1
$\begingroup$

I am interested in taking the differentiation of an integral representation containing the fundamental solution of the heat equation, hence the Greens function.

The equation of interest I want to differentiate is given as:

$f(x)=\int_0^t \rho(\tau) K(-x,t,y,\tau)d \tau,$

with $K(-x,t,y,\tau)=\frac{1}{2\sqrt{\pi(t-\tau)}}e^{-\frac{(-x-y)^2}{4(t-\tau)}}=\frac{1}{2\sqrt{\pi(t-\tau)}}e^{-\frac{(x+y)^2}{4(t-\tau)}}.$

I want to evaluate the spatial derivative of this equation at the boundary $x=s(t)$, hence I am interested in $\frac{\partial f(x)}{\partial x}(x=s(t))$.

I know of a Lemma [(2.1) on page 501 of [Fr 1959] A. Friedman, Free boundary problems for parabolic equations I. Melting of solids. J. Math. Mech. 8 (1959), 499-517.] for evaluating derivatives of integral representations containing $K(x,t,y,\tau)=\frac{1}{2\sqrt{\pi(t-\tau)}}e^{-\frac{(x-y)^2}{4(t-\tau)}}$, which states that I cannot simply differentiate the right-hand-side in respect of $x$ at that boundary.

The Lemma reads as:

$\lim\limits_{x \rightarrow s(t)-0}\int_0^t {\frac{\partial}{\partial x}\rho(\tau)K(x,t,s(\tau),\tau)d \tau} = \frac{1}{2} \rho(t)+\int_0^t \rho(\tau)\Bigr[\frac{\partial}{\partial x}K(x,t,s(\tau),\tau) \Bigr](x=s(t)) d \tau $

With this knowledge how do I however differentiate

$\int_0^t {\frac{\partial}{\partial x}\rho(\tau)K(-x,t,y,\tau)d \tau}$ and evaluate it at $x=s(t)$?

Are there similar rules for terms containing $K(-x,t,y,\tau)$ as for those with $K(x,t,y,\tau)$?

I highly appreciate any help!

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.