How to find the right combination of general solutions to fulfill nonhomogeneous boundary conditions when solving heat equation in spherical polars

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This question below is a deeper dive to the problem described here, but it should stand alone on its own as well.


I am trying to solve $$\dot{u} = \alpha \,\Delta u,$$ where$$u=u(r,t) \text{ and } R \geq r \geq 0$$ in spherical polar coordinates with initial contition $$u(r \leq R,t=0) = f(r)$$ & boundary condition $$u(r=R,t>0) = T_f.$$

I planned to use separation of variables: $u=T(t)\rho(r)$. After some calculations (which is described in this answer), I arrive at:

$$T(t)=Ae^{\lambda \alpha t}$$

and

$$\rho(r) = \frac{c_0}{\sqrt {\lambda} r} \sinh(\sqrt\lambda r).$$

So, a general $u$ can be written in this form:

$$\frac{c_0}{\sqrt {\lambda} r} \sinh(\sqrt\lambda r)e^{\lambda \alpha t}$$

My understanding is that I can make $c_0$ to be a function of $\lambda$ & add these different general solutions up to fulfill the IC & BC.

If $\lambda$ could take discrete values only, I would try to fulfill the IC (where $t=0$, so the $e^{\lambda \alpha t}$ factor is just $1$) by writing $u(r,t=0)$ this way:

$$\sum_{n=0}^{\infty}\frac{c_0(\lambda_n)}{\sqrt {\lambda_n} r} \sinh(\sqrt{\lambda_n} r) = f(r)$$

Since I see no reason why $\lambda$ would need to take discrete values only, to fulfill the IC, I think I need to have:

$$\int_{-\infty}^{\infty}\frac{c_0(\lambda)}{\sqrt {\lambda} r} \sinh(\sqrt\lambda r) \,d\lambda = f(r)$$

Using this, $c_0(\lambda)$ could be determined. After finding $c_0(\lambda)$, my solution for $u(r,t)$ would be:

$$u(r,t) = \int_{-\infty}^\infty \frac{c_0(\lambda)}{\sqrt {\lambda} r} \sinh(\sqrt\lambda r) e^{\lambda \alpha t} \,d\lambda$$

Question

Is this a right way to tackle the problem, or am I misunderstanding something? If this is the right way, I don't see how can I calculate $c_0(\lambda)$ witout a computer, but that just seems an ugly way of doing it, so I am hoping for a better solution.

But then... How do I make sure that the BC is satisfied as well? For simplicity, lets say $f(R)=T_f$, but I still fail to see how that will remain satisfied for $t>0$. I might be doing something seriously wrong.

I have some memories that Green's functions might have to do something with this, but I don't know how I can use them here.