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| * Chap. 12 and the Definition 12.8 of the even extension on p.56 should read x∈[0,L] instead of x∈[0,π] | * Chap. 12 and the Definition 12.8 of the even extension on p.56 should read x∈[0,L] instead of x∈[0,π] | ||
| - | * Example 15.4 should read 0<x<π instead of 0<x<2. | + | * Example 15.4 should read 0<x<π instead of 0<x<2. //(Courtesy DH)// |
| + | * Eqn (16.7) should have λ2 instead of λ on the equation for G. //(Courtesy DH)// | ||
| * Theorem 18.3 should read the initial conditions of u(x,0)=f(x) and ut(x,0)=g(x). //(Courtesy RA)// | * Theorem 18.3 should read the initial conditions of u(x,0)=f(x) and ut(x,0)=g(x). //(Courtesy RA)// | ||
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| * PS3: Index notation needs to be removed from the main solutions in 19-20' delivery since the topic has been moved to the Appendix. Students are still free to use the technique if they learn it (as noted in lectures). | * PS3: Index notation needs to be removed from the main solutions in 19-20' delivery since the topic has been moved to the Appendix. Students are still free to use the technique if they learn it (as noted in lectures). | ||
| * PS4 Q4 dx and dy transposed | * PS4 Q4 dx and dy transposed | ||
| + | * PS6 Q1. The function sin(x)exp(−cos(x2)) is indeed not periodic but not for the reasons stated in the solutions. The point here is that cos(x2) is not a periodic function. You can verify this either by checking whether it's possible that (x+L)2=x2+nπ independent of x, or simply by plotting the cos(x^2)$ and observing its behaviour, particularly near the origin. | ||
| + | * **PS9 Q2:** There is a missing factor of p on the bottom here. See Lecture 29. | ||
| + | * **PS9 Q4:** Looks like axes were doubled here. //(Courtesy RA, HC)// | ||
| + | * **PS10 Q2:** Looks like there are a few κs missing on the right hand side (I count three); **Q3** In the redefinition of the energy (above "So now you are back to..." | ||
| + | ==== Lectures ==== | ||
| + | * Correction to {{ : | ||
| + | |||
| + | ==== Past exams ==== | ||
| + | * 2017-18 exam Q4d. The index should start from n=1 once the 2n+1 shift is done. | ||