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HA9 / Re: HA9-P2
« on: November 14, 2015, 05:08:57 PM »
solution to b)
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HA8 / Re: HA8-P3
« on: November 12, 2015, 03:56:06 PM »
Thank you Emily, I know (10) and (11), but it doesn't meantion the replacement. 
. Maybe it from lectures.
. Maybe it from lectures.
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HA8 / Re: HA8-P3
« on: November 10, 2015, 08:34:45 PM »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 it，so 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
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HA7 / Re: HA7-P7
« on: November 05, 2015, 05:08:01 PM »
hmmmm...... I think the final answer from Yumeng is still not correct.
should be something about cosh(|y|(1-y)) exists.
Furthermore, e−|k|y+e|k|y=2cosh(|k|y) instead of 2cosh(|k|)
should be something about cosh(|y|(1-y)) exists.
Furthermore, e−|k|y+e|k|y=2cosh(|k|y) instead of 2cosh(|k|)
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HA6 / Re: hm6 Q1
« on: October 28, 2015, 07:32:13 PM »(d)YU MENG,
I think in the step By Foundamental Theorem of Calculus, it should be d(-Xn'Xm + XnXm')/dx
the signs of your solution is wrong, but anyway, it doesn't influence the right answer.
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Test 1 / Re: TT1-P2
« on: October 22, 2015, 05:43:13 PM »
Xi Yue, We have different answers. I'm not sure which one is correct.
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HA5 / Re: HA5-P7
« on: October 20, 2015, 07:30:15 PM »
For part a),
I add some proof details to show that it's actually decrease unless U(x,t) = constant.
Case A: If Ux= 0, then we are done.
Case B: If Ux = 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 Ut = KUxx, we can have ft(t) = 0, then f(t) = Constant, so we totally done.
I add some proof details to show that it's actually decrease unless U(x,t) = constant.
Case A: If Ux
Case B: If Ux = 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 Ut = KUxx, we can have ft(t) = 0, then f(t) = Constant, so we totally done.
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HA5 / Re: HA5-P3
« on: October 17, 2015, 09:25:59 PM »
I think the e) solution posted by Emily is right! thank you
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