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Consider the following first-order ODE:

$$y'(x) = f'(x) y(x);$$

this has solution $y(x) = C e^{f(x)}$. Now, consider taking

$$f_\lambda(x) = \frac{\alpha}{2}\left(1+\tanh(\lambda x)\right);$$

in the limit $\lambda \to \infty$, $f_\lambda(x) \to \alpha\Theta(x)$ (where $\Theta(x)$ is the Heaviside step function), and the solution becomes $y(x) = C e^{\alpha \Theta(x)}$, which is discontinuous at $x = 0$. Presumably, there is a sense in which one can think of this as some kind of solution to the ODE with delta-function coefficient

$$y'(x) = \alpha \delta(x) y(x).$$

Now, the discontinuity in $y$ at $x=0$ is

$$\Delta y \equiv y(0^+) - y(0^-) = C(e^\alpha - 1).$$

My question is the following: is there a way to extract $\Delta y$ from the singular ODE directly without first finding a solution with smooth $f(x)$ and then taking an appropriate limit? For instance, in analogy with what one does with second-order ODEs with delta function coefficients, one could try to integrate it:

$$\Delta y = \lim_{\epsilon \to 0} \int^\epsilon_{-\epsilon} y' dx = \lim_{\epsilon \to 0} \int^\epsilon_{-\epsilon} \alpha \delta(x) y(x) dx = \alpha y(0),$$

but this has the problem that $y(0)$ is not defined due to the discontinuity in $y(x)$.

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  • $\begingroup$ One can't multiply by a distribution which is not given by integration against a function, which is why this doesn't really make any sense. $\endgroup$ – Ian Jan 23 '16 at 13:56
  • $\begingroup$ I agree that the last step doesn't make sense; that's exactly my problem. I suggested it as a straw man argument. My question in whether or not there is some well-defined way to extract $\Delta y$ directly from the differential equation without needing to solve it explicitly. $\endgroup$ – Sebastian Jan 23 '16 at 22:35
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A friend of mine figured out a solution, at least to this toy problem. As commenters pointed out, the problem with naively integrating the ODE to get an expression for $\Delta y$ is that it is ill-defined to integrate a delta function against a discontinuous function. The solution is to first divide through by $y$ to get

$$\frac{y'(x)}{y(x)} = \alpha\delta(x).$$

This equation can now be integrated across the delta function to yield

$$\Delta \ln y \equiv \ln y(0^+) - \ln y(0^-) = \alpha,$$

thus

$$\frac{y(0^+)}{y(0^-)} = e^\alpha,$$

which indeed agrees with the $f(x) \to \alpha \Theta(x)$ limit of the solution for general $f(x)$.

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A cool way of solving this is using "symbolic derivatives".

Some mathematical background

Symbolic functions

Let $\phi(x) \in C_0^\infty (I)$ (a continuous function with continuous derivatives of all orders, all vanishing outside a finite interval $I$, also known as a test function). This set forms a vector space and, in this space, we say a sequence of testing functions $\phi_n(x)$ converges to zero if the functions $\phi_n(x)$ and all their derivatives converge uniformly to zero and if all the functions $\phi_n(x)$ vanish identically outside the same finite interval.

We sy $F(\phi)$ is a linear functional on the space of testing functions if to every testing function $\phi(x)$ a real or complex number is assigned such that \begin{align} F(\phi_1 + \phi_2) &= F(\phi_1) + F(\phi_2), \\ F(\lambda \phi) &= \lambda F(\phi), \end{align} where $\lambda$ is a scalar. A functional $F(\phi)$ is said to be continuous if the sequence of numbers $F(\phi_n)$ converges to zero whenever the sequence of testing functions $\phi_n(x)$ converges to zero in our sense. Here are some examples of continuous linear functionals: \begin{align} F_1(\phi) &= \phi'(0),\\ F_2(\phi) &= \int_0^1 \phi(x)\,dx. \end{align} Laurent Schwartz, in his famous work Théorie des distributions, calls any continuous linear functional on the space of testing functions a distribution.

