Consider the following problem:
\begin{equation}
\begin{aligned}
\frac{\partial}{\partial t} u(x, t)-\frac{\partial^2}{\partial x^2} u(x, t) & =1, \quad x \in \mathbb{R}, t>0 \\
u(x, 0) & =x e^{-x^2 / 2}, \quad x \in \mathbb{R} .
\end{aligned}
\end{equation}Decide $u(x,t)$ such that $\lim _{|x| \rightarrow \infty} \frac{\partial}{\partial x} u(x, t)=0$.
The solution provided is
Taking the fourier transform with respect to x on both sides of the partial differential equation yields
\begin{equation}
\frac{\partial}{\partial t} \hat{u}(\omega, t)+\omega^2 \hat{u}(\omega, t)=2 \pi \delta(\omega).
\end{equation}Solving with integrating factor yields
\begin{equation}
\frac{\partial}{\partial t}\left(\hat{u}(\omega, t) e^{\omega^2 t}\right)=2 \pi \delta(\omega) e^{\omega^2 t},
\end{equation}and we therefor have
\begin{equation}
\begin{aligned}
\hat{u}(\omega, \tau) & =e^{-\omega^2 \tau} \hat{u}(\omega, 0)+\int_0^\tau 2 \pi \delta(\omega) e^{-\omega^2(\tau-t)} \mathrm{d} t \\
& =-\sqrt{2 \pi} e^{-\omega^2 \tau} \mathrm{i} \omega e^{-\omega^2 / 2}+\int_0^\tau 2 \pi \delta(\omega) \mathrm{d} t \\
& =-\sqrt{2 \pi} \mathrm{i} \omega e^{-\omega^2(\tau+1 / 2)}+2 \pi \tau \delta(\omega).
\end{aligned}
\end{equation}Inverse transform yields our solution
\begin{equation}
u(x, \tau)=-\frac{\partial}{\partial x} \frac{e^{-x^2 /(4(\tau+1 / 2))}}{\sqrt{2(\tau+1 / 2)}}+\tau=\frac{x}{(2 \tau+1)^{3 / 2}} e^{-x^2 /(4(\tau+1 / 2))}+\tau.
\end{equation}
The property $\lim _{|x| \rightarrow \infty} \frac{\partial}{\partial x} u(x, t)=0$ gives us uniqueness of our solution.
How does
\begin{equation}
\frac{\partial}{\partial t}\left(\hat{u}(\omega, t) e^{\omega^2 t}\right)=2 \pi \delta(\omega) e^{\omega^2 t},
\end{equation}
imply
\begin{equation}
\begin{aligned}
\hat{u}(\omega, \tau) & =e^{-\omega^2 \tau} \hat{u}(\omega, 0)+\int_0^\tau 2 \pi \delta(\omega) e^{-\omega^2(\tau-t)} \mathrm{d} t ?
\end{aligned}
\end{equation}
In my initial calculations I got
\begin{equation*}
\begin{aligned}
\frac{\partial}{\partial t}\left(\hat{u}(\omega, t) e^{\omega^2 t}\right) & =2 \pi \delta(\omega) e^{\omega^2 t}, \\
\hat{u}(\omega, t) e^{\omega^2 t} & =\int 2 \pi \delta(\omega) e^{\omega^2 t} d t+A(\omega), \\
\hat{u}(\omega, t) &= e^{-\omega^2t}(2 \pi \delta(\omega)\frac{e^{\omega^2 t}}{\omega^2}) + e^{-\omega^2 t} A(\omega), \\
\hat{u}(\omega, t) &= 2 \pi \frac{\delta(\omega)}{\omega^2} + e^{-\omega^2 t} A(\omega),
\end{aligned}
\end{equation*}
where $A(\omega)$ is a constant of integration.
Anyone know what's going on here?
Best Answer
The notation in this solution is slightly confusing. Begin with $$\frac{\partial}{\partial t}\left(\hat{u}(\omega,t)e^{\omega^2 t}\right) = 2\pi\delta(\omega)e^{\omega^2t}$$ We now proceed as you did and integrate the equation with respect to $t$. However, I will perform a definite integral from $0$ to $t$ with the use of the dummy variable $\tau$, which will eliminate the arbitrary constant of integration and make the dependence of the solution on the initial conditions explicit. This will result in $t$ and $\tau$ being swapped from the solution you provided, but it is equivalent. This is much less confusing to me personally and is the way I teach it. $$\begin{aligned} \frac{\partial}{\partial t}\left(\hat{u}(\omega,t)e^{\omega^2 t}\right) &= 2\pi\delta(\omega)e^{\omega^2t} \\ \implies \left.\hat{u}(\omega,\tau)e^{\omega^2\tau}\right|^{\tau = t}_{\tau = 0} &= \int_0^t2\pi\delta(\omega)e^{\omega^2\tau} \ \mathrm{d}\tau \\ \implies \hat{u}(\omega,t)e^{\omega^2t} - \hat{u}(\omega,0) &= \int_0^t2\pi\delta(\omega)e^{\omega^2\tau} \ \mathrm{d}\tau \\ \implies \hat{u}(\omega,t) &= e^{-\omega^2t}\left[\hat{u}(\omega,0) + \int_0^t2\pi\delta(\omega)e^{\omega^2\tau} \ \mathrm{d}\tau\right] \\ &= e^{-\omega^2t}\hat{u}(\omega,0) + \int_0^t2\pi\delta(\omega)e^{\omega^2(\tau-t)} \ \mathrm{d}\tau \\ &= e^{-\omega^2t}\hat{u}(\omega,0) + \int_0^t2\pi\delta(\omega)e^{-\omega^2(t-\tau)} \ \mathrm{d}\tau. \end{aligned}$$
Notice that in the last step we were able to bring the factor of $e^{-\omega^2t}$ inside the integral since the integration was with respect to the dummy variable $\tau$.
This form of writing the solution to a nonhomogeneous linear equation is preferred in some settings because it separates the dynamics due to the initial condition and the forcing term and relates them both back to the propogation of the original linear operator ($-\omega^2I$ in this case).