Problem: Let $~u(x,t)~$ be a solution of the heat equation $~u_t=u_{xx}~$ in a rectangle $~[0,\pi]\times[0,T]~$ subject to the boundary conditions $~u(0,t)=u(\pi,t)=0,~~0\le t\le T~$ and the initial condition $~u(x,0)=\phi(x),~~0\le x\le \pi~.$ If $~f(x)=u(x,T),~$ then which of the following is true for a suitable kernel $~k(x,y)~?$ $$1.~~~\int_0^\pi k(x,y)~\phi(y)~dy=f(x),~0\le x\le \pi~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~$$ $$2.~~~\phi(x)+\int_0^\pi k(x,y)~\phi(y)~dy=f(x),~0\le x\le \pi~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~$$ $$3.~~~\int_0^x k(x,y)~\phi(y)~dy=f(x),~0\le x\le \pi~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~$$ $$4.~~~\phi(x)+\int_0^x k(x,y)~\phi(y)~dy=f(x),~0\le x\le \pi~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~$$
My approach: Solution of heat equation can be written in the form of integral by using Green's function.
As if $~G(x,y,t)~$ be the Green's function of given differential equation, then solution is $$u(x,t)=\int_0^\pi G(x,y,t)~\phi(y)~dy~.$$
At $~t=T~,$ $$u(x,T)=\int_0^\pi G(x,y,T)~\phi(y)~dy$$
$$\implies f(x)=\int_0^\pi K(x,y)~\phi(y)~dy$$where $~K(x,y)=G(x,y,T)~.$
Hence option $(1)$ is the only correct option.
Now my question is: Is my way to solve the problem correct ? Is there any other method to solve the problem ? Please verify the approach and let me know whether is it okay or not ?