Burgers' equation - issues with final form
- Thread starter Fun Coolo
- Start date
Hi, I don't understand what you mean with original statement of the problem, sorry (I am studying fluid dynamics, in particular the part relating to turbulent flows, and this equation is inserted in this context).
For the other 2 questions:
2. I suppose the solution has the following form: I am making a hypothesis that I will see confirmed if that solution, if substituted in the equation, will solve it and under what conditions (I know it will because whoever wrote this part tells me (it's an excerpt from a book)).
3. I know that always from the material at my disposal (very little, these are slides extrapolated I don't know from which book, I'm trying to build the reasoning that led to writing the final equation, which I suppose true), the author of the slides gets to write that equation.
Unfortunately the material at my disposal is little, I can't tell you anything else, my problem is only to demonstrate that final step that you can see in the pic.
I don't know if this is a series development of some kind, or if there is (probably) some preliminary step on summation.
Unfortunately I am unable to provide you with other significant material, sorry.
For the other 2 questions:
2. I suppose the solution has the following form: I am making a hypothesis that I will see confirmed if that solution, if substituted in the equation, will solve it and under what conditions (I know it will because whoever wrote this part tells me (it's an excerpt from a book)).
3. I know that always from the material at my disposal (very little, these are slides extrapolated I don't know from which book, I'm trying to build the reasoning that led to writing the final equation, which I suppose true), the author of the slides gets to write that equation.
Unfortunately the material at my disposal is little, I can't tell you anything else, my problem is only to demonstrate that final step that you can see in the pic.
I don't know if this is a series development of some kind, or if there is (probably) some preliminary step on summation.
Unfortunately I am unable to provide you with other significant material, sorry.
blamocur
Elite Member
- Joined
- Oct 30, 2021
- Messages
- 3,140
I assumed that this was a part of homework, but my assumption was wrong. Here are my notes on questions 2 and 3:
2. I believe it is not unusual to represent the solution as Fourier series, although it is not clear why there are no cosines but only sines.
3. I think I can see where that transformation comes from. I've started it below -- please let me know if it makes sense or if you see any errors in it:
[math]u \frac{\partial u}{\partial x} = \left(\sum_i A_i \sin ix \right) \left(\sum_j j A_j \cos jx\right) = \sum_{ij} j A_i A_j \sin ix \cos jx[/math]
[math]= \frac{1}{2} \sum_{ij} jA_i A_j \left(\sin (i+j)x + \sin (i-j)x \right)[/math]
[math]= \frac{1}{2} \sum_k \sum_{i+j=k} jA_i A_j \sin kx + \frac{1}{2} \sum_k \sum_{i-j=k} jA_i A_j \sin kx - \frac{1}{2} \sum_k \sum_{i-j=-k} jA_i A_j \sin kx[/math]
[math]= \frac{1}{2} \sum_k \sin kx \left( \sum_{i+j=k} jA_i A_j + \sum_{i-j=k} jA_i A_j - \sum_{i-j=-k} jA_i A_j \right)[/math]
Replace [imath]j[/imath] with [imath]m[/imath] and [imath]i[/imath] with corresponding expression:
[math]= \frac{1}{2} \sum_k \sin kx \left( \sum_m mA_{k-m} A_m + \sum_m mA_{k+m} A_m - \sum_m mA_{m-k} A_m \right)[/math]
blamocur
Elite Member
- Joined
- Oct 30, 2021
- Messages
- 3,140
Summing by i,j can be done in different orders. For the first term I take (1,1) for i+j=2=k. then (1,2) and (2,1) for i+j=3, then (1,3),(2,2) and (3,1) for i+j=4, etc. The second term (the one with [imath]\sin(i-j)x[/imath]) is broken into two parts, one part for positive [imath]i-j[/imath] and one part for negative [imath]i-j[/imath].
blamocur
Elite Member
- Joined
- Oct 30, 2021
- Messages
- 3,140
No need to apologize -- I've signed up to this forum to help and explain to the best of my abilities.
As for the properties of summation I am only using the standard properties of commutativity and associativity.
