Find the functions $f:\mathbb{R}\rightarrow\mathbb{R}$, that have primitives, such that $F(x-1)+f(1-x)=x^3$

Question

Find the functions $f:\mathbb{R}\rightarrow\mathbb{R}$, that have primitives $F:\mathbb{R}\to\mathbb{R}$, such that $F(x-1)+f(1-x)=x^3$ for any $x\in\mathbb{R}$. For $1-x$ instead of $x$, $F(-x)+f(x)=(1-x)^3$, $F(x)+f(-x)=(1+x)^3$ (1). \begin{align*} \left(\frac{F(x)-F(-x)}{e^x}\right)' &=\frac{(f(x)+f(-x))-(F(x)+F(-x))}{e^x}\\ &=\left(\frac{F(x)}{e^x}\right)'-\left(\frac{F(-x)}{e^x}\right)'. \end{align*} From here on however I don't know how to solve it. I feel that the two relations from (1) or similar ones can be used.

Answer 1

So, we're solving for differentiable $g$ such that $g(-x) + g'(x) = (1 - x)^3$. Note that $g'(x) = (1 - x)^3 - g(-x)$, and hence $g$ is twice differentiable with $$g''(x) = -3(1 - x)^2 + g'(-x) = -3(1 - x)^2 + (1 + x)^3 - g(x)$$ That is, $g$ must satisfy the differential equation $$y'' + y = -3(1 - x)^2 + (1 + x)^3.$$ This differential equation has the general solution $$y = A \cos(x) + B \sin(x) + x^3 + 3x - 2.$$ But, of course, you're looking for $f = g'$, so the possible solutions we have are $$f(x) = -A \sin(x) + B \cos(x) + 3x^2 + 3,$$ with corresponding primitive as above.

Do all such functions work? We have \begin{align*}f(x) + F(-x) &= -A\sin(x) + B\cos(x) + 3x^2 + 3 + A\cos(-x) + B\sin(-x) + (-x)^3 + 3(-x) - 2 \\ &= (1 - x)^3 + (A + B)\cos(x) - (A + B)\sin(x). \end{align*} Therefore, we only get a solution with $A = -B$. Thus, the general solution is $$f(x) = A(\sin(x) + \cos(x)) + 3x^2 + 3,$$ with corresponding primitive $$F(x) = A(\sin(x) - \cos(x)) + x^3 + 3x - 2.$$

Answer 2

As you note, the function $f$ with primitive $F$ satisfies $F(x-1)+f(1-x)=x^3$ if and only if $$F(x)+f(-x)=(x+1)^3.$$ The primitive $F(x)$ uniquely determines $f(x)$, its derivative, so we'll focus solely on $F(x)$. If $F(x)$ is twice differentiable, or equivalently, if $f$ is differentiable, then differentiating the functional equation $$F(x)+F'(-x)=(x+1)^3 \qquad\text{ yields }\qquad F(x)-F''(-x)=3(x+1)^2.$$ Plugging $-x$ into the latter and subtracting the two equations yields $$F(x)+F''(x)=(x+1)^3-3(-x+1)^2=x^3+9x-2.$$ This is a second order linear non-homogeneous ODE, and the standard method yields $$F(x)=c_1\cos(x)+c_2\sin(x),$$ as the solution to the homogeneous ODE. For the non-homogeneous part, the form suggests that taking $F(x)$ a cubic polynomial will yield a solution. Setting $F(x)=ax^3+bx^2+cx+d$ yields $$x^3+9x-2=F(x)+F''(x)=ax^3+bx^2+(6a+c)x+(2b+d),$$ which shows that $a=1$, $b=0$, $c=3$ and $d=-2$. Hence every such $F$ is of the form $$F(x)=c_1\cos(x)+c_2\sin(x)+x^3+3x-2.$$ Plugging this back into the original equation yields \begin{eqnarray*} F(x)+f(-x) &=&(c_1\cos(x)+c_2\sin(x)+x^3+3x-2)\\ &\ &+(-c_1\sin(-x)+c_2\cos(-x)+3x^2+3)\\ &=&(c_1+c_2)\cos(x)+(c_1+c_2)\sin(x)+x^3+3x^2+3x_1)\\ &=&(c_1+c_2)(\cos(x)+\sin(x))+(x+1)^3. \end{eqnarray*} This shows that the function equation is met if and only if $c_1=-c_2$, and so \begin{eqnarray*} F(x)&=&c(\cos(x)-\sin(x))+x^3+3x-2\\ f(x)&=&-c(\cos(x)+\sin(x))+3x^2+3. \end{eqnarray*} About nondifferentiable solutions $f$ I have no idea at all.

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Edit: As Theo Bendit notes in the comments, and shows in his answer, the fact that $$f(x)=(1-x)^3-F(-x),$$ implies that $f$ is differentiable. So the solutions above are all solutions.