How a solving a nonlinear PDE with Self-similar Solution?I want to solve the problem :∂u∂t=|∇u|−2p(∂2u∂ξ2+(1−p)∂2u∂η2)We search for a self-similarity solution, the general form of which is as followsu(x,y,t)=f(ξ),&nbsp;&nbsp;with&nbsp;&nbsp;ξ=d(x2+y2)na(t)from which we obtainαξ=(1−p)(2n)−2p+2((12n(1−p)+2n−12n)(ddfdξ)−2p+ξ(ddfdξ)−2p−1dd2fdξ2)Now, I am very confused on how to solve the above equation and find the exact solution of&nbsp;f(ξ), thereby finding the exact solution of&nbsp;u(x,y,t). So my question is, does anyone know how to solve this ordinary differential equation?

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How a solving a nonlinear PDE with Self-similar Solution?I want to solve the problem :∂u∂t=|∇u|−2p(∂2u∂ξ2+(1−p)∂2u∂η2)We search for a self-similarity solution, the general form of which is as followsu(x,y,t)=f(ξ),&amp;nbsp;&amp;nbsp;with&amp;nbsp;&amp;nbsp;ξ=d(x2+y2)na(t)from which we obtainαξ=(1−p)(2n)−2p+2((12n(1−p)+2n−12n)(ddfdξ)−2p+ξ(ddfdξ)−2p−1dd2fdξ2)Now, I am very confused on how to solve the above equation and find the exact solution of&amp;nbsp;f(ξ), thereby finding the exact solution of&amp;nbsp;u(x,y,t). So my question is, does anyone know how to solve this ordinary differential equation?

Abdullah Avery · Accepted Answer

Explanation:αξ=(1−p)(2n)−2p+2((12n(1−p)+2n−12n)(ddfdξ)−2p+ξ(ddfdξ)−2p−1dd2fdξ2)A bunch of constants that can be written as where I replaced ξ=z.Cfz2p+1=azfz+zfzz.Define g=fz, thenCg2p+1=azg+zgz

How a solving a nonlinear PDE with Self-similar

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