Suppose that $f_1$ and $f_2$ are two functions defined on $\mathbb{R}\times(0,+\infty)$ and satisfy:

$$\mathcal{L}_1 f_1 = 0\quad\mathrm{and} \quad\mathcal{L}_2 f_2 = 0$$

where $\mathcal{L}_1$ and $\mathcal{L}_2$ are two linear operators. Does the convolution in space $f_1*f_2$ defined as:

$$\quad (f_1*f_2)(x,t) = \int_\mathbb{R}f_1(x-y,t)f_2(y,t)dy$$

satisfy another explicit PDE ?

Thank you very much!

  • $\begingroup$ If $\mathcal{L}_1$ and the integral are interchangeable, then $\mathcal{L}_1(f_1*f_2)(x, t) = 0$ $\endgroup$ Commented Jun 11 at 8:45
  • 1
    $\begingroup$ Thank you @HyperbolicPDEfriend! $\endgroup$
    – NancyBoy
    Commented Jun 11 at 12:27
  • $\begingroup$ Just for clarity: $\mathcal{L}_1$ and $\mathcal{L}_2$ are differential operators, right? $\endgroup$ Commented Jun 11 at 13:47
  • 1
    $\begingroup$ Yes, they are typically differential operators of order 1 and 2 w.r.t. time and space variables $\endgroup$
    – NancyBoy
    Commented Jun 11 at 16:48


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