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 $f_0(y)$


  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 $\sum_{k=0}^{\infty}$ ?


1 Answer 1


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 $|f_k(y)| \leq C_y k^t(\ln{k})^{-1-\epsilon}$, $\epsilon > 0$, $t \leq 1$).

  • $\begingroup$ 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. $\endgroup$
    – Avrana
    Commented Nov 21, 2019 at 13:48
  • 1
    $\begingroup$ 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 $T$. $\endgroup$
    – Aphelli
    Commented Nov 21, 2019 at 14:11
  • $\begingroup$ 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 ? $\endgroup$
    – Avrana
    Commented Nov 21, 2019 at 16:04
  • $\begingroup$ 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. $\endgroup$
    – Aphelli
    Commented Nov 22, 2019 at 7:39

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