I have a problem on how to obtain the derivative of an integral representation which includes the fundamental solution of the heat equation.
The equation of interest is given as:
$ f(x,t)=\sqrt{\alpha_1}\int_0^t T_y^1(g(s),s) \Bigl[K(g(s),\alpha_1 t, x, \alpha_1 s)+K(g(s),\alpha_1 t, -x, \alpha_1 s)\Bigr]ds$, where the fundamental solution of the heat equation is used as:
$ K(x,t,y,s)=\frac{1}{2\sqrt{\pi(t-s)}}e^{-\frac{(x-y)^2}{4(t-s)}}$
From the exponent $e^{-\frac{(x-y)^2}{4(t-s)}}$ it can be seen that the fundamental solution is invariant for $K(x,t,y,s)=K(-x,t,-y,s)=K(y,t,x,s).$
Thus the function of interest can be rewritten as:
$ f(x,t)=\sqrt{\alpha_1}\int_0^t T_y^1(g(s),s) \Bigl[K(x,\alpha_1 t, g(s), \alpha_1 s)+K(-x,\alpha_1 t,g(s), \alpha_1 s)\Bigr]ds$
I am now interested in the spatial derivative of the function of interest $\frac{\partial f(x,t)}{\partial x}(x=g(t))$ evaluated at the boundary $x=g(t)$.
Here, however I struggle due to the fact, that I cannot simply differentiate the right-hand-side in respect of $x$ at that boundary as stated in the 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., which reads as:
$\lim\limits_{x \rightarrow g(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=g(t)) d \tau $
Thus, the derivative of $f(x,t)$ can only be obtained by using the Lemma above.
However, it is only stated how I would build the derivative of the Term including $K(x,t,y,s)$, no information about the differentiation of $K(-x,t,y,s)$ is given. A potential idea to apply the Lemma on the problem at hand reads as:
$ \frac{\partial f(x,t)}{\partial x}(x=g(t))=\lim\limits_{x \rightarrow s(t)-0}{\int_0^t T_y^1(g(\tau))\Bigl[\frac{\partial}{\partial x}(K(x,t,s(\tau),\tau)+\frac{\partial}{\partial x}\Bigl[K(-x,t,s(\tau),\tau))\Bigr]\Bigr]d \tau}$
In the following the chail-rule for differentiation is being used as:
$\frac{\partial}{\partial x}\Bigl[K(-x,t,s(\tau),\tau))\Bigr]\Bigr]=-\frac{\partial}{\partial x}K(-x,t,s(\tau),\tau))\Bigr]$, leading to:
$\frac{\partial f(x,t)}{\partial x}(x=g(t))=\lim\limits_{x \rightarrow s(t)-0}{\int_0^t T_y^1(g(\tau))\Bigl[\frac{\partial}{\partial x}(K(x,t,s(\tau),\tau)-\frac{\partial}{\partial x}K(-x,t,s(\tau),\tau))\Bigr]d \tau}$
Now applying the Lemma above would lead to:
$\frac{\partial f(x,t)}{\partial x}(x=g(t))=\alpha_1 \Bigl[\frac{1}{2}T_y^1(g(t),t)+\int_0^t T_y^1(g(s),s) \frac{\partial K(x,t,s(\tau),\tau) }{\partial x}(x=g(t)) d \tau\Bigr]\\ -\alpha_1 \Bigl[\frac{1}{2}T_y^1(g(t),t)+\int_0^t T_y^1(g(s),s) \frac{\partial K(x,t,s(\tau),\tau) }{\partial x}(x=-g(t))d \tau \Bigr]$
$\rightarrow \frac{\partial f(x,t)}{\partial x}(x=g(t))=\alpha_1 \Bigl[\int_0^t T_y^1(g(s),s) \frac{\partial K(x,t,s(\tau),\tau) }{\partial x}(x=g(t)) d \tau -\int_0^t T_y^1(g(s),s) \frac{\partial K(x,t,s(\tau),\tau) }{\partial x}(x=-g(t)) d \tau\Bigr]$
However I am not sure if I am allowed to apply the Lemma on the fundamental solution $K(-x,t,y,s)$ in the same manner I am using on $K(-x,t,y,s)$ at the boundary $x=g(t)$, especially due to the limes definition, when I change the sign of $x$ to $-x$, I guess I should also change the limes? Is the Lemma then still applicable?
I would highly appreciate any advice on how to obtain derivatives of integral representations including the shown Lemma. Any advice on literature on derivatives of integral representation at a boundary could help me a lot.
Many thanks and best regards.