# ODE's involving distributions or radon measures

Well I would like to know an approach (method) to solve "singular" ODE's in a non-formal way. I seek for a method other (simpler) than the parameters variation, Laplace transform.

The equation would be $$Ly=\delta$$ Suppose that we have solutions of the homogeneous problem.

• uh, explain your symbols? From your title I assume that $\delta$ is the Dirac $\delta$ on the line? But what is $L$? – Willie Wong Jun 6 '13 at 15:09
• If you are only interested in the source term being the Dirac $\delta$ and its derivatives, and $L$ is a linear operator, you should search the internet for "ode impulse response". – Willie Wong Jun 6 '13 at 15:13
• Yes Willie that is the case, L is an ordinary differential operator , perhaps of constant coefficients. And $\delta$ is the Dirac atomic measure. – checkmath Jun 6 '13 at 16:29

Consider a linear ODE (the coefficients don't have to be constant, but I assume they are nice enough to not cause trouble): $$Ly:=y^{(n)}+a_{n-1}y^{(n-1)}+\dots + a_0 y = \delta \tag1$$ The delta-function always comes from the highest derivative. For $y^{(n)}$ to contain a delta-function at $0$, the next-to-highest derivative $y^{{n-1}}$ must jump by $1$ at zero. Formally, $$y^{(n-1)}(0 +)=y^{(n-1)}(0-)+1 \tag2$$ All lower order derivatives must be continuous at $0$, otherwise $y^{(n)}$ will contain a higher order singularity that we don't want. $$y^{(k)}(0 +)=y^{(k)}(0-),\quad k=0,\dots,n-2 \tag3$$
The general solution of homogeneous equation $Ly=0$ on the interval $(-\infty,0)$ involves $n$ undetermined coefficients $b_1,\dots,b_n$. The general solution of $Ly=0$ on the interval $(0,\infty,0)$ also involves $n$ undetermined coefficients, say $c_1,\dots,c_n$. The conditions (2)-(3) are $n$ linear relations between these coefficients. You can use them to eliminate $c_1,\dots,c_n$ from your general solution, leaving it in terms of $b_1,\dots,b_n$.
An example: $$y''+y'-6y=\delta \tag4$$ The general solution of homogeneous equation is $y=b_1e^{2x}+b_2 e^{-3x}$ for $x<0$, and $y=c_1e^{2x}+c_2 e^{-3x}$ for $x>0$. The relations (2) and (3) take the form $$\begin{split}2c_1-3c_2&=2b_1-3b_2+1 \\ c_1+c_2 &= b_1+b_2 \end{split}$$ Hence $c_1=b_1+1/5$ and $c_2=b_2-1/5$. The general solution of (4) is $$y=b_1e^{2x}+b_2 e^{-3x}+\frac15(e^{2x}- e^{-3x})\,H(x) \tag5$$ where $H$ is the Heaviside function.