Heat equation with non homogeneous BCs

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SUMMARY

The discussion centers on solving the heat equation with non-homogeneous boundary conditions using the variable substitution $$\theta(r,z) = T(r,z)-T_{\infty}$$. The resulting equation is $$\frac{1}{r}\frac{\partial }{\partial r}(r\frac{\partial \theta}{\partial r})+\frac{\partial^2 \theta}{\partial z^2}=0$$, with specific boundary conditions outlined. The user expresses difficulty in determining the function $$w(r,z)$$ that satisfies the boundary conditions and mentions the method of separation of variables as a potential solution approach. The conversation also clarifies a typo regarding the boundary condition at $$x=0$$ and discusses the applicability of Sturm-Liouville theory in this context.

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jackkk_gatz
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Homework Statement
$$\frac{1}{r}\frac{\partial }{\partial r}(r\frac{\partial T}{\partial r})+\frac{\partial^2T}{\partial z^2}=0$$

$$\left.-k\frac{\partial T}{\partial r}\right\rvert_{r=R}=h[T(R,z)-T_{\infty}]$$

$$\left.k\frac{\partial T}{\partial z}\right\rvert_{z=H}+h[T(r,H)-T_{\infty}]=q_s$$

$$\left.-k\frac{\partial T}{\partial z}\right\rvert_{z=0}=0$$

$$\left.-k\frac{\partial T}{\partial r}\right\rvert_{r=0}=0$$

$where \ T_0 \ ,T_{\infty} \ and \ q_s \ are \ constants$
Relevant Equations
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I did a change of variable $$\theta(r,z) = T(r,z)-T_{\infty}$$ which resulted in

$$\frac{1}{r}\frac{\partial }{\partial r}(r\frac{\partial \theta}{\partial r})+\frac{\partial^2 \theta}{\partial z^2}=0$$

$$\left.-k\frac{\partial \theta}{\partial r}\right\rvert_{r=R}=h\theta$$

$$\left.k\frac{\partial \theta}{\partial z}\right\rvert_{z=H}+h\theta=q_s$$

$$\left.-k\frac{\partial \theta}{\partial z}\right\rvert_{z=0}=0$$

$$\left.-k\frac{\partial \theta}{\partial r}\right\rvert_{r=0}=0$$

After that I proposed $$\theta(r,z)=v(r,z)+w(r,z)$$ where w(r,z) should must satisfy

$$\left.-k\frac{\partial w}{\partial r}\right\rvert_{r=R}=hw(R,z)$$

$$\left.k\frac{\partial w}{\partial z}\right\rvert_{z=H}+hw(r,H)=q_s$$

$$\left.-k\frac{\partial w}{\partial z}\right\rvert_{z=0}=0$$

$$\left.-k\frac{\partial w}{\partial r}\right\rvert_{r=0}=0$$

I already tried interpolation, doesn't work. I don't know how w(r,z) should look like in order to satisfy the above equations. Is there an easier method?
In short, everything I have been trying has failed and I don't know what to do anymore, I have looked for books on PDEs, all the ones I have found deal with very simple cases, which are of no use to me. I have almost no experience solving this kind of equations to know what to do or to guess how w might look like
 
Last edited:
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I just noticed I asked my question in the wrong section 💀
 
Have you tried obtaining the general solution by the method of separation of variables? It looks like it will work in this case but I have not solved the problem.

What is ##x=0## doing in the boundary condition ##~\left.-k\dfrac{\partial T}{\partial r}\right\rvert_{x=0}=0~##. Is it a typo?

I reported this thread and it should be moved to the Advanced Homework forum by a mentor at some point in time.
 
kuruman said:
Have you tried obtaining the general solution by the method of separation of variables? It looks like it will work in this case but I have not solved the problem.

What is ##x=0## doing in the boundary condition ##~\left.-k\dfrac{\partial T}{\partial r}\right\rvert_{x=0}=0~##. Is it a typo?

I reported this thread and it should be moved to the Advanced Homework forum by a mentor at some point in time.
Yes it was a typo, fixed it already. And yes I have tried to get the the general solution by the method of separation of variables, the thing is I know how to apply it but with homogeneous BC where I do some things with Sturm-Liouville. The thing is Sturm-Liouville only works with homogeneous BCs, I know a method to transform the non homogeneous BCs to homogeneous, which is the one I already wrote where I have to guess the form of w(r,z)

And thanks for helping to move my question to the right section
 
jackkk_gatz said:
I have tried to get the the general solution by the method of separation of variables,
Wolfram gives me Bessel functions.
 

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