How to solve $ (e^{x^4} -1) u' = 0 $ in the distributional sense?

Question

How to solve $ (e^{x^4} -1) u' = 0 $ on $ \mathbb{R} $ in the distributional sense? Since $ (e^{x^4} -1) = 0 $ for $ x = 0 $, we cannot divide the equation by this expression, so we somehow have to work with it. But how? I would be grateful for any clues!

Answer 1

A somewhat sketchy solution:

Around $x=0$ we have $e^{x^4} - 1 = O(x^4)$. This means that we practically have the equation $x^4 u' = 0$ at $x=0$. This has solutions $u' = a\delta + b\delta' + c\delta'' + d\delta'''$ for some constants $a,b,c,d$. But away from $x=0$ we can divide by $e^{x^4} - 1$. The full solution should thus be $$u' = a\delta + b\delta' + c\delta'' + d\delta''',$$ for some constants $a,b,c,d$.

Answer 2

Hint Since $(e^{x^4}-1)u'=0$ we have $\mbox{supp}(u') \subseteq \{ 0 \}$.

This gives that there exists some $N$ such that $$u'=\sum_{k=0}^n a_k D^{K}$$ where $D^k$ is $D^k(f)=f^{(k)}(0)$ (or if you preffer $D^k(f)=(-1)^kf^{(k)}(0)$, the choice is irrelevant since $(-1)^k$ can go into $a_k$).

Hint 2 $$0=(e^{x^4}-1)u'=\sum_{k=0}^n a_k (e^{x^4}-1)D^{K}$$ and you can calculate explicitely $(e^{x^4}-1)D^{K}$. You are going to discover that this is $0$ for the first few values of $k$, and non-zero for the rest. From there, the rest is easy.

Hint 3 Once you get explicitely $u'$, you can calculate $u$.

Answer 3

Expand the definition ($D$ the test function space). $$\begin{align} (\mathrm{e}^{x^4}-1)u' &=0 \\ \forall_{\phi:D}\langle u',(\mathrm{e}^{x^4}-1)\phi \rangle &=0 \\ \forall_{\phi:D}\langle u,((\mathrm{e}^{x^4}-1)\phi)' \rangle &=0\text{.} \\ \end{align}$$

Next, divide the equation by $$f(x)=\int_0^1\mathrm{e}^{tx^4}\mathrm{d}t\text{.}$$ To "divide" means to substitute $\phi\leftarrow \tfrac{1}{f}\phi$ for the quantified variable $\phi$, which is legitimate because $f$ is an invertible linear map $f\cdot(\quad): D\to D$ and consequently does not change the domain of quantification $D$. Thus

$$\forall_{\phi:D}\langle u,(x^4\phi)' \rangle = 0\text{.}$$

That is, if we define $L:D\to D$ to be $L\phi=(x^4\phi)'$, we are being asked to find a cokernel

$$\mathrm{coker}\,L:D \to \mathrm{cod}\, \mathrm{coker}\,L$$

the vector space of necessary and sufficient conditions on a test function $\psi$ for it to be of the form $\psi = L\phi = (x^4\phi)'$ for some $\phi:D$.

Now, if such a $\phi$ exists, then it must be the case that

$$\phi(x)=\frac{1}{x^4}\int_{-\infty}^x\psi(x)\mathrm{d}x\text{.}$$ However, this $\phi$ does not belong to $D$ in general. Recall that a function belongs to $D$ iff it is compactly supported and smooth. For $\phi$ to be compactly supported, it is necessary and sufficient for to have $$\int_{-\infty}^{\infty}\psi(x)\mathrm{d}x =0\text{.}$$

As for smoothness: expand

$$\frac{1}{x^4}\int_{-\infty}^x\psi(x)\mathrm{d}x=\frac{1}{x^4}\int_{-\infty}^0\psi(x)\mathrm{d}x +\frac{\psi(0)}{x^3}+\frac{\psi^{(1)}(0)}{2x^2}+\frac{\psi^{(2)}(0)}{6x}+\frac{1}{6}\int_0^1 \psi^{(3)}(tx)(1-t)^3 \mathrm{d}t\text{.}$$ Then the necessary and sufficient conditions for $\phi$ to be smooth are $$\begin{align} \int_{-\infty}^0\psi(x)\mathrm{d}x&=0 & \psi(0) &= 0 & \psi^{(1)}(0) &= 0 & \psi^{(2)}(0) &= 0 \end{align}$$ and we conclude that a cokernel of $L$ is given by $$\begin{align} \mathrm{coker}\,L&:D \to \mathbb{R}^{\oplus 5} & \mathrm{coker}\,L[\psi]&= \int_{-\infty}^{\infty}\psi(x)\mathrm{d}x\oplus\int_{-\infty}^0\psi(x)\mathrm{d}x \oplus \psi(0) \oplus \psi'(0)\oplus \psi''(0)\text{.} \end{align}$$ The span of the linear functionals on the right side of this definition are the solution to the original differential equation.