- #1
MathematicalPhysicist
Gold Member
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I have this PD equation:
[tex]
T_t=\nabla ^2 T^4
[/tex]
where T=T(r,t) the above laplacian is spherical symmetrical (i.e only the spherical radial coordinate of the operator should be taken into account).
and [tex]Q_0=\int_{0}^{\infty}T(r,t=0)dr[/tex].
So I tried solving it by separation of variables but I get a tough ODE of the radial part.
Here's what I got
[tex]T(r,t)=F(r)G(t) \newline \frac{\frac{dG}{dt}}{G^4}=\frac{\nabla ^2 F^4}{F}=\lambda [/tex]
Now after some manipulations I get for the radial equation the next equation:
[tex]F''(r)+3(F'(r))^2/F(r)+2F'(r)/r-\lambda/(2F(r))^2=0[/tex]
And this is where I am stuck, any suggestion as to how to untie this equation, is even possible?
[tex]
T_t=\nabla ^2 T^4
[/tex]
where T=T(r,t) the above laplacian is spherical symmetrical (i.e only the spherical radial coordinate of the operator should be taken into account).
and [tex]Q_0=\int_{0}^{\infty}T(r,t=0)dr[/tex].
So I tried solving it by separation of variables but I get a tough ODE of the radial part.
Here's what I got
[tex]T(r,t)=F(r)G(t) \newline \frac{\frac{dG}{dt}}{G^4}=\frac{\nabla ^2 F^4}{F}=\lambda [/tex]
Now after some manipulations I get for the radial equation the next equation:
[tex]F''(r)+3(F'(r))^2/F(r)+2F'(r)/r-\lambda/(2F(r))^2=0[/tex]
And this is where I am stuck, any suggestion as to how to untie this equation, is even possible?