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
Best Answer
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.