# Solid-Fluid coupled Thermal Boundary Value Problem [**UPDATED**]

I have been trying to solve a coupled solid-fluid heat transfer problem. I took help from the Math Stack community in the linked question Partio-Integral Differential Equation for a Heat Sink.

I write the basic equations describing the case followed by my attempt

$$\alpha,\beta,\gamma$$ are constants $$\underbrace{\frac{\partial T_f}{\partial x} + \alpha (T_f - T(x,y))=0}_{FLUID} \Rightarrow T_f=e^{-\alpha x}\int e^{\alpha x} T \mathrm{d}x \\ \Rightarrow T_f=\alpha e^{-\alpha x} \Bigg[\int_0^x e^{\alpha s}T(s,y)\mathrm{d}s+\frac{T_{fi}}{\alpha}\Bigg] \tag 1$$ $$T_f(x=0)=T_{fi}$$ is a known quantity. $$\underbrace{\Bigg(\frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2}\Bigg)T-\beta(T-T_f)=0}_{SOLID} \tag 2$$ Substituting from (1) in (2): $$\nabla^2 T - \beta T + \beta\Bigg[\alpha e^{-\alpha x} \Bigg(\int_0^x e^{\alpha s}T(s,y)\mathrm{d}s+\frac{T_{fi}}{\alpha}\Bigg)\Bigg]=0 \tag 3$$ (3) is dictated by the following boundary conditions: $$\frac{\partial T}{\partial x} \vert_{x=0} = \frac{\partial T}{\partial x} \vert_{x=L} = \frac{\partial T}{\partial y} \vert_{y=d} = 0 , \frac{\partial T}{\partial y} \vert_{y=0}=\gamma$$

Attempt Using the ansatz: $$T(x,y)=\sum_{k=0}^{\infty}f_k(y)\cos(\frac{k\pi x}{L})=f_0(y)+\sum_{k=1}^{\infty}f_k(y)\cos(\frac{k\pi x}{L})$$ The final expression after substituting the ansatz in $$(3)$$ is: $$f'_0(y)+\sum_{k=1}^{\infty}\Bigg(f''_k(y)-f_k(y)(\frac{k\pi}{L})^2-\beta f_k(y)\Bigg)\cos(\frac{k\pi x}{L})+\\ \beta e^{-\alpha x}(T_{fi}-f_0(y))+\\ \sum_{k=1}^{\infty}\frac{(\alpha L)\beta f_k(y)}{(\alpha L)^2 + (k\pi)^2}\Bigg[(\alpha L) \cos(\frac{k\pi x}{L})-(\alpha L)e^{-\alpha x}+(k\pi)\sin(\frac{k\pi x}{L})\Bigg]=0 \tag 4$$

Multiplying $$(4)$$ with $$\sin(\tfrac{n\pi x}{L})$$ and integrating over the $$x$$-domain $$f'_0(y)\frac{L}{n\pi}(1-\cos(n\pi))+\sum_{k=1}^{\infty}\Bigg[\Bigg(f''_k(y)-f_k(y)(\frac{k\pi}{L})^2-\beta f_k(y)\Bigg)+\frac{(\alpha L)^2\beta f_k(y)}{(\alpha L)^2 + (k\pi)^2}\Bigg]\color{red}{I_1}+\\ \beta(T_{fi}-f_0(y))\frac{L(n\pi)}{(\alpha L)^2 + (n\pi)^2}(1-e^{-\alpha L}\cos(n\pi))+\frac{(n\pi)(\alpha L^2)\beta f_n(y)}{2((\alpha L)^2 + (n\pi)^2)}- \\ \sum_{k=1}^{\infty}\frac{(\alpha L)^2\beta f_k(y)}{(\alpha L)^2 + (k\pi)^2}\Bigg(\frac{(n\pi)L}{(\alpha L)^2 + (n\pi)^2}(1-e^{-\alpha L}\cos(n\pi))\Bigg)=0 \tag A$$

Multiplying $$(4)$$ with $$\cos(\tfrac{n\pi x}{L})$$ and integrating over the $$x$$-domain $$\Bigg(f''_k(y)-f_k(y)(\frac{k\pi}{L})^2-\beta f_k(y)\Bigg)\frac{L}{2}+\frac{(\alpha L)\beta f_n(y)}{(\alpha L)^2 + (n\pi)^2}\frac{L}{2}+\\+\beta(T_{fi}-f_0(y))\frac{\alpha L^2 }{(\alpha L)^2 + (n\pi)^2}(1-e^{-\alpha L}\cos(n\pi))+\\ \sum_{k=1}^{\infty}\frac{(\alpha L)(k\pi)\beta f_k(y)}{(\alpha L)^2 + (k\pi)^2}\color{blue}{I_2}-\\ \sum_{k=1}^{\infty}\frac{(\alpha L)^2 \beta f_k(y)}{(\alpha L)^2 + (k\pi)^2} \Bigg(\frac{\alpha L^2}{(\alpha L)^2 + (n\pi)^2}(1-e^{-\alpha L}\cos(n\pi))\Bigg)=0 \tag B$$

