Late time Theis equation solution

[Guest]

(∂h(r,t))/∂t=S/T (∂^2 h(r,t))/(∂r^2 )

[Guest]

S and T are constants?

If so then I would try a solution of the form h(r, t)= u(r)v(t). That is, u is a function of r only and v is a function of t only.

Then [tex]\frac{\partial^2 h}{\partial r^2}= v(t)\frac{d^2u}{dr^2}[/tex] and [tex]\frac{\partial h}{\partial t}= u(r)\frac{dv}{dt}[/tex].

The equation becomes [tex]u\frac{dv}{dt}= (S/T)v \frac{d^2u}{dr^2}[/tex].

Divide both sides by uv: [tex]\frac{1}{v}\frac{dv}{dt}= (S/T)\frac{1}{u}\frac{d^2u}{dr^2}[/tex].

The left side depends only on t while the right side depends only on r, If we were to change t while leaving r unchanged, the right side would not change. Since the left side is equal to it, the right side must be a constant and the left side must be the same constant.

That is [tex]frac{1}{v}\frac{dv}{dt}= C[/tex] and [tex](S/T)\frac{1}{u}\frac{d^2u}{dr^2}= C[/tex].'

So [tex]\frac{dv}{dt}= Cv[/tex], [tex]\frac{dv}{v}= Cdt[/tex]. Integrating, [tex]ln(v)= Ct+ D[/tex]. Taking the exponential, [tex]v(t)= e^{Ct+ D}= D'e^{Ct}[/tex] where [tex]D'= e^D[/tex].

And [tex]\frac{d^2u}{dr^2}= (CS/T)u[/tex]. That is a second order, linear equation, with constant coefficients with characteristic equation [tex]r^2= (CS/T)[/tex] with characteristic roots [tex]r= \sqrt{CS/T}[/tex] and [tex]r= -\sqrt{CS/T}[/tex]. The general solution to the differential equation is [tex]u(r)= Ae^{r\sqrt{CS/T}}+ Be^{-r\SQRT{CS/T}}[/tex].

Putting those together, [tex]h(r, t)= A'e^{Ct+ r\sqrt{CS/T}}+ B' e^{Ct- r\sqrt{CS/T}}[/tex].

C is undecided so the general solution is either the sum over all C if C is discrete, or the integral over all C if C is continuous. Which is correct is usually determined by the boundary conditions.