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In the formulation of the partial differential heat equation

$$\frac{\partial \theta}{\partial t}=\frac{\partial^2 \theta}{\partial x^2}, \hspace{1 cm}0\le x \le D$$

there is an incompatibility between the initial condition

$$t=0:\hspace{1 cm} \theta=0 \hspace{1 cm} 0\le x \le D$$

and the boundary condition at $x=0$

$$x=0:\hspace{1 cm} \theta=1 \hspace{1 cm} t\ge 0$$

that renders the analytical Fourier Series solution invalid at very small times from the initial condition. This incompatibility can be resolved by posing the initial condition as

$$t=0:\hspace{1 cm} \theta=H(-x)$$

where $H(x)$ is the Unit Heaviside Step Function. Plugging in this initial condition into the Fourier Series solution and skipping details, I end up having to evaluate the following integral

$$\int_0^D H(-x)\cos\lambda x dx$$

which following integration by parts and noting that the derivative of the Heaviside step Function is the Dirac-Delta function, reduces to

$$\frac{1}{\lambda}\int_0^D \delta(x)\sin\lambda x dx$$

and this is where I'm stuck. I do know that

$$\int_{-\infty}^\infty \delta(x)f(x) dx=f(0)$$

but I don't think that this property would be preserved if the integral is broken up into sums of integrals from $-\infty$ to $0_-$, $0_+$ to $D_-$ and $D_+$ to $\infty$.

Any suggestions on how to proceed would be deeply appreciated. Thanks

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  • $\begingroup$ well, $H(-x)=0$ for $x>0$, so the original integral is zero, assuming that it is an integral of Lebesgue $\endgroup$ – Masacroso Apr 20 '19 at 20:15
  • $\begingroup$ Why care about the corners? $\endgroup$ – md2perpe Apr 20 '19 at 21:29
  • $\begingroup$ Well, because the very short term solutions are bogus and it takes a while for the correct transient profile to evolve. $\endgroup$ – Sharat V Chandrasekhar Apr 21 '19 at 13:32
  • $\begingroup$ @Masacroso, it's actually posed above as a Riemann integral, but with a redefinition of the measure, it could be cast as a Lebesque integral. $\endgroup$ – Sharat V Chandrasekhar Apr 21 '19 at 13:39
  • $\begingroup$ @SharatVChandrasekhar ok, but the answer is the same, because $H(-x)=0$ for $x>0$, so the integral is zero. You are mixing things here, note that the Dirac delta cannot be stated inside a integral of Riemann because it is not a function, at most a measure or a distribution $\endgroup$ – Masacroso Apr 21 '19 at 13:41
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Yes it is fine to say that $\frac{1}{\lambda}\int_0^D \delta(x)\sin(\lambda x) = \sin(0) = 0$.

You can think of the dirac delta as just being an infinite mass on the point $0$. Any other parts of the integral become irrelevant.

Also, you have to choose which condition to use at $t=0, x=0$ surely.

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  • $\begingroup$ WELL! Then this is not good news, because in this case, my supposedly carefully crafted attempt at resolving the incompatibility bombs and the resulting solution degenerates to the classic solution with the incompatibility. Thanks for the clarification nonetheless. $\endgroup$ – Sharat V Chandrasekhar Apr 20 '19 at 20:07
  • $\begingroup$ Haha well if anyone asks it wasn't me $\endgroup$ – fGDu94 Apr 20 '19 at 20:16
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I think that I may be trying to square the circle here (almost literally!). The initial condition is a Heaviside Step Function. Therefore, any attempt to represent this as a finite term Fourier Series is bound to result in errors. What I have noticed however is that if in reality, the boundary condition is actually a function of time such that the incompatibility is eliminated, then I get the evolution of a nice smooth solution with the use of Duhamel's Principle.

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