$$\color{red}{I_1=\int_0^L \cos(\frac{k\pi x}{L})\sin(\frac{n\pi x}{L})}$$ $$\color{blue}{I_2=\int_0^L \sin(\frac{k\pi x}{L})\cos(\frac{n\pi x}{L})}$$ I want to use $$A$$ and $$B$$ to find $$f_k(y)$$ and $$f_0(y)$$

Questions

1. What will be the integral $$I_1$$ and $$I_2$$? I know that it amounts to zer0 on the full period $$x\in[0,2L]$$. When I substitute the integral $$\color{red}{I_1}$$ in $$(A)$$ how is the summation going to behave? Can we say something about which terms will vanish and which would remain ?

2. Are $$\color{red}{I_1}$$ and $$\color{blue}{I_2}$$ identical under the $$\sum_{k=0}^{\infty}$$ ?

From $$\cos(b)\sin(a)=\frac{1}{2}(\sin(a+b)+\sin(a-b))$$, it follows that:

$$2I_1=\int_0^L{\sin{\frac{(n+k)\pi x}{L}}+\sin{\frac{(n-k)\pi x}{L}}}=\frac{L}{\pi}\left(\int_0^{\pi}{\sin((n+k)x)}+\int_0^{\pi}{\sin((n-k)x)}\right).$$

Therefore, by a variable change, $$I_1=0$$ if $$k+n$$ is even, and $$\frac{2\pi}{L}I_1=\frac{2}{n+k}+\frac{2}{n-k}=\frac{2n}{n^2-k^2}$$, ie $$I_1=\frac{Ln}{\pi(n^2-k^2)}$$ is $$n+k$$ is odd.

Let me repeat: the terms that vanish are the ones where $$k$$ and $$n$$ have same parity.

In other words, your sequence of $$I_1$$, when $$k$$ varies, is $$\ell^p$$ for exactly all $$p> 1$$, so the rest in (A) relies on decay assumptions about the sequence $$f_k$$, as well as the precise meaning you want to give to the summation (pointwise? Almost-everywhere? Locally uniformly? In $$L^2$$?).

If you want something locally uniform, you need locally uniform (in $$y$$) estimates $$|f_k’’(y)| \leq C_yk^t(\ln{k})^{-1-\epsilon}$$ for some $$\epsilon > 0$$, $$t \leq 1$$, and a locally normal convergence for $$\sum_{f_k(y)}$$ because of the term in $$f_k(y)k^2\pi^2/L^2 I_1$$.

It’s easy to see that $$I_2$$ and $$I_1$$ arw the same when you switch variables: $$I_2=0$$ if $$k+n$$ is even, and $$I_2=\frac{Lk}{\pi(k^2-n^2)}$$ else.

When you look at (B), the condition again for a locally normal convergence is that $$|f_k(y)|/k^2$$ be locally in $$y$$ uniformly integrable (so eg $$|f_k(y)| \leq C_y k^t(\ln{k})^{-1-\epsilon}$$, $$\epsilon > 0$$, $$t \leq 1$$).

• Thanks for this insight. I wanted to know further how should I handle the summation in $(A)$ and $(B)$. Usually, in such problems , we normally take the value of $n=k$ and remove the summation but here since the integrals $I_1$ and $I_2$ will be depending on $n+k$, how can I simplify my equations. This said, your answer obviously explains the questions I asked. Nov 21, 2019 at 13:48
• My suggestion is that your decomposition is sort of wasteful. You have a perfectly good $L^2$ basis of $[0,L]$ with the functions $\cos{k\pi x}{L}$ with even $k$. Of course, one stumbles on difficulties because of noncontinuous behavior at the border: so why not write a series with only even $k$ or $k=1$? This is “just as expressive” but it’s easier to write down and the coordinates are better defined from $T$. Nov 21, 2019 at 14:11
• Ok that is really interesting. But won't leaving the $k=0$ term affect the solution ? If its not too much to ask, could you add a few lines to your answer regarding what you are suggesting ? Nov 21, 2019 at 16:04
• I’ve been thinking about it and I think my previous comment was wrong. It would, of course, have been really nice to get an orthogonal basis (which was the goal) but my suggestion is wrong. Nov 22, 2019 at 7:39