Timeline for Wave equation on 2D semi-infinite plane
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15 events
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| May 19, 2023 at 1:11 | comment | added | Willie Wong | As far as I can see, $\Phi$ and $\Psi$ are necessarily coupled, and you cannot isolate them separately. (You could solve with general boundary conditions: namely set $\partial^2_{xz}\Phi(x,0,t) = k(x,t) = - \partial^2_{xx}\Psi(x,0,t) + \partial^2_{zz} \Psi(x,0,t)$ and then match the two solutions at the boundary; but I am not sure if that will make any simplification over just solving the equations for $u$ to start with.) | |
| May 18, 2023 at 22:10 | comment | added | basketas | @WillieWong I decomposed $w$ into longitudinal and transverse potential too, so $f$ and $g$ only have the appropriate corresponding directions. But I get what you're saying with regard to the boundary conditions. Does this then imply that $\Phi$ and $\Psi$ cannot be solved for independently? I would have to find the general solutions, calculate back to $\mathbf{u}$ and then apply the boundary condition? Or is there maybe another more elegant way that I'm failing to see? Thank you for your help! | |
| May 18, 2023 at 22:03 | comment | added | Willie Wong | Basically: you started with two boundary conditions; if you arbitrarily split this up into 4 separate boundary conditions, it is not too hard to see intuitively that this can make your problem overdetermined. | |
| May 18, 2023 at 22:02 | comment | added | Willie Wong | Going to your final sentence in your edit: no; you cannot assume that the conditions are satisfied separately. The functions $\Phi$ and $\Psi$ are not independent. For starters, when you rewrote $u = v + w$ the chosen function $w$ generally has components both in the longitudinal and transverse directions, so the functions $f$ and $g$ that appears in the wave equation for $\Phi$ and $\Psi$ are coupled. | |
| May 18, 2023 at 21:45 | comment | added | basketas | @WillieWong I added to the post the original thermoelastic equations and how I got to the equation for the potential $\Psi$. Hopefully this will make it clearer where my problem comes from. | |
| May 18, 2023 at 21:43 | history | edited | basketas | CC BY-SA 4.0 |
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| May 18, 2023 at 20:09 | comment | added | Willie Wong | You see this already when looking at the separation of variables: as you observed, the only homogeneous solutions are of the form $$ (A \sin(\sqrt{2} q tc ) + B \cos(\sqrt{2} q c t) \cos(q z) ( C \sin(qx) + D \cos(qx) ) $$ This is too few degrees of freedom to capture arbitrary initial data. // You may wish to consider showing the original thermoielastic equations for $u$ and how you performed the reduction to something for a scalar function $\Psi$, which may provide some clues. | |
| May 18, 2023 at 19:56 | comment | added | Willie Wong | Combining the boundary condition $\partial^2_{xx}\Psi = \partial^2_{zz} \Psi$ on $\{z = 0\}$ with the original PDE, you see that restricted to $\{z = 0\}$ the function $\Psi$ solves the PDE $$- \frac{1}{c^2} \partial^2_{tt}\Psi(x,0,t) + 2 \partial^2_{xx} \Psi(x,0,t) = f(0,t) H(t) \sin(q_0 x)$$ This means that your boundary conditions force $\Psi |_{\{z = 0\}}$ to be completely prescribed. Together with the Neumann type boundary condition $\partial^2_{xz} \Psi = 0$, you are simultaneously prescribing dirichlet and Neumann data for a wave equation, and that is overdetermined. | |
| May 18, 2023 at 17:37 | comment | added | basketas | @WillieWong (1), (2) I edited the last paragraph to add details. (3) Is having second order conditions a problem, and if so, why? My math on the theory of differential equations is a bit weak. This problem is actually a part of a more complicated physical problem, where I'm solving a second order thermoelastic equations for the displacement of the material $u$. The boundary conditions for $u$ are of the first order. The equations can be simplified by taking $u=\nabla \Phi + \nabla\times \Psi$ and one gets two equation like the one above. The boundary conditions are then of the second order. | |
| May 18, 2023 at 17:30 | history | edited | basketas | CC BY-SA 4.0 |
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| May 18, 2023 at 14:50 | comment | added | Willie Wong | (1) You are prescribing initial data, so your solution should be unique. So I am not sure what you mean about "getting the homogeneous solution". (2) Maybe it would be clearer if you provide the details of how you "found the particular solution using Fourier transform". (3) You are prescribing second order boundary conditions on $\{z = 0\}$, are you sure your problem is not overprescribed? | |
| May 18, 2023 at 12:42 | history | edited | YCor |
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| May 18, 2023 at 4:44 | history | edited | Daniele Tampieri | CC BY-SA 4.0 |
Minor Math Jaxing
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| S May 17, 2023 at 23:19 | review | First questions | |||
| May 18, 2023 at 0:33 | |||||
| S May 17, 2023 at 23:19 | history | asked | basketas | CC BY-SA 4.0 |