Let's try a simpler example with short sums and simple expressions:
[math]S = \sum_{i=1}^3 \sum_{j=1}^3 b_{ij} = b_{11} +b_{12} +b_{13} +b_{21} +b_{22} +b_{23} + b_{31} +b_{32} +b_{33}[/math]I can represent [imath]S[/imath] as a sum of "sub-sums": [imath]S = S_2 + S_3 + S_4 + S_5 + S_6[/imath], where [math]\begin{array}{lll} S_2 &=& b_{11} \\ S_3 &=& b_{12} + b_{21} \\ S_4 &=& b_{13} + b_{22} + b_{31}\\ S_5 &=& b_{23} + b_{32} \\ S_6 &=& b_{33} \\ \end{array}[/math]As can be seen [imath]\;\;S_k = \sum_{i+j=k} b_{ij}\;\;[/imath] (where [imath]2 \leq k \leq 6[/imath]) and thus [imath]\;\;S = \sum_{k=2}^6 \sum_{i+j=k} b_{ij}[/imath]
Did this help?
As for the properties of summation I am only using the standard properties of commutativity and associativity.
Let's try a simpler example with short sums and simple expressions:
[math]S = \sum_{i=1}^3 \sum_{j=1}^3 b_{ij} = b_{11} +b_{12} +b_{13} +b_{21} +b_{22} +b_{23} + b_{31} +b_{32} +b_{33}[/math]I can represent [imath]S[/imath] as a sum of "sub-sums": [imath]S = S_2 + S_3 + S_4 + S_5 + S_6[/imath], where [math]\begin{array}{lll} S_2 &=& b_{11} \\ S_3 &=& b_{12} + b_{21} \\ S_4 &=& b_{13} + b_{22} + b_{31}\\ S_5 &=& b_{23} + b_{32} \\ S_6 &=& b_{33} \\ \end{array}[/math]As can be seen [imath]\;\;S_k = \sum_{i+j=k} b_{ij}\;\;[/imath] (where [imath]2 \leq k \leq 6[/imath]) and thus [imath]\;\;S = \sum_{k=2}^6 \sum_{i+j=k} b_{ij}[/imath]
Did this help?
I think I understand, sorry if I didn't answer before, I saw if it suited my case and yes, I adapted it when the sum starts from zero (there are also the terms with zero subscript) and it's ok (even if some terms they repeat themselves, but being null they don't worry me, so the discourse holds). Thanks very much for your precious help, appreciated?
blamocur
Elite Member
- Joined
- Oct 30, 2021
- Messages
- 3,140
Not sure I get the question, but I'll guess that the switch from [imath]\sum_{i+j=k} jA_iA_j[/imath] to
[imath]\sum_m mA_{k-m}A_m[/imath] is a good candidate. I've replaced [imath]j[/imath] with [imath]m[/imath] purely to match your notation, so let's consider this transformation: [imath]\sum_{i+m=k} m A_i A_m \Longrightarrow \sum_m mA_{k-m}A_m[/imath]. Since [imath]i+m=k[/imath] we get [imath]i=k-m[/imath] and thus [imath]\sum_m m A_{k-m}A_m[/imath]
[imath]\sum_m mA_{k-m}A_m[/imath] is a good candidate. I've replaced [imath]j[/imath] with [imath]m[/imath] purely to match your notation, so let's consider this transformation: [imath]\sum_{i+m=k} m A_i A_m \Longrightarrow \sum_m mA_{k-m}A_m[/imath]. Since [imath]i+m=k[/imath] we get [imath]i=k-m[/imath] and thus [imath]\sum_m m A_{k-m}A_m[/imath]
Maybe I explained myself wrong, I meant that I did not understand the following index exchange
Since it is a condition (having to add up all terms such that i + j = k), shouldn't there be another condition as a substitute, rather than the index m?
Since it is a condition (having to add up all terms such that i + j = k), shouldn't there be another condition as a substitute, rather than the index m?
blamocur
Elite Member
- Joined
- Oct 30, 2021
- Messages
- 3,140
In the first some the equation [imath]i+m=k[/imath] describes all pairs of [imath](i,m)[/imath] for which the summation is performed. You can vary [imath]m[/imath] and compute [imath]i[/imath] from the equation, or you can vary [imath]i[/imath] and compute [imath]m[/imath], as in [imath]\sum_i (k-i) A_i A_{k-i}[/imath]. Either way you will get the same pairs of indices for A's.
For example, if [imath]k=4[/imath] in post #8 we get a set of pairs (1,3), (2,2) and (3,1). Here [imath]m[/imath] goes through the set 3,2,1 and i=4-m. You can rewrite the set of pairs as (4-3,3), (4-2,2) and (4-1,1).