frac{n \pi}{L})^2kt}sin(\frac{n\pi x}{L}) $
Let $ k=k, L = \pi, b_1 = 1$ and $n=1$
$b_1e^{-(\frac{n \pi}{\pi})^2kt}sin(\frac{n\pi x}{\pi}) $
$e^{-kt}sin(x)$
Now, let $ n=2$
$b_2e^{-(\frac{2 \pi}{\pi})^2kt}sin(\frac{2 \pi x}{\pi}) $
$\frac{1}{2}e^{-4kt} sin(2x) $
$u(x,t) = e^{-kt}sin(x)+\frac{1}{2}e^{-4kt} sin(2x) $
$u_t = ke^{-kt}sin(x)-2ke^{-4kt}sin(2x)$
$f(x) = sin(x) + \frac{1}{2} sin(2x)$
$f'(x) = cos(x) + cos(2x)$
We need to find an angle that would produce a zero.
$f'(0) = cos(0) + cos(0)= 1+1 =2$[increasing]
$f'(\frac{\pi}{6}) = cos(\frac{\pi}{6}) + cos(\frac{\pi}{3})= \frac{\sqrt{3}}{2} \frac{1}{2} =\frac{\sqrt{3}+1}{2}$[increasing]
$f'(\frac{\pi}{3}) = cos(\frac{\pi}{3})+cos(\frac{2 \pi}{3}) = cos(60)+cos(120) = \frac{1}{2}-\frac{1}{2} = 0$
The line of the graph will be decreasing if the angle is beyond 60 degrees.
$f(x) = sin(x) + \frac{1}{2} sin(2x)$
$sin(60) + \frac{1}{2}sin(60)$
$\frac{\sqrt{3}}{2} + (\frac{1}{2})\frac{\sqrt{3}}{2}$
$\frac{2 \sqrt{3}+\sqrt{3}}{4} = \frac{3\sqrt{3}}{4}$
- Suppose the Joe adds the term $ \frac{1}{10}x^3(\pi-x)^3$ to the initial temperature distribution in Problem 3. Assuming that he can find a solution of this new problem, show that this solution differs from the solution of the original problem by at most $\frac{\pi^6}{640}$ at any $0 \leq x \leq \pi$ and $ t \geq 0$
Hint. Apply Theorem 3 noting that $\mid f_1(x)-f_2(x) \mid = \mid \frac{1}{10}x^3(\pi-x)^3 \mid$
Theorem 3 (Continuous Dependence on I.C and B.c) states that we let $u_1(x,t)$ and $u_2(x,t)$ be $C^2$ solutions of the respective problems $(0 \leq x \leq L, t \geq 0)$
D.E $u_t=ku_{xx}$
B.C $u(0,t) = a_1(t)$
$(L,t) = b_1(t)$
I.C $u(x,0)=f_1(x)$
and
D.E $u_t=ku_{xx}$
B.C $u(0,t) = a_2(t)$
$(L,t) = b_2(t)$
I.C $u(x,0)=f_2(x)$
If for some $\epsilon \geq 0$ we have $\mid f_1(x) -f_2(x) \mid \leq \epsilon \forall x , 0 \leq x \leq L$ and $\mid a_1(x) -a_2(x) \mid \leq \epsilon $ and $\mid b_1(x) -b_2(x) \mid \leq \epsilon$ $\forall t , 0 \leq t \leq T$ then $\mid u_1(x,t) -u_2(x,t) \mid \leq \epsilon$...$\forall x,t$ where $0 \leq x \leq L, 0 \leq t \leq T$
The original proof is that we let $v(x,t) = u_1(x,t) -u_2(x,t)$. Then, $v_t=kv_{xx}$ and we have $ \mid v(x,0) \mid = \mid f_1(x)-f_2(x) \mid \leq \epsilon$, for $0 \leq x leq L$. So now we have $ \mid v(0,t) \mid = \mid a_1(t)-a_2(t) \mid \leq \epsilon$ and $ \mid v(L,t) = b_1(t)-b_2(t) \mid \leq \epsilon$ for $ 0 \leq t \leq T$.
I can only attempt part of this because I have to find the $\epsilon$ and I'm stuck on that part, but I do know that there is
$\mid v(x,0) \mid = \mid \frac{1}{10}x^3(\pi-x)^3 \mid \leq \epsilon$ for $0 \leq x \leq \pi$
and $ \mid v(0,t) \mid = \mid 0-0 \mid \leq \epsilon$ and $ \mid v(L,t) \mid =\mid 0-0 \mid \leq \epsilon$ for $0 \leq t \leq T$