Timeline for Generalized Fuchsian-type PDE
Current License: CC BY-SA 4.0
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| Apr 24 at 2:28 | history | edited | Math2024 | CC BY-SA 4.0 |
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| Mar 7 at 4:34 | answer | added | Willie Wong | timeline score: 3 | |
| Mar 7 at 4:19 | history | edited | Math2024 | CC BY-SA 4.0 |
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| Mar 7 at 4:10 | history | edited | Math2024 | CC BY-SA 4.0 |
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| Mar 7 at 4:04 | history | edited | Math2024 | CC BY-SA 4.0 |
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| Mar 7 at 4:02 | comment | added | Math2024 | Yes, the simplified PDE has such scaling invariance (but the original PDE does not have this property). For the simplified PDE, one can first solve it order by order in the $t \to 0$ limit and observe the n-th order solution $\sim (x^3 t)^n$. | |
| Mar 6 at 22:33 | comment | added | Bob Terrell | $x^3t$ can be seen to be special in the simplified equation by a scaling argument. $A(kx,lt)$ is also a solution when $k^3=l^{-1}$. So if it happens that all the functions $A(kx,k^-3t)$ are the same that suggests looking for a function of $x^3t$. | |
| Mar 6 at 19:12 | history | edited | Math2024 | CC BY-SA 4.0 |
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| Mar 6 at 16:22 | history | edited | Math2024 | CC BY-SA 4.0 |
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| Mar 6 at 16:09 | comment | added | Math2024 | To clarify, I added the additional constraints (used also in the simpler PDE) in the post, $i.e. a_i(0)=0$. | |
| Mar 6 at 16:07 | history | edited | Math2024 | CC BY-SA 4.0 |
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| Mar 6 at 15:54 | comment | added | Math2024 | Thanks yes – that leads to the equation form I posted in mathoverflow.net/questions/465993/a-4th-order-linear-pde. I realised that writing the PDE in terms of $A(x,t)$ might be more useful to talk about BCs or make a potential connection to Fuchsian-type equations.. | |
| Mar 6 at 15:46 | comment | added | Peter Kravchuk | The simpler PDE is then $\partial_t \partial_y^3 F(y,t)=F(y,t)$. In either case, I think this form highlights that your equation may not have a unique solution since without additional constraints, $\partial_y^3$ is not invertible. | |
| Mar 6 at 15:43 | history | edited | Math2024 | CC BY-SA 4.0 |
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| Mar 6 at 15:40 | comment | added | Peter Kravchuk | FWIW setting $A(x,t) = (x^2t)^{-1}F(1-x,t)$ gives a much simpler-looking equation for $F(y,t)$: $y^3\partial_t \partial_y^3 F(y,t)=F(y,t)$. | |
| Mar 6 at 15:17 | history | edited | Math2024 | CC BY-SA 4.0 |
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| Mar 6 at 14:25 | history | edited | Math2024 | CC BY-SA 4.0 |
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| Mar 6 at 13:04 | history | edited | Math2024 | CC BY-SA 4.0 |
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| Mar 6 at 12:53 | history | edited | Math2024 | CC BY-SA 4.0 |
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| Mar 6 at 12:15 | answer | added | Igor Khavkine | timeline score: 6 | |
| Mar 6 at 9:51 | answer | added | Daniele Tampieri | timeline score: 3 | |
| Mar 6 at 8:17 | history | edited | Daniele Tampieri | CC BY-SA 4.0 |
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| Mar 6 at 2:51 | history | edited | Math2024 | CC BY-SA 4.0 |
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| Mar 6 at 2:31 | history | edited | Math2024 | CC BY-SA 4.0 |
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| Mar 6 at 2:25 | history | edited | Math2024 | CC BY-SA 4.0 |
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| Mar 6 at 2:12 | history | edited | Math2024 | CC BY-SA 4.0 |
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| Mar 6 at 2:07 | history | edited | Math2024 | CC BY-SA 4.0 |
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| Mar 5 at 23:54 | history | edited | Math2024 | CC BY-SA 4.0 |
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| Mar 5 at 23:38 | history | asked | Math2024 | CC BY-SA 4.0 |