# Finding Weak Solutions to ODEs

I'm wondering if anyone has a reference to a good set of notes on finding weak (distributional) solutions to ODEs, or has any tips or tricks. For example,

$$xy^\prime=0$$ has a classical solution $y(x)=C$, but it also has a distributional solution $y(x)=H(x)$, the Heaviside step function, since (symbolically) $H^\prime(x)=\delta(x)$ and so $xH^\prime=x\delta(x)=0$. We can prove that this is a weak solution by acting on test functions, i.e. show that $\langle xH^\prime,\varphi\rangle=0$ for all suitable $\varphi$.

For the above ODE, it was fairly simple to guess that $H(x)$ is a solution. The set of lecture notes I have only goes this far - they "guess" the solution then prove it works. My question is: how do we go about finding weak solutions for a general homogeneous ODE of the form $Ly=0$? Is "guess and check" the best we can do? Am I guaranteed a certain number of weak solutions (i.e. for a second order equation, there are 2 weak solutions and 2 strong solutions...?)

Thanks!

am I guaranteed a certain number of weak solutions (i.e. for a second order equation, there are 2 weak solutions and 2 strong solutions...?)

No, you can have more or fewer. For nice ODE, every weak solution is a classical solution: you get nothing new by allowing weak solutions. Let's take $$y'=2y\tag1$$ as a simple example. Its weak form is $$\int (y\phi'+2y\phi)=0\tag2$$ for every test function $\phi$. Rewrite it as $$\int e^{-2x}y(e^{2x}\phi)'=0\tag3$$ Given any test function $\psi$, we can let $\phi=e^{-2x}\psi$ and conclude from (3) that $$\int e^{-2x}y \psi'=0\tag4$$ for all test functions $\psi$. From (4) you can conclude that $e^{-2x}y$ is in fact a constant function. Indeed, every test function that integrates to $0$ over $\mathbb R$ is of the form $\psi'$ for some test function $\psi$. Therefore, the integral $\int e^{-2x}y \varphi$ depends only on the value of $\int \varphi$. By linearity of $\int e^{-2x}y \varphi$ there is $c\in\mathbb R$ such that $\int e^{-2x}y \varphi=c\int \varphi$ for all test functions $\varphi$. But this means precisely that $e^{-2x}y=c$ as distributions.

More generally, for any constant-coefficient linear ODE every weak solution is a classical solution. One way to see this is to take the Fourier transform (I ignore the existence issue): then the equation $\mathcal L(y)=0$ becomes $p\hat y=0$ where $p$ is a polynomial (the symbol of $L$). Therefore, $\hat y$ is supported on the zero set of $p$, in particular it's compactly supported. Therefore, $y$ is real-analytic.

For linear ODE with variable smooth coefficients everything works about the same
as long as the coefficient of the highest derivative does not vanish. This is because we can rewrite the ODE as $y'=F(x,y)$ with vector-valued $y$, where $F$ is smooth in $x$ and linear in $y$. Instead of giving a proof, I'll just say that $F(x,y)$ is as regular as $y$ is, which makes $y=\int F(x,y)$ more regular than that: it self-improves ad infinitum.

But your question is motivated by an ODE where the coefficient of the highest derivative (call it $\alpha(x)$) vanishes. What can we do then? The set $\{x:\alpha(x)\ne 0\}$ is a union of open intervals. On each such intervals the above applies: weak solutions are classical solutions. The question is how/if they can be glued together at the zeros of $\alpha$, called the singular points of the equation. This is where we have to look at the particular form of equation, because a lot of different things can happen at singular points.

Let's generalize your example to $\alpha(x)y'=0$ where $\alpha$ is a smooth function with $n$ zeros on the real line, denoted $b_1< \dots <b_n$. On each of $n+1$ open intervals $(-\infty,b_1)$, $(b_1,b_2)$, ... , $(b_n,\infty)$ the function $y$ must be constant. This may be an $(n+1)$-dimensional space of weak solutions, but we must check that they are indeed weak solutions. It turns out that they are: the weak form of equation is $$\int (\alpha y\phi' +\alpha'y \phi)=0$$ which holds because on each interval we can move the constant $y$ out of the integral, leaving
$$\int_{b_k}^{b_{k+1}} (\alpha \phi'+\alpha '\phi) = (\alpha \phi)\bigg|_{b_k}^{b_{k+1}}=0 \tag5$$ where the last step is based on $\alpha \phi$ being zero at $b_1,\dots,b_n$.

Thus, we have a first-order equation with $(n+1)$-dimensional space of weak solutions. In your example with $\alpha(x)=x$ there is only one zero, and therefore the solution space is $2$-dimensional.

On the other extreme, if $\alpha(x)=\sin x$, the solution space becomes infinite-dimensional.

Let's consider two more examples. $$xy'=2y \tag6$$ Outside of $x=0$, the solutions are $y=cx^2$. The constant can be different for $x<0$ and for $x>0$. The solutions nicely patch up at $0$, forming a $2$-dimensional space of classical solutions.

$$xy'=-2y \tag7$$ Outside of $x=0$, the solutions are $y=cx^{-2}$. The constant can be different for $x<0$ and for $x>0$. But they only solution that is locally integrable near $0$ is $y\equiv 0$. Thus, we have only one weak solution on $\mathbb R$.

Upshot: a lot of things can happen at singular points, where the leading coefficient vanishes. Power series methods are pretty good at finding the asymptotic behavior of solutions there: see Regular singular point, Frobenius method.

• Very useful, thanks! Can you recommend any books or lecture notes on the topic so I can learn more? – icurays1 Aug 6 '13 at 5:56
• I think you mean "every classical solution is also weak solution" in the begining of your nice answer. – dmtri Nov 21 '18 at 10:48