# Example solutions for PDE

examples-counterexamplespartial differential equations

$$\newcommand{\R}{\mathbb{R}} \newcommand{\1}{\mathbf{1}} \newcommand{\id}{\mathbb{1}} \newcommand{\allone}{\mathbb{A}}$$

Which smooth funcion $$f: \mathbb{R}^n \rightarrow \mathbb{R}$$ solves the PDE

$$\tag{1} \label{PDE} \text{d}f(x) \cdot x = f(x) + \lambda \, x \cdot \1$$

or in coordinates

$$\sum_{i=1}^n \partial_if(x) \, x_i = f(x) + \lambda \, \sum_{i=0}^nx_i$$

for some $$\lambda \in \mathbb{R}$$? I denote $$\1 = (1, \cdots, 1)$$.

Some solutions I found already are

• Any 1-homogeneous function, with $$\lambda = 0$$, by Euler's theorem for homogeneous functions;

• $$f(x) = x \cdot \ln{x} = \sum_{i=0}^n x_i \ln{x_i}$$, with $$\lambda = 1$$

To provide some context, this PDE appears in the study of the Hamiltonian structure of Riemannian Evolutionary Games; more precisely it is the condition that the potential of a Hessian metric should satisfy for the corresponding system to be Hamiltonian with respect to a constant coefficient Poisson structure.

Edit: I have not been clear in stating the problem. The answer of Sal provides solution $$f_{\lambda}(x)$$ to $$\eqref{PDE}$$ for any fixed $$\lambda$$. I am looking for functions $$f$$ such that there exists $$\lambda$$ such that $$\eqref{PDE}$$ holds true. For example it is easy to check that in $$n$$ dimensions

$$\tag{2} \label{f} f(x) = a Mx \cdot \ln{x} + b \cdot x$$
admits values of $$\lambda$$ such that $$\eqref{PDE}$$ holds true, whenever $$M$$ is a matrix such that
$$(Mx) \cdot \1 \text{ is proportional to } x \cdot \1$$

Indeed if $$f$$ is in the form \eqref{f} then
$$x \cdot df(x) = f(x) + a \, (Mx) \cdot \1$$

So the question would now be whether there exists other forms beyond \eqref{f} such that \eqref{PDE} holds true for some $$\lambda$$.

Let me demonstrate how to solve the specific case $$n=2$$ using the method of characteristics. The case arbitrary $$n$$ is below. The PDE for $$f(x_1,x_2)$$ is

$$\tag{1} x_1\partial_1f+x_2\partial_2f=f+\lambda (x_1+x_2)$$

Together with initial condition $$f(x_1,g(x_1))=\psi(x_1)$$ along some curve in the $$x_1,x_2$$ plane which is the graph of $$x_2=g(x_1)$$. Parametrize the characteristics with $$z$$ and differentiate

$$\tag{2} \frac{df}{dz}=\frac{dx_1}{dz}\partial_1f+\frac{dx_2}{dz}\partial_2f:=\text{RHS of (1)}$$

On the characteristic curves we have the following coupled ODEs

$$\tag{3} \frac{dx_1}{dz}=x_1 \\ \frac{dx_2}{dz}=x_2 \\ \frac{df}{dz}=f+\lambda (x_1+x_2)$$

We choose $$z=0$$ along the initial data curve $$f(a_1,g(a_1))=\psi(a_1)$$. The first two equations in (3) may be integrated directly

$$\tag{4} x_1(z,a_1)=a_1e^z \\ x_2(z,a_1)=g(a_1)e^z$$

We can substitute these into the last equation in (3)

$$\frac{df}{dz}-f=\lambda e^z(a_1+g(a_1))$$

This is a linear, inhomogeneous, first order equation that we may solve directly (eg. using integrating factor or homogeneous plus particular solution), with the initial condition $$f(z=0)=\psi(a_1)$$

$$\tag{5} f(z,a_1)=\psi(a_1)e^z+\lambda z e^z (a_1+g(a_1))$$

What remains is to invert the co-ordinate transformation defined in (4) back to $$x_1,x_2$$ from $$z,a_1$$. For example, if $$g(a)=a^2$$, then the solution is

$$f(x_1,x_2)=\frac{x_1^2}{x_2}\psi\left(\frac{x_2}{x_1}\right)+\lambda(x_1+x_2)\ln\left(\frac{x_1^2}{x_2}\right)$$

Which you can check does solve (1) for arbitrary $$\psi$$. Now you can generate all kinds of specific solutions by choosing particular forms for $$\psi$$.

Looking over the calculations, the only change for arbitrary $$n$$ is the initial condition: $$f(x_1,x_2,\dots ,x_{n-1},g)=\psi(x_1,x_2,\dots ,x_{n-1})$$

Where $$g=g(x_1,x_2,\dots ,x_{n-1})$$. The co-ordinate transformation in (4) becomes

$$x_i(z,a_1,\dots,a_{n-1})=a_ie^z \qquad, \qquad i\neq n \\ x_n(z,a_1,\dots,a_{n-1})=g e^z$$

And the solution is

$$\tag{6} f(z,a_1,\dots,a_{n-1})=\psi(a_1,\dots,a_{n-1})e^z+\lambda z e^z (a_1+\cdots+a_{n-1}+g)$$

Where again the solution is to be expressed in the original co-ordinates after inverting the co-ordinate transform. If we choose $$g=a_1^2$$ (for simplicity) then the inversion may be done neatly for arbitrary $$n$$

$$\tag{7} e^z=\frac{x_1^2}{x_n} \\ a_i=\frac{x_i x_n}{x_1^2}$$

Substituting (7) into (6)

$$f(x)= \frac{x_1^2}{x_n} \psi(a)\big|_{a=a(x)}+\lambda (x_1+\cdots+x_n) \ln\left( \frac{x_1^2}{x_n}\right)$$