16

**HA9 / Re: HA9-P2**

« **on:**November 14, 2015, 05:08:57 PM »

solution to b)

This section allows you to view all posts made by this member. Note that you can only see posts made in areas you currently have access to.

18

Thank you Emily, I know (10) and (11), but it doesn't meantion the replacement. . Maybe it from lectures.

19

b) General solution is given by http://www.math.toronto.edu/courses/apm346h1/20159/PDE-textbook/Chapter6/S6.4.html#mjx-eqn-eq-6.4.11. Fourier coefficients are the same as before, except we replace $A_n$ with $B_n$, $C_n$ with $D_n$, and $a^{-n}$ with $a^{n}$. The final solution isI have question about itï¼Œso under this conditions given by Question 3, we can apply that replacing $A_n$ with $B_n$, $C_n$ with $D_n$, and $a^{-n}$ with $a^{n}$? where is it from?

$$u(r,\theta)=\frac{4}{\pi}\sum_{n\geq 1, odd}\frac{r^{-n}a^{n}}{n}\sin(n\theta)$$

Thank you! Fei Fan

20

hmmmm...... I think the final answer from Yumeng is still not correct.

should be something about*cosh(|y|(1-y))* exists.

Furthermore, eâˆ’|k|y+e|k|y=2cosh(|k|y) instead of 2cosh(|k|)

should be something about

Furthermore, eâˆ’|k|y+e|k|y=2cosh(|k|y) instead of 2cosh(|k|)

22

(d)YU MENG,

I think in the step By Foundamental Theorem of Calculus, it should be

the signs of your solution is wrong, but anyway, it doesn't influence the right answer.

23

Xi Yue, We have different answers. I'm not sure which one is correct.

24

For part a),

I add some proof details to show that it's actually decrease unless*U(x,t)* = constant.

Case A: If*U*_{x} ~~=~~ 0, then we are done.

Case B: If*U*_{x} = 0, then *U(x,t)* = *f(t)* or Constant *C*.

Case 1 of B: If*U(x,t) = C*, then we are done.

Case 2 of B: If*U(x,t) = f(t)*, we can plug it into *U*_{t }= KU_{xx}, we can have *f*_{t}(t) = 0, then *f(t)* = Constant, so we totally done.

I add some proof details to show that it's actually decrease unless

Case A: If

Case B: If

Case 1 of B: If

Case 2 of B: If

25

I think the e) solution posted by Emily is right! thank you

27

29