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Mathematics
Differential Equations
How to solve simple 2D space-time PDE numerically
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[QUOTE="pasmith, post: 6548063, member: 415692"] The Mathematica result is basically correct; it's just made the schoolboy error of using [itex]\int 1/u\,du = \log u + C[/itex] rather than [itex]\int 1/u\,du = \log|u| + C[/itex]. However I don't think your result is correct. The method of characteristics gives that [itex]\eta[/itex] is constant on curves [itex](x(t),t)[/itex] where, for [itex]x(0) \neq 0, L/2, L[/itex], [tex] \int_{x(0)}^{x(t)} \csc(2\pi x/L)\,dx = t[/tex] or [tex] \left[\ln|\csc(2\pi x/L) + \cot(2\pi x/L)|\right]_{x(0)}^{x(t)} = \left[\log |\cot(\pi x/L)|\right]_{x(0)}^{x(t)}= -2\pi t/L.[/tex] Solving this yields [tex] \cot(\pi x(t)/L) = \cot(\pi x(0)/L) e^{-2\pi t/L},[/tex] irrespective of the sign of[itex]\cot(\pi x/L)[/itex]. Applying the initial condition then gives [tex] \eta(x,t) = \sin\left( 2 \operatorname{arccot}\left( \cot(\pi x/L) e^{2\pi t/L} \right) \right).[/tex] This holds also for [itex]x(0) = 0, L/2, L[/itex] where the characteristic is [itex](x,t) = (x(0),t)[/itex] and the initial condition is [itex]\eta(x_0,0) = 0[/itex] provided one identifies [itex]\operatorname{arccot}(\infty) = 0[/itex]. Simplifying the Mathematica expression will, for positive arguments of [itex]\log[/itex], reduce to the above. [/QUOTE]
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How to solve simple 2D space-time PDE numerically
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