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I just want to check if this is true, my rationale being that, by calling the LHS $F$, the $uu_{x}$ term ensures that $\lambda F(t,x,u,u_t,u_x)\neq F(t,x,\lambda u,\lambda u_t,\lambda u_x)$ (so non-linear), and since the zero function solves the PDE then is homogeneous?

maths54321
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  • Yes, that is fine – Paul May 22 '20 at 15:59
  • It's correct that the $uu_x$ term makes the PDE non-linear. But I wouldn't call it homogeneous, since there's no $\alpha$ such that $F(t, x, \lambda u, \lambda u_t, \lambda u_x) = \lambda^\alpha F(t, x, u, u_t, u_x).$ https://math.stackexchange.com/questions/1363664/can-a-differential-equation-be-non-linear-and-homogeneous-at-the-same-time – md2perpe May 22 '20 at 16:03
  • This is confusing, since I recall I had a homework problem with the ODE $\frac{du}{dx}+\frac{x}{u}=0$, and the solutions stated that the equation was both non-linear and homogeneous? – maths54321 May 22 '20 at 16:22

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