A singular differential equation

Question

In a neighbourhood of $0$ in $\mathbb{R}^n$ a smooth function $h=h(x)$, $h(0)=0$, is given. Take arbitrary real numbers $w,\lambda_1,\dots,\lambda_n\in\mathbb{R}$.

The problem is to find a smooth function $u=u(x)$ around $0$ such that $$\sum_i\lambda_i\cdot x^i\,\frac{\partial u}{\partial {x^i}}(x)=w\cdot u(x)+h(x)\,.$$

I think that there are solutions of this problem only for special values of $w,\lambda_1,\dots,\lambda_n\in\mathbb{R}$. In the simplest case $h=0$, for most of $w,\lambda_1,\dots,\lambda_n\in\mathbb{R}$ there are even no formal solutions, but maybe $u$ could be flat at $0$.

Any ideas about solving this problem?

Answer 1

I solved the problem.

We can write our PDE as $X(u)=w\cdot u+h$, where $X$ is the vector field $X=\sum_i\lambda_i\cdot x^i\,\partial_{x^i}$. Note that $X$ induces a concept of $\textit{weight}$ of a function: $f$ is homogeneous of weight $w$ if $X(f)=w\cdot f$. In particular, the coordinate $x^i$ is of weight $\lambda_i$ and any monomial $x^m$, where $m=(m_1,\dots,m_n)\in\mathbb{N}^n$ is a multi-index, is homogeneous of weight $|m|=\sum_im_i\,\lambda_i$.

Checking now for formal solutions, we get for $u(x)=\sum_ma_mx^m$ that $X(u)=\sum_m|m|a_m x^m$ that leads to some constraints for coefficients of the Taylor expansion of $h$, so a necessary condition. If this condition is fulfilled, i.e. there is a formal solution, we can reduce our problem to the case when $h$ is flat at $0$. But this always has a solution.

We can reduce to the case when all $\lambda_i\ne 0$ and some of them, say $\lambda_1,\dots,\lambda_k$, are $>0$. Let us consider functions $y_i=(x^i)^{1/w_i}$, $i=1,\dots,k$, and $z_j=(x^j)^{\lambda_1}\cdot(x^1)^{-\lambda_i}$, $j=k+1,\dots,n$. Obviously, $y_i$ are of weight $1$ and $z_j$ are of weight $0$. These functions vanish at $0$ and are generally not coordinates, but as $h$ is flat at $0$, it is also a flat at $0$ smooth function in variables $(y_i,z_j)$. Moreover, in these variables, $X=\sum_iy^i\partial_{y^i}$. Put $$u(y,z)=\left(\sum_i(y^i)^2\right)^{w/2}\cdot u_1(y,z)\,.$$ We have $$X(u)=w\cdot u+\left(\sum_i(y^i)^2\right)^{w/2}\cdot X(u_1)\,,$$ so that it is enough to find $u_1$ satisfying $$X(u_1)(y,z)=\left(\sum_i(y^i)^2\right)^{-w/2}\cdot h(y,z)\,.$$ As $h$ is flat at $0$, the right hand side of the above equation is a flat at $0$ smooth function in variables $(y,z)$, say $h_1$. In consequence, we reduced to the equation $\sum_iy^i\cdot\frac{\partial u_1}{\partial {y^i}}(y,z)=h_1(y,z)$. The function $h$ vanishes at $0$, so it can be written in the form $h_1=\sum_iy^i\cdot g_i$, where $$g_i(y,z)=\int_0^1\frac{\partial h_1}{\partial y^i}(ty,z)\rm d t\,.$$ Now, it is enough to solve the system of differential equations $\frac{\partial u}{\partial y^i}=g_i$. The integrability condition is $$\frac{\partial g_i}{\partial y^j}=\frac{\partial g_j}{\partial y^i}\quad\text{for}\quad i\ne j\,,$$ which, due to the form of $g_i$, is clearly satisfied.

Answer 2

Note: This is only a partial answer, and actually I am not sure that I understood the problem correctly.

