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Let
$$ \gamma(s,t)=f(t \cdot g(s))+h(t) $$ where $\gamma$ is a known function of $s \in \mathbb{R}$ and $t \in \mathbb{R}$ while $f$, $g$, and $h$ are unknown functions. Assume $f(0)=f'(0)=h(0)=h'(0)=0$ and $f''(0)=1$. Find all functions $f$, $g$, and $h$ that solve this equation.

Edit: I would be interested if it's even possible to solve