We know that any continuous linear functional on a vector space having a scalar product can be expressed as a scalar product $[1]$, i.e. $$ F(\phi) = \langle s(x), \phi(x) \rangle. $$ With the natural choice of scalar product $$ \langle \phi(x),\psi(x) \rangle = \int_{-\infty}^\infty \phi \psi\,dx, $$ some linear functionals can be expressed as a scalar product and some not (this is because with this definition of scalar product, the space of testing functions is not complete). For example, $$ F(\phi) = \int_0^1 \phi(x)\,dx = \int_{-\infty}^\infty s(x)\phi(x)\,dx, $$ where $s(x) = 1$, $0 < x < 1$, and $s(x) = 0$ otherwise. There are other functionals that cannot be expressed this way; for example, $F(\phi) = \phi(0)$ or $F(\phi) = \phi'(0)$. However, we can introduce the $\delta(x)$ symbol such that $$ F(\phi) = \phi(0) = \int_{-\infty}^\infty \delta(x) \phi(x)\,dx. $$ In this context, the $\delta$-function is a symbolic function. In general, given a continuous linear functional $F(\phi)$ on the space of testing functions, we introduce a symbol, say $s(x)$, and write $$ \int_{-\infty}^\infty s(x) \phi(x)\,dx = F(\phi). $$ We say that $s(x)$ is a symbolic function. Note that a symbolic function need not have values. It produces values only when multiplied by a testing function and then integrated. A final observation is that every integrable function is a symbolic function (why?).

Derivatives of symbolic functions

To define the concept of the derivative of a symbolic function in such a way that it will be valid, we can use integration by parts. We say that $s'(x)$ is the derivative of $s(x)$ if $$ \int_{-\infty}^\infty s'(x) \phi(x)\,dx = - \int_{-\infty}^\infty s(x)\phi'(x)\,dx $$ for every testing function $\phi(x)$. Since the right-hand of this equation always exists, the left-hand side can be used to define the derivatives of $s(x)$. For example, $\delta'(0)$ is defined by $$ \int_{-\infty}^\infty \delta'(x) \phi(x)\, dx = -\int_{-\infty}^\infty \delta(x) \phi(x)\,dx = -\phi'(0). $$ Extending this argument, it is easy to show that $\delta^{(n)}(x)$ is such that $$ \int_{-\infty}^\infty \delta^{(n)}\phi(x)\,dx = (-1)^n \frac{d^n \phi}{d x^n}\Bigg|_{x=0}. $$

The problem at hand

Of course, the relation $$ \frac{y'(x)}{y(x)} = a \delta(x) $$ is formally nonsence, but if we write it in the context above, i.e., $$ \int_{-\infty}^\infty \frac{y'(x)}{y(x)} \phi(x)\,dx = a \int_{-\infty}^\infty \delta(x) \phi(x)\,dx = \phi(0), $$ everything is well stated.

Before we proceed, let's derive a fantastic relation between the $\delta$-function and the Heaviside function. Given $$ \int_{-\infty}^\infty H(x) \phi(x)\,dx = \int_0^\infty \phi(x)\,dx, $$ we have $$ \int_{-\infty}^\infty H'(x) \phi(x)\,dx = -\int_{-\infty}^\infty H(x) \phi'(x)\,dx = - \int_0^\infty \phi'(x)\,dx = \phi(0). $$ Consequently $H'(x) = \delta(x)$ in the symbolic framework!

With this amazing result (for $y(x) > 0$), \begin{align} 0 &= \int_{-\infty}^\infty \left[\frac{y'(x)}{y(x)} - a\delta(x)\right]\phi(x)\,dx \\ \\ &= \int_{-\infty}^\infty \frac{d}{dx} \left[\log y(x) - a H(x)\right]\phi(x)\,dx \\ \\ &= -\int_{-\infty}^\infty \left[\log y(x) - a H(x)\right] \phi'(x)\,dx \end{align} for every test function $\phi(x)$. Finally, using the du-Bois-Reymond theorem, the function $$ y(x) = e^{a}H(x) + H(-x) $$ satisfies the differential equation in the symbolic sense.

Nice, huh!

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