Solution of $u_t=\mathcal{F}u$

Question

I tried to solve the equation $u_t=\mathcal{F}u$, where $\mathcal{F}$ denotes the Fourier transform, with initial data $u(x,0)=u_0(x)$. The solution should be given by $$ u(x,t)=e^{\mathcal{F}t}u_0=\left(\sum_{j\geq0}\frac{\mathcal{F}^jt^j}{j!}\right)u_0(x) $$ and I looked for an explicit expression of it in terms of $u_0$. Since $\mathcal{F}^2=(\cdot)^\check{}$ (the operator that maps $v(x)\mapsto\check{v}(x):=v(-x)$), we have that $\mathcal{F}^3=\mathcal{F}^{-1}$ and $\mathcal{F}^4=id$. Suppose $u_0$ is a Schwartz function so all of this makes sense. Then we have $$ u(x,t)=\left(\sum_{j\in4\mathbb{N}}\frac{t^j}{j!}+\sum_{j\in1+4\mathbb{N}}\frac{t^j}{j!}\mathcal{F}+\sum_{j\in2+4\mathbb{N}}\frac{t^j}{j!}(\cdot)\check{}+\sum_{j\in3+4\mathbb{N}}\frac{t^j}{j!}\mathcal{F}^{-1}\right)u_0(x) $$
Next i wrote the function $u_0=\phi+\psi$, where $\phi$ is even and $\psi$ is odd, namely $$ \phi(x)=\frac{1}{2}(u_0(x)+u_0(-x))\quad\text{and}\quad\psi(x)=\frac{1}{2}(u_0(x)-u_0(-x)) $$ In this way (recalling that Fourier transform of an even function is odd and viceversa) one computes $$ u(x,t)=\sum_{j\in\mathbb{N}}\frac{t^{2j}}{(2j)!}\phi(x)+\sum_{j\in\mathbb{N}}(-)^j\frac{t^{2j}}{(2j)!}\psi(x)+\sum_{j\in\mathbb{N}}(-)^j\frac{t^{2j+1}}{(2j+1)!}\mathcal{F}\phi(x)+\sum_{j\in\mathbb{N}}\frac{t^{2j+1}}{(2j+1)!}\mathcal{F}\psi(x) $$ which equals $$ u(x,t)=\cosh(t)\phi(x)+\cos(t)\psi(x)+\sin(t)\mathcal{F}\phi(x)+\sinh(t)\mathcal{F}\psi(x). $$ The problem is that now, trying to check $$ u_t(x,t)=\sinh(t)\phi(x)-\sin(t)\psi(x)+\cos(t)\mathcal{F}\phi(x)+\cosh(t)\mathcal{F}\psi(x) $$ and $$ \mathcal{F}u(x,t)=\cosh(t)\mathcal{F}\phi(x)+\cos(t)\mathcal{F}\psi(x)+\sin(t)\phi(x)-\sinh(t)\psi(x) $$ they are not equal. Is my calculation wrong? Every help is extremely appreciated.

Answer 1

As you noted, $\mathcal{F}^4=I$. So $$ (1-\lambda^4)I=(\mathcal{F}^4-\lambda^4 I)=(\mathcal{F}-\lambda I)(\mathcal{F}^3+\lambda\mathcal{F}^2+\lambda^2\mathcal{F}+\lambda^3I) $$ which gives $$ (\lambda I-\mathcal{F})^{-1}=\frac{1}{\lambda^4-1}(\mathcal{F}^3+\lambda \mathcal{F}^2+\lambda^2\mathcal{F}+\lambda^3 I) $$ So, the following is evaluated by computing residues at $1,i,-1,-i$: $$ e^{t\mathcal{F}}=\frac{1}{2\pi i}\oint_{C}e^{\lambda t}(\lambda I-\mathcal{F})^{-1}d\lambda \\ =\frac{1}{2\pi i}\oint_{C}\frac{e^{\lambda t}(\mathcal{F}^3+\lambda\mathcal{F}^2+\lambda^2\mathcal{F}+\lambda^3I)}{(\lambda-1)(\lambda-i)(\lambda+1)(\lambda+i)}d\lambda $$