I have arrived at the following partio-integral Differential equation while solving the coupled heat transfer between a solid which is heated from the bottom over which a liquid flows. The solid temperature is denoted by ${\tilde{T}(x,y)}$ while the fluid temperature is $T_f$ for the domain, $x\in[0,L]$ and $y\in[0,d]$. There is no fluid temperature in the equation I write below because it has been incorporated in the equation (integral term) by expressing it in the form of ${\tilde{T}(x,y)}$.

$$\bigg(\frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2}\bigg)\tilde{T}(x,y) + \beta \alpha e^{-\alpha x}\int e^{\alpha x} \tilde{T}(x,y)\mathrm{d}x - \beta \tilde{T}(x,y) = 0$$

The boundary conditions are:

$$\frac{\partial \tilde{T}(0,y) }{\partial x} = \frac{\partial \tilde{T}(L,y) }{\partial x} = \frac{\partial \tilde{T}(x,d) }{\partial y} = 0$$

$$\frac{\partial \tilde{T}(x,0) }{\partial y} = \gamma$$

Here $\alpha,\beta$ and $\gamma$ are constants. This is a Partio-Integral Differential equation and I have no experience whatsoever in handling such problems. Any help or guidance is appreciated.

Attempt Using some known constants like the fluid inlet temperature $T_{fi}$, the PIDE can be written as:

$$\bigg(\frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2}\bigg)\tilde{T}(x,y) + \beta \alpha e^{-\alpha x}\bigg[\int_0^x e^{\alpha s} \tilde{T}(s,y)\mathrm{d}s + \frac{T_{fi}}{\alpha}\bigg] - \beta \tilde{T}(x,y) = 0$$

UPDATE $$f_n''(y)=\Big((\tfrac{n\pi}{L})^2+\beta\Big)f_n(y)$$

Solving this gives:

$$ f_k(y)=C_1 e^{\sigma y} + C_2 e^{-\sigma y} $$ where $$ \sigma = \sqrt{(\frac{n\pi}{L})^2 + \beta} $$ Utilising the homogeneous condition along $y$ i.e. at $y=d$ gives

$$ C_1 e^{\sigma d} - C_2 e^{-\sigma d} = 0 \\ \Rightarrow C_2=C_1 e^{2\sigma d} $$ The problem comes when I try to apply the non-homogeneous condition at $(x,y=0)$

$$ \sum_{k=0} \sigma(C_1 e^{\sigma y} - C_2 e^{-\sigma y})\cos(\frac{k\pi x}{L})=\gamma \\ @y=0 $$ But this approach just cancels out all the constants. If I multiply both sides of the above equation with $\cos(\frac{k\pi x}{L})$ and integrate over the $x$ domain from $0$ to $L$, the RHS would just vanish.

  • $\begingroup$ Could you please change the integration variable to something thats not $x$ and add the lower and upper limit? $\endgroup$
    – maxmilgram
    Nov 11, 2019 at 10:11
  • $\begingroup$ @maxmilgram Thanks for the suggestion. I have edited my original question by adding an Attempt section where I have incorporated your suggestion. Have a look. $\endgroup$
    – Avrana
    Nov 11, 2019 at 10:52

1 Answer 1


Using the following Ansatz: $$ \tilde{T}(x,y)=\sum_{k=0}^{\infty}f_k(y)\cos(\tfrac{k\pi x}{L}) $$ (which incorporates the boundary conditions for $x$) leads to $$ 0=\sum_{k=0}^\infty\Big(f_k''(y)-(\tfrac{k\pi }{L})^2f_k(y)-\beta f_k(y)\Big)\cos(\tfrac{k\pi x}{L})+e^{-\alpha x}\Big(\alpha\int_0^x\sum_{k=0}^{\infty}f_k(y)\cos(\tfrac{k\pi s}{L})e^{\alpha s}ds+T_{fi}\Big) $$ Now calculating the integral yields $$ e^{-\alpha x}\Big(\alpha\int_0^x\sum_{k=0}^{\infty}f_k(y)\cos(\tfrac{k\pi s}{L})e^{\alpha s}ds+T_{fi}\Big)=T_{fi}e^{-\alpha x}+\sum_{k=0}^\infty\frac{-\alpha^2 L^2 e^{-\alpha x}f_k(y)+\pi \alpha k L \sin(\tfrac{k\pi x}{L})f_k(y)+\alpha^2 L^2 \cos(\tfrac{k\pi x}{L})f_k(y)}{\pi^2 k^2 + \alpha^2 L^2}$$ Now if we multiply everything by $\sin(\tfrac{k\pi x}{L})$ resp. $\cos(\tfrac{k\pi x}{L})$ and integrate from $0$ to $L$ we achieve ODEs for the $f_k(y)$. Let me know if you need further help!

Next steps: If we multiply with $\sin(\tfrac{n\pi x}{L})$ and integrate over the $x$-domain, the first sum is zero due to $\int_0^L\sin(\tfrac{n\pi x}{L})\cos(\tfrac{k\pi x}{L})dx=0$ and we end up with $$ 0=L\pi n\frac{1-(-1)^ne^{-\alpha L}}{\pi^2 n^2 + \alpha^2 L^2}\Big(T_{fi}-\sum_{k=0}^\infty\frac{\alpha^2 L^2 f_k(y)}{\pi^2 k^2 + \alpha^2 L^2}\Big)+\frac{L}{2}\frac{\pi\alpha n Lf_n(y)}{\pi^2n^2+\alpha^2L^2}\\ \Rightarrow0=\Big(1-(-1)^ne^{-\alpha L}\Big)\Big(T_{fi}-\sum_{k=0}^\infty\frac{\alpha^2 L^2 f_k(y)}{\pi^2 k^2 + \alpha^2 L^2}\Big)+\frac{L\alpha f_n(y)}{2} $$ And the same for cosine: $$ \frac{L}{2}(f_n''(y)-(\tfrac{n\pi}{L})^2f_n(y)-\beta f_n(y))+\alpha L^2\frac{1-(-1)^ne^{-\alpha L}}{\pi^2 n^2 + \alpha^2 L^2}\Big(T_{fi}-\sum_{k=0}^\infty\frac{\alpha^2 L^2 f_k(y)}{\pi^2 k^2 + \alpha^2 L^2}\Big)+\frac{L}{2}\frac{\alpha^2 L^2 f_n(y)}{\pi^2 n^2 + \alpha^2 L^2} $$ Combining the equations to get rid of the nonlocal term: $$ 0=\frac{L}{2}(f_n''(y)-(\tfrac{n\pi}{L})^2f_n(y)-\beta f_n(y))-\frac{\alpha L^2}{2}\frac{L\alpha f_n(y)}{\pi^2 n^2 + \alpha^2 L^2}+\frac{L}{2}\frac{\alpha^2 L^2 f_n(y)}{\pi^2 n^2 + \alpha^2 L^2}\\ \Rightarrow f_n''(y)=\Big((\tfrac{n\pi}{L})^2+\beta\Big)f_n(y) $$ Now, to be completely honest, I have not understood the magic that is happening here. Most likely it is a mistake in the calculation but I havent found one.

  • $\begingroup$ I followed up on all of the steps. So subsequent from here I should multiply the equation with $\sin(\frac{k\pi x}{L})$ and $\cos(\frac{k\pi x}{L})$ to use their orthogonality ? What I confuse here is what are the coefficients I would be determining ? Could you elaborate a bit more? Really appreciate your help here. $\endgroup$
    – Avrana
    Nov 11, 2019 at 16:07
  • $\begingroup$ I tried out your steps and have updated the attempt section with it. Am I doing this allright ? $\endgroup$
    – Avrana
    Nov 12, 2019 at 4:52
  • $\begingroup$ Hi again. I re-did all the steps and reached the exact same equation for $f(y)$ as you did. The problem now comes from the boundary conditions. The one at $x,y=0$. I tried to use the orthogonality property for the non-homogeneous condition. I have added my attempt. Can you have a look and suggest some workaround ? I really appreciate the attention you have spent on this single problem. $\endgroup$
    – Avrana
    Nov 12, 2019 at 13:12
  • $\begingroup$ If you could comment If there is something wrong with the boundary conditions, let me know. $\endgroup$
    – Avrana
    Nov 13, 2019 at 11:35
  • $\begingroup$ I have been with this problem for the last few days and I find that the expression $ \int_0^L \cos(\frac{k\pi x}{L})\sin(\frac{n\pi x}{L})\mathrm{d}x=0 $ is not always true for the half period $x\in[0,L]$ while it is true for $x\in[0,2L]$. I have posted my recent attempt in math.stackexchange.com/questions/3435036/… question. Have a look. $\endgroup$
    – Avrana
    Nov 17, 2019 at 6:04

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .