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I'm trying to find the class of differential equations $\frac{\mathrm dx}{\mathrm dt} = f(t,x)$ with $f$ continuous that are invariant under the change of variables $s=2t$ and $y=-x$.

I have come up with this functional equation: $$-\frac{1}{2}f(t,x) = f(2t,-x) \text.$$

Letting $f(t,x) = \frac{q(x)}{p(t)}$ with $p$ and $q$ multiplicative and $p(t) \ne 0$ for all $t$, we get a family of solutions; because $$ f(2t, -x) = \frac{q(-x)}{p(2t)} = \frac{-q(x)}{2p(t)} = -\frac{1}{2} f(t,x) \text. $$ $f(t,x) = \log_a\left(k^{\frac{q(x)}{p(t)}}\right)$ with $k \in \mathbb{R}^+$ gives another type of solutions; because $$ f(2t, -x) = \log_a\left(k^{\frac{-q(x)}{2p(t)}}\right) = -\frac{1}{2} \log_a\left(k^{\frac{q(x)}{p(t)}}\right)= -\frac{1}{2} f(t,x) \text. $$

However, are they all the solutions? Is there a standard method to solve this?

Mohsen Shahriari
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Ricardo
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  • There are very few multiplicative continuous functions. You are impossing a very strong conidtion on $p$ and $q$. – Kavi Rama Murthy Aug 22 '21 at 11:44
  • Don't you mean "homogeneous (of degree $ 1 $)" instead of "multiplicative"? – Mohsen Shahriari Aug 22 '21 at 19:15
  • Also, note that the second kind of functions you've mentioned are in fact the same as the first, as $$ \log _ a \left( k ^ { \frac { p ( x ) } { q ( t ) } } \right) = \left( \log _ a k \right) \frac { p ( x ) } { q ( t ) } = \frac { r ( x ) } { q ( t ) } \text , $$ where $ r $ is another homogeneous (?) function defined with $ r ( x ) = ( \log _ a k ) p ( x ) $. – Mohsen Shahriari Aug 22 '21 at 19:24
  • Additionally, even for your family of examples, the only requred properties for $ p $ and $ q $ are $ q $ being odd and $ p $ satisfying $ p ( 2 x ) = 2 p ( x ) $ for all $ x $. – Mohsen Shahriari Aug 22 '21 at 19:37
  • And finally, your example doesn't work! That's because $ p ( 2 \times 0 ) = 2 p( 0 ) $ implies $ p ( 0 ) = 0 $, and that can't appear in the denominator. Even if you only assume that $ p $ is continuous at $ 0 $ and $ p ( 2 x ) = 2 p ( x ) $ is satisfied for $ x \ne 0 $, you can show that $ p \left( \frac x { 2 ^ n } \right) = \frac { p ( x ) } { 2 ^ n } $ for any nonnegative integer $ n $, which taking the limit as $ n $ tends to infinity, implies $ p ( 0 ) = 0 $. Note that as you've required $ f $ to be continuous, $ p $ being continuous at $ 0 $ is not a very harsh condition. – Mohsen Shahriari Aug 22 '21 at 19:47

1 Answers1

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The only continuous solution to the functional equation $$ - \frac 1 2 f ( t , x ) = f ( 2 t , - x ) \tag 0 \label 0 $$ is the constant zero function (which obviously is a solution). To see that, substitute $ \frac t 2 $ for $ t $ and $ - x $ for $ x $ in \eqref{0} to get $$ f \left( \frac t 2 , - x \right) = - 2 f ( t , x ) \text . \tag 1 \label 1 $$ Again, substitute $ \frac t 2 $ for $ t $ and $ - x $ for $ x $ in \eqref{1} and use \eqref{1} itself to see that $$ f \left( \frac t 4 , x \right) = - 2 f \left( \frac t 2 , - x \right) = 4 f ( t , x ) \text . \tag 2 \label 2 $$ Now, using \eqref{2} and mathematical induction, you can prove $$ f \left( \frac t { 4 ^ n } , x \right) = 4 ^ n f ( t , x ) \tag 3 \label 3 $$ for any nonnegative integer $ n $. The left-hand side of \eqref{3} has the limit $ f ( 0 , x ) $ as $ n $ tends to infinity, by continuity of $ f $ at $ ( 0 , x ) $. But if $ f ( t , x ) \ne 0 $, then the right-hand side of \eqref{3} becomes unbounded as $ n $ tends to infinity, and cannot have a finite limit. Therefore, we must have $ f ( t , x ) = 0 $. As this could be argued for arbitrary $ t $ and $ x $, the value of $ f $ must be constantly zero.

Mohsen Shahriari
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