- #1

erobz

Gold Member

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Would method of separation of variables lead to a solution to the following PDE?

$$ \frac{1}{r} \frac{ \partial}{\partial r} \left( kr \frac{ \partial T}{ \partial r}\right) = \rho c_p \frac{\partial T }{ \partial t }$$

This would be for the transient conduction of a hollow cylinder, of wall thickness ##r##, and thermal conductivity ##k##, with temperature distribution ##T(r,t)##

How do we tell if it will work out?

I imagine expanding like:

$$ \frac{k}{r} \left( \frac{ \partial T}{\partial r} + r \frac{ \partial^2 T}{\partial r^2} \right) = \rho c_p \frac{ \partial T}{\partial t}$$

Looking for a solution:

$$T(r,t) = f(r)g(t)$$

That would yield the following equation:

$$ \frac{1}{f(r)} \frac{k}{r} \left( f'(r) + r f''(r) \right) = \lambda= \rho c_p \frac{g'(t)}{g(t)} $$

You would get a damped oscillator for ##f(r)##, and exponential function for ##g(t)## from.

$$f''(r) + \frac{1}{r} f'(r) - \frac{ \lambda}{k} f(r) = 0 \tag{2}$$

$$ g'(t) - \frac{ \lambda}{\rho c_p}g(t) = 0 \tag{3} $$

When would I find out whether or not "it makes sense or works"

If at all possible, keep this as low level explanation. I'm on paper thin ice in any type of foundational understanding of PDE's ( or perhaps I have already unknowingly fallen through?)

$$ \frac{1}{r} \frac{ \partial}{\partial r} \left( kr \frac{ \partial T}{ \partial r}\right) = \rho c_p \frac{\partial T }{ \partial t }$$

This would be for the transient conduction of a hollow cylinder, of wall thickness ##r##, and thermal conductivity ##k##, with temperature distribution ##T(r,t)##

How do we tell if it will work out?

I imagine expanding like:

$$ \frac{k}{r} \left( \frac{ \partial T}{\partial r} + r \frac{ \partial^2 T}{\partial r^2} \right) = \rho c_p \frac{ \partial T}{\partial t}$$

Looking for a solution:

$$T(r,t) = f(r)g(t)$$

That would yield the following equation:

$$ \frac{1}{f(r)} \frac{k}{r} \left( f'(r) + r f''(r) \right) = \lambda= \rho c_p \frac{g'(t)}{g(t)} $$

You would get a damped oscillator for ##f(r)##, and exponential function for ##g(t)## from.

$$f''(r) + \frac{1}{r} f'(r) - \frac{ \lambda}{k} f(r) = 0 \tag{2}$$

$$ g'(t) - \frac{ \lambda}{\rho c_p}g(t) = 0 \tag{3} $$

When would I find out whether or not "it makes sense or works"

If at all possible, keep this as low level explanation. I'm on paper thin ice in any type of foundational understanding of PDE's ( or perhaps I have already unknowingly fallen through?)

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