Let $u(x,t)$ be a solution to the heat equation problem $u_{t}=u_{xx}$ in $[0,π]×[0,L]$ subject to $u(0,t)=u(π,t)=0$ on $ [0,L] $ and $ u(x,0)= \phi (x)$ on $[0,π]$. If $u(x,L)=f(x)$ then which of the following is true for a suitable kernel $k(x,y)$
- $\int_{0}^{π} k(x,y) \phi (y) dy = f(x)$ for $0 \leq x \leq π$
- $\phi (x)+ \int_{0}^{π} k(x,y) \phi (y) dy = f(x)$ for $0 \leq x \leq π$
- $\int_{0}^{x} k(x,y) \phi (y) dy = f(x)$ for $0 \leq x \leq π$
- $\phi (x) + \int_{0}^{x} k(x,y) \phi (y) dy = f(x)$ for $0 \leq x \leq π$.
My approach
Let $u(x,y)= X(x)T(t)$ then the pde is $\dfrac{X''}{X}= \dfrac{T'}{T}= \lambda$ with boundary conditions $X(0)=X(π)=0$ and $T(0)=c_{1}$ and $T(L)=c_{2}$.
Let $\lambda <0$ say $\lambda = -k, k>0$ $X(x)=a\cos(\sqrt{k} x) + b\sin(\sqrt{k }x)$ and $T=e^{\lambda t}$.
Using boundary conditions for $X$, $a=0$ and $b$ is arbitrary with $\sqrt{k} =n$ implies $\lambda = -n^{2}$ where $n \in \mathbb{Z}$.
Thus $u(x,t) = b \sin (nx) c_{1}c_{2} e^{-n^{2}t}$
Here the differential equations are boundary value problems which give fredholm integral equations upon integration therefore option $3$ and $4$ are not true. Can anyone help me with $1$ and $2$?
You're almost there.
The separable solutions are of the form $u(x,t) = b \sin(nx)e^{-n^2t}$. So the full solution is of the form $$u(x,t) = \sum_{n=1}^{\infty} b_n \sin(nx)e^{-n^2t}$$ Hence more specifically $\phi(x) = \sum_n b_n \sin(nx)$ and $f(x) = \sum_n b_n \sin(nx)e^{-n^2L}$.
You need the lemma $$\int_0^\pi \sin(ny)\sin(my) dy = \delta_{nm} \frac \pi 2$$ which is true for all $n,m \in \mathbb Z_{>0}$.
From there: $$b_n = \int_0^\pi \phi(y) \sin(ny) dy \cdot \frac 2\pi$$ $$f(x) = \sum_n \left(\int_0^\pi \phi(y) \sin(ny) dy \cdot \frac 2\pi\right) \sin(nx)e^{-n^2L}$$ $$f(x) = \int_0^\pi \frac 2\pi \sum_n \sin(ny)\sin(nx)e^{-n^2L} \phi(y) dy$$
So for (1.) $k_1(x,y) = \frac 2\pi \sum_n \sin(ny)\sin(nx)e^{-n^2L}$.
You need to check for convergence, but the exponential decay makes sure everything's fine there.
For (2.) you could take $k_2(x,y) = k_1(x,y) - \delta(x-y)$, provided that counts as a proper kernel. Otherwise I don't think there's a solution.