I have come across the following two integral equations: \begin{align} a&=f(1)+f(2)+\int_{0}^{3}t f(t)\mathrm{d}t\\ b&=\int_{0}^{3}t^2f(t)\mathrm{d}t. \end{align} Now, my question is can there exists at least one piecewise continuous function $f(t)$ on $[0, 3]$ which satisfies the above equations for any given real numbers $a$ and $b?$
Any help is appreciated.
Thank you.
You should be able to do this with a polynomial $f$. Indeed, first assume that $f(t) = 4At + 2B$ for some constants $A,B$. Then $$\int^3_0 tf(t) dt = \int^3_0 (4At^2 + 2Bt) dt = 36A + 9 B$$ and $$\int^3_0 t^2 f(t) dt = \int^3_0 (4At^3 + 2Bt^2)dt = 81A + 18B.$$ Thus the desired equations read \begin{align*} a &= (4A + 2B) + (8A + 2B) + 36A + 9B = 48A +13B,\\ b &= 81A + 18B. \end{align*} You can write this in matrix vector form as $$\begin{pmatrix} 48 & 13 \\ 81 & 18 \end{pmatrix} \begin{pmatrix}A \\ B \end{pmatrix} = \begin{pmatrix}a \\ b \end{pmatrix}.$$ This has a unique solution $\binom A B$ for any $\binom a b$ since the matrix $\left(\begin{smallmatrix} 48 & 13 \\ 81 & 18 \end{smallmatrix}\right)$ is invertible, and thus the corresponding $f(t) = 4At + 2B$ will solve the equation.
Yes, you can fit a linear function $f(t) = ct+d$ into your system and solve for $c$ and $d$. The integrals are simple and you get a system of $2$ linear equations to solve for $c$ and $d$
I came up with $$ 11c + 13/2 d =a \\ 81/4c + 9 d = b $$ which does have a solution for $c$ and $d$ in terms of $a$ and $b$
