Solving $y' + ay = \delta$ in $S'$ space

Question

I am solving ordinary differential equation in $S'$ (dual to Schwartz space) given as:

$y' + ay = \delta$, where $\delta$ is a Dirac delta function.

The general solution of homogenous equation is $Ce^{-ax}$, where $C$ is a constant.

I actually started solving it via Fourier transform, but it is not probably efficient and I got for $x \lt 0$ a zero solution. But according to my textbook the solution is:

$y(x) = \begin{cases} (C+1)e^{-ax}, & x \gt 0 \\[2ex] Ce^{-ax}, & x \lt 0 \end{cases}$

And no matter how long I am staring at it, I don't understand. My textbook solves it via fundamental solution of the equation given as this in general: $Lu =f$, where $L$ is an ordinary differential operator. And then I suppose is used the gluing of the solution (which I don't know how to proceed, nor I found any good example on the internet).

Can anyone help me to understand this?

Answer 1

The solution to $$y'+ay=0$$ is in the following form: $$y(x)=C\exp(-ax)$$ So we could solve it for both $x<0$ and $x>0$: $$y_{+}(x)=C_{+}\exp(-ax)$$ $$y_{-}(x)=C_{-}\exp(-ax)$$ Now let's apply the $\int_{-\epsilon}^{\epsilon}\mathrm{d}x$ operator on both sides of the differential equation (with $\epsilon > 0$): $$\int_{-\epsilon}^{\epsilon}\mathrm{d}x\,\left(y'(x)+ay(x)\right)=\int_{-\epsilon}^{\epsilon}\mathrm{d}x \delta(x)$$ $$y_{+}(\epsilon)-y_{-}(-\epsilon)+a\int_{-\epsilon}^{\epsilon}\mathrm{d}x \,y(x)=1$$ $$C_{+}\exp(-a\epsilon)-C_{-}\exp(a\epsilon)+a\int_{-\epsilon}^{\epsilon}\mathrm{d}x \,y(x)=1$$ And if we let $\epsilon \to 0+0$, we get that $$C_{+}-C_{-}=1$$ Alternatively, we can apply fourier transform to both sides: $$(i \omega) \hat{f}(\omega)+a \hat{f}(\omega)=\frac{1}{\sqrt{2 \pi}}$$ $$\hat{f}(\omega)=\frac{1}{\sqrt{2 \pi}} \frac{1}{a + i \omega}$$ And applying inverse fourier transform we get that: $$f(x)=\exp(-ax)u(x)$$ And I think this is what you get. And this is correct, because it's just a particular solution, and you can also add the general solution to it to get that $$y(x)=\exp(-ax)(C+u(x))$$

Answer 2

Here is a solution fully inside the theory of distributions:

First we multiply on both sides with the smooth and everywhere nonzero integrating factor $e^{ax}$: $$y' e^{ax} + y \, a e^{ax} = e^{ax} \delta.$$

Now, the left hand side can be written as a derivative: $$y' e^{ax} + y \, a e^{ax} = (y e^{ax})'$$

And the right hand side can be simplified to just $\delta,$ since it is generally valid that $f \delta = f(0) \delta.$

Thus we have the equation $(y e^{ax})' = \delta.$

Taking the antiderivative gives us $y e^{ax} = H + C,$ where $H$ is the Heaviside function and $C$ is a constant.

Finally, multiplying with $e^{-ax}$ gives us $y$: $$y = e^{-ax}(H + C).$$

Answer 3

Let $Y(s)=L\{y(t)\}$ be the Laplace transform of $y$. Using $$ L\{y'(t)\}=sY(s)-y(0), L\{\delta\}=1, $$ one has $$ sY(s)-y(0)+aY(s)=1 $$ from which, $$ Y(s)=\frac{y(0)+1}{s+a}. $$ Hence $$ y(t)=L^{-1}\{Y(s)\}=(y(0)+1)L^{-1}\{\frac{1}{s+a}\}=(y(0)+1)e^{-at}u(t) $$ where $u(t)$ is the unit step function.

Answer 4

"I am solving ordinary differential inhomogeneous equation in S′ (dual to Schwartz space) given as: $$y' + ay = \delta(x)$$ -where $\delta(x)$ is a Dirac delta function. The general solution of homogenous equation is $y_{1}(x)=Ce^{-ax}$, where $C$ is a constant." General solution of the inhomogeneous equation: $y(x)=y_{1}(x)+y_{2}(x)$, where $y_{2}(x)$-partial solution of an inhomogeneous equation. A particular solution is found by the method of variation of the constant (Lagrange method): $$y_{2}(x)=e^{-ax}\int_{-\infty }^{x}\delta (t)e^{at}dt=e^{-ax}\int_{-\infty }^{x}\delta (t)e^{a0}dt=e^{-ax}\int_{-\infty }^{x}\delta (t)dt=e^{-ax}\theta (t)|_{-\infty }^{x} =e^{-ax}\theta (x)$$ Where - $\theta \left ( x \right )=\left\{\begin{matrix}1,& x>0 \\ 0 ,& x<0\end{matrix}\right.$ -Heaviside function, $t$ - integration variable. Consequently $$y(x)=y_{1}+y_{2}=Ce^{-ax}+\theta (x)e^{-ax}=\left (\theta (x)+C \right )e^{-ax}$$