Late time Theis equation solution
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Please share your work/thoughts about this assignment.
What are S & T?
Could you assume;
h(r,t) = f(r) * g(t)
Hi there,
This is to solve the Theis equation in hydrogeology.
S in this case would be Storativity and T transmissivity of an aquifer.
I have unfortunately not touched calculus in years and wouldn't know whether you could assume that f(r)* g(t), but suppose you do how do I go about finding a solution to this problem?
I have not attempted it myself as I don't know where to begin. Any direction would be helpful.
This is to solve the Theis equation in hydrogeology.
S in this case would be Storativity and T transmissivity of an aquifer.
I have unfortunately not touched calculus in years and wouldn't know whether you could assume that f(r)* g(t), but suppose you do how do I go about finding a solution to this problem?
I have not attempted it myself as I don't know where to begin. Any direction would be helpful.
HallsofIvy
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What do you mean by "the" solution? Such an equation will have infinitely many solutions. In fact, I would expect to have three arbitrary functions, determined by two "boundary conditions" and one "initial condition".
As Subhotosh Kahn suggests, you can try "separation of variables": let h(r, t)= U(r)V(t), a product of a function of r only and a function of t only. Then the equation becomes \(\displaystyle U\frac{dV}{dt}= \frac{S}{T}V\frac{d^2U}{dr^2}\). Divide both sides by UV to get \(\displaystyle \frac{1}{V}\frac{dV}{dt}= \frac{S}{T}\frac{1}{U}\frac{d^2U}{dr^2}\).
Now, the left side is a function of t only while the right side is a function of r only. The only way they can be equal for all t and r is if each is a constant: \(\displaystyle \frac{1}{V}\frac{dV}{dt}= \frac{S}{T}\frac{1}{U}\frac{d^2U}{dr^2}= \alpha\) for \(\displaystyle \alpha\) any constant. That we can write as two separate equations, \(\displaystyle \frac{dV}{dt}= \alpha V\) and \(\displaystyle \frac{d^2U}{dr^2}= \frac{S}{T}\alpha U\).
Now, how we would solve those equations, indeed, what \(\displaystyle \alpha\) could be, depend strongly upon the boundary conditions on U and the initial condition on V.
As Subhotosh Kahn suggests, you can try "separation of variables": let h(r, t)= U(r)V(t), a product of a function of r only and a function of t only. Then the equation becomes \(\displaystyle U\frac{dV}{dt}= \frac{S}{T}V\frac{d^2U}{dr^2}\). Divide both sides by UV to get \(\displaystyle \frac{1}{V}\frac{dV}{dt}= \frac{S}{T}\frac{1}{U}\frac{d^2U}{dr^2}\).
Now, the left side is a function of t only while the right side is a function of r only. The only way they can be equal for all t and r is if each is a constant: \(\displaystyle \frac{1}{V}\frac{dV}{dt}= \frac{S}{T}\frac{1}{U}\frac{d^2U}{dr^2}= \alpha\) for \(\displaystyle \alpha\) any constant. That we can write as two separate equations, \(\displaystyle \frac{dV}{dt}= \alpha V\) and \(\displaystyle \frac{d^2U}{dr^2}= \frac{S}{T}\alpha U\).
Now, how we would solve those equations, indeed, what \(\displaystyle \alpha\) could be, depend strongly upon the boundary conditions on U and the initial condition on V.
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"I have not attempted it myself as I don't know where to begin. Any direction would be helpful."
Yeah, that's not really good enough. Why are you asking about this if you have NO CHANCE to find a solution? That's not really how learning works. That's more like "doomed to failure". Why not give us something - anything - so that we can have a clue where you are. So far, we're just throwing darts, and not at a close target.
...slightly edited for typo: "life" ==> "like".
Yeah, that's not really good enough. Why are you asking about this if you have NO CHANCE to find a solution? That's not really how learning works. That's more like "doomed to failure". Why not give us something - anything - so that we can have a clue where you are. So far, we're just throwing darts, and not at a close target.
...slightly edited for typo: "life" ==> "like".
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D
Deleted member 4993
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Please study:
Parabolic P.D.E.
These are common PDE s - in heat-transfer problems. If I remember correctly, there is no closed form solution. The solution takes the form a "series solution". Generally these are "approximated" through numerical methods
Parabolic P.D.E.
These are common PDE s - in heat-transfer problems. If I remember correctly, there is no closed form solution. The solution takes the form a "series solution". Generally these are "approximated" through numerical methods
HallsofIvy
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In what sense is this a "transformation"? What is being transformed?
D
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That is a "dangerous" change of variables in PDE. If you change \(\displaystyle \lambda\) - you can change both r and t.
D
Deleted member 4993
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If you are not given boundary and/or initial conditions, you cannot "solve" the problem.