Solution to $\dot u(x,t)=\int_0^{1-x} u(y,t)\,dy$ I would like to solve the integro-PDE
$$\dot u(x,t)=\int_0^{1-x} u(y,t)\,dy$$
on the domain $x\in[0,1]$, with the initial condition $u(x,0)=u_0(x)$. Differentiating in $x$, this is equivalent to
$$u_{xt}(x,t)=-u(1-x,t)$$
with the conditions $u(x,0)=u_0(x)$ and $\dot u(1,t)=0$. Using either of these forms, is there a way to arrive at an analytical solution?
 A: The functions
$$U_k(x):=\cos{(2k+1)\pi x\over 2}\qquad(k\geq0)$$
have the property that
$$\int_0^{1-x} U_k(y)\>dy\equiv{2(-1)^k\over (2k+1)\pi}U_k(x)\ ,$$
in other words, they are eigenfunctions of the integral operator appearing in your problem. It follows that the functions
$$g_k(x,t):=U_k(x)\exp\left({2(-1)^k\over (2k+1)\pi}\,t\right)\qquad(k\geq0)$$
are solutions of the given equation
$$u_t(x,t)=\int_0^{1-x}u(y,t)\>dy\ ,$$
and so is any convergent expansion $u(x,t):=\sum_{k=0}^\infty a_k\>g_k(t)$.
Now the $U_k(x)$ appear as "odd cosine terms"  in Fourier expansions of functions of period $4$. The  initial function $x\mapsto u_0(x)$ is defined only on $[0,1]$. Extend it to a function on $[0,2]$ which is odd with respect to the point $1$, and then to an even function on $[{-}2,2]$. Compute the Fourier expansion of this extended function. It will only contain odd cosine terms, and no sine terms. In other words, it is possible to write
$$u_0(x)=\sum_{k=0}^\infty a_k\>U_k(x)\qquad(0<x<1)$$
with suitable $a_k\in{\mathbb R}$. Therewith we have obtained a complete "analytic" solution of the given problem:
$$u(x,t)=\sum_{k=0}^\infty a_k\>g_k(x,t)\ .$$
