Fundamental solution of a first order distributional equation

Question

Which is a solution of $u^{'}+\alpha u=\delta_0$?
What about $u^{'}+f(x)u=\delta_0$?
(Where $u$ is a distribution over an open set $\Omega$ with $\Omega\subseteq\mathbb{R}$, $\alpha$ a real constant and $f\in C^{\infty}(\Omega)$ and $\delta_0$ is the Dirac's delta in $0$).
I think i have to use the Fourier's transform but i am not sure about the calculations can someone please give me some advice?

Answer 1

Solving $u' + fu = \delta$

Multiply the equation with the integrating factor $e^F,$ where $F'=f$: $$ e^F (u'+fu) = e^F \delta = e^{F(0)} \delta \\ (e^F u)' = e^{F(0)} \delta \\ e^F u = e^{F(0)} H + C, $$ where $H$ is the Heaviside step function and $C$ is a constant.

Finally, multiply with $e^{-F}$ to get $$ u = e^{-F}(e^{F(0)} H+C) = e^{-(F-F(0))}(H + C'), $$ where $C' = C e^{-F(0)}.$

Answer 2

"Dirac's delta function" is the generalized function (not a true function) that is 0 for all x except x= 0 and such that the integral over any interval containing x= 0 is 1.

I presume that if you are given a problem like this you are taking or have taken a course on differential equations. Then you should know that you can solve "non-homogeneous equations" like these by first finding the general solution to the associated homogeneous equation the adding to that solution any one solution to the entire equation.

For $u'+ \alpha u= \delta_0(x)$ the associated homogeneous equation is $u'+ \alpha u= 0$. That can be written as $u'= \frac{du}{dx}= -\alpha u$, $\frac{du}{u}= -\alpha dx$. Integrating, $ln(u)= -\alpha x+ C$ so $u= e^{-\alpha x+ C}= C'e^{-\alpha x}$.

For $u'+ f(x)u= \delta_0(c)$ the associated homogeneous equation is $u'+ f(x)= 0$. That can be written as $u'= \frac{du}{dx}= -f(x)u$, $\frac{du}{u}= -f(x)dx$. Integrating, $ln(u)= -\int f(x)dx+ C$ so $ e^{-\int f(x)dx+ C}= C'e^{-\int f(x)dx}$.

Now to find a single solution to the entire equation, I would use "variation of parameters". That means that we look for a solution of the form $y= u(x)e^{-\int f(x)dx}$. That is, we let the "parameter" (constant), C, vary as the parameter u (f(x)= 1 in the first equation). Putting that in the equation, $y'+ f(x)y= u'e^{\int f(x)dx}- u\left(f(x)e^{\int f(x)dx}\right)+ u\left(f(x)e^{\int f(x)dx}\right)= u'e^{\int f(x)dx}= \delta_0(x)$.

So $u'(x)= \delta_0(x)e^{\int f(x)dx}$. The integral of $\delta_0(x)F(x)$ for any function F is F(0) so u(x) is some constant, C, and y= Ce^{-\int f(x)dx}. In the case that $f(x)= \alpha$, that constant is $e^{alpha}$ so $u= e^{\alpha- x}$

Answer 3

As the other two answers show, for a first-order linear equation it is not necessary to think in terms of Fourier transforms, and the method applies broadly, not just for your coefficient function a constant.

Still, it may be worthwhile to think about how to legitimately apply Fourier transform. First, this will work best just for the constant-coefficient case. And we must be asking for solutions that are tempered distributions. In particular, for the constant $\alpha$ not purely imaginary, the corresponding homogeneous equation has solutions $e^{-\alpha x}$ that are not tempered distributions (although are otherwise-reasonable classical functions). Thinking in these terms, it is not a-priori clear that there need be tempered-distribution solutions even when there are classical solutions.

But we can proceed, and discover constraints as we go. Assuming $u'-\alpha u=\delta$ for a tempered distribution $u$, taking Fourier transform (with some choice of normalization) gives $(2\pi i x-\alpha)\widehat{u}=1$, and then $\widehat{u}={1\over 2\pi ix-\alpha}$. The right-hand side is in $L^2$ for $\alpha$ not purely imaginary, but not in $L^1$, so the inverse Fourier transform needs the Plancherel extension. (For purely imaginary $\alpha$, the associated principal value integral is a reasonable tempered distribution...) For $\Re(\alpha)<0$, $u(x)$ is $0$ for $x<0$ and is $e^{\alpha x}$ for $x>0$. For $\Re(\alpha)>0$, $u(x)$ is $0$ for $x>0$ and is $e^{\alpha x}$ for $x<0$. The case $\Re(\alpha)=0$ is similar...

EDIT: prompted by @md2perpe's remarks, to clarify: the above discussion was aimed at showing how Fourier transform gives a solution to the inhomogeneous equation. For $\alpha$ not purely imaginary, in fact there is a unique tempered distribution solving the equation. For $\alpha$ purely imaginary, however, the original homogeneous equation and its Fourier transformed version both have non-trivial tempered solutions. E.g., on the Fourier transform side, $(x-n)\widehat{u}=0$ has solution $\delta(x-n)$. In that case, the procedure sketched above still does succeed in producing a solution to the inhomogeneous equation, if (for example) we interpret $(x-n)\widehat{u}=1$ as giving solution $\widehat{u}$ equal to the principal value integral attached to $1/(x-n)$. Then its inverse Fourier transform is a step function twisted by an exponential... but is ambiguous by the exponentials obtained as Fourier transforms of $\delta(x-n)$.

In fact, even for arbitrary complex $\alpha$, if we allow Fourier transforms to map arbitrary distributions into the dual of the Paley-Wiener space (the latter being the image of Fourier transforms of test functions), we can obtain the general solution to the equation.

But perhaps the most daunting task is the initial one of obtaining a single solution to the inhomogeneous equation, since we do know how to solve the homogeneous equation by completely elementary methods (not mentioning distributions).