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If $x_i(t) = \alpha_i e^{\lambda_i t}$ is the characteristic of the equation, then the left-hand side of the equation can be written as $\tfrac{d}{dt} u(x(t)$, and the equation becomes $$ \tfrac{d}{dt} u(x(t)) = w u(x(t)) + h(x(t)) . \tag{$\spadesuit$}$$ This, of course, has a unique global solution for any given $u(x(0))$.

Thus, for every function $u_0$ on a hyper-surface that intersects every characteristic at a single point, there is a unique corresponding solution $u$ in $\mathbb R^n \setminus \{0\}$ equal to $u_0$ on that surface. If $u_0$ is smooth, then $u$ is smooth, too, so it remains to see whether $u$ extends to a smooth function at the origin.

For $h$ constant zero, the picture seems to be rather clear, although a bit technical to describe. First, we have $u(x(t)) = e^{w t} u(x(0))$. Now the answer depends on the signs of $\lambda_i$ and $w$:

• If all $\lambda_i$ are positive and $w < 0$, then $x(t)$ goes to zero as $t \to -\infty$ and hence $u$ is either constant zero or unbounded near zero.
• If all $\lambda_i$ are positive and $w > 0$, then again $x(t)$ goes to zero as $t \to -\infty$ and hence $u$ is continuous at zero (with $u(0) = 0$). Note that along each characteristic, either $u$ is constant zero or $u(x) \sim c |x|^{\lambda_j}$ for some $j$ (namely, $j$ corresponding to the minimal value of $\lambda_j$ among indices with $x_j(0) \ne 0$). In particular, no non-constant solution $u$ is flat at $0$. Furthermore, by Taylor's theorem, as $t \to -\infty$ we have $$e^{w t} = u(x(t)) = \sum_{k_1 + \ldots + k_n \leqslant N} c_{k_1,\ldots,k_n} \exp \biggl( \sum_{i = 1}^n k_i \lambda_i t \biggr) + o(|x(t)|^N) . $$ Since $|x(t)| \approx \exp(\mu t)$ with $\mu = \min \{\lambda_1, \ldots, \lambda_n\}$, considering $N$ large enough we see that all non-vanishing coefficients $c_{k_1,\ldots,k_n}$ necessarily correspond to groups with a given sum $\sum k_i \lambda_i$, and the sum of the coefficients in each group is either zero (when $\sum k_i \lambda_i \ne w$) or one (if $\sum k_i \lambda_i = w$). So no no non-trivial solutions exist if $w$ is not of the form $\sum_{i = 1}^n \lambda_i k_i$, and if the solution exists, it is unique only when $\lambda_i$ are linearly independent over $\mathbb Q$.
• If $\lambda_i$ have different signs, no characteristic ever touches the origin, and a different description is needed. This is no simpler than in the previous item, and unfortunately I do not have time now to work this out (if anyone out there has, please do edit this answer).

I do not have a clear intuition how will non-constant $h$ affect this picture.

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Edit: The above answer deals with general solutions for $h = 0$. If one asks for a smooth particular solution for a given $h$, the situation is much simpler. Indeed, the ODE ($\spadesuit$) is: $$ \tfrac{d}{dt} (e^{-w t} u(x(t))) = e^{-w t} h(x(t)) , $$ and so, if all $\lambda_i$ are positive, a particular solution with $u(0) = 0$ is given by $$ u(x(t)) = \int_{-\infty}^t e^{w (t - s)} h(x(s)) ds . $$ In other words, if we choose $x(0) = x$, then $$ u(x) = \int_0^\infty e^{w s} h(e^{-\lambda_1 s} x_1, \ldots, e^{-\lambda_n s} x_n) ds . $$ This is well-defined if, say, for some $\mu > w$ we have $$ |h(x)| \leqslant C (|x_1|^{\mu/\lambda_1} + \ldots + |x_n|^{\mu/\lambda_n}) $$ in the neighbourhood of $0$, and hence in particular if $h$ is flat at $0$. In the latter case it is easy to see that $u$ is flat at zero, too, and this, in turn, almost immediately gives the smoothness of $u$. For non-flat $h$, one has to consider the Taylor expansion of $u$ separately (if possible, of course), and so the answer seems much more involved, but since in your answer you seem to not to worry too much about this case, I'll stop here.