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According to Debnath and Bhatta 2015 , the solution of heat diffusion/conduction equation \ref{a} on a half line with Neumann boundary condition can be found by using Fourier Cosine Transform method. $$ \frac{\partial h}{\partial t} =D\frac{{\partial}^2 h}{\partial {x}^2} \label{a}\tag{1} $$ $D$ represents material property. The initial condition \begin{align} \begin{split} h(x,0) &=0 \end{split} \end{align} and boundary conditions \begin{align} \begin{split} h(\infty,t) &=0 \end{split} \end{align} B.C at left side $$ \frac{dh(0,t)}{dx} = f(t) \label{b}\tag{2} $$ The solution is $$ h(x,t) =-\sqrt[]{\frac{D}{\pi}} \int_0^t \frac{f( \tau )}{t-\tau}\exp\{- \frac{x^2}{4D(t-\tau)}\}d\tau \label{e}\tag{3} $$ Now lets consider the following situation where I want to include the low conductive area on left side of this 1D coil,

Conceptual Model

To include this low conductive area, I have modified the equation \ref{b} to $$ \frac{dh(0,t)}{dx} =\frac{1}{a} \left[ g(t)-h(0,t) \right] \label{d}\tag{4} $$

Where 'a' is responsible for the 'extra resistance' from the low conductive layer.$h(0,t)$ represents input which will be a time series and let's suppose it has a shape of sine wave. $g(t)$ represents the damped sine wave at the end of low conductive layer.

With this slight modification in B.C we can write solution \ref{e} as $$ h(x,t) =-\sqrt[]{\frac{D}{\pi}} \int_0^t \frac{\frac{1}{a}g(\tau)-h(0,\tau)}{t-\tau}\exp\{- \frac{x^2}{4D(t-\tau)}\}d\tau \label{c}\tag{5} $$

By increasing the value of $a$, the calculated signal $h(x,t)$ at 'external observation point' should get damped and by reducing its value the signal $h(x,t)$ should increase in amplitude.

Problem

The problem with the above solution is that when I model the above scenario by writing code for equation \eqref{c}, the values of observed signal $h(x,t)$ -by reducing value of $a$ - get higher and higher and even go higher than the input signal $h(0,t)$ which is incongruous with the physical observation.

Extra rise in observed signal (h(x,t)) by reducing value of $a$

Question:

Is the conceptualization of low conductive layer as done in equation \eqref{d} is correct? Definitely not, because by making its value equal to $0$, results in division by zero, while it should mean that there is no 'low conductive area' in our model. How can this B.C be made realistic?

How can I include (model) more parameters of low conductive layer, at least its length by making changes in the equation \eqref{d}.

Using Modified Fourier Transform:

An alternative approach would be to use modified fourier transform as defined by Churchil_1972 $$ \Gamma[f(x)] = \int_0^{\infty}f(x)[\alpha \cos (\alpha x)+h\sin (\alpha x)]dx $$

The operational property of this transform when applied to second derivative of space is $$ \Gamma [\frac{d^2 f}{dx^2}] = - \alpha^2 f(\alpha) - \alpha [\frac{df}{dx}\mid_{x=0}-hf\mid_{x=0}]\label{f}\tag{6} $$ Equation \ref{f} can be made use of by changing the above conceptual model a little to following New Conceptual Model We can note that now, b.c \ref{d} can be used by operational property of modified fourier transform \ref{f}.

Reference

Debnath, Lokenath; Bhatta, Dambaru, Integral transforms and their applications, Boca Raton, FL: Chapman & Hall/CRC (ISBN 1-58488-575-0). 104 p. (2006). ZBL1113.44001.

R.V. Churchill, Operational Mathematics, Chap. 13, Fourier Transforms on the Half Line, (third edition). (1972).

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    $\begingroup$ Condition 4 is no longer a Neumann boundary condition, instead it is a Robin boundary condition. It seems to me that the cosine transform won't suffice. $\endgroup$ – Gribouillis Nov 2 '17 at 8:58
  • $\begingroup$ @Gribouillis Yes you are right. The question would be, does this change invalidates the solution in equation 5 because to arrive at equation 5, I just followd the procedure which is used by Debnath and Bhatta to arrive at equation 3! Can you describe elaborate that why 'cosine transform' will not suffice? $\endgroup$ – Ather Cheema Nov 2 '17 at 10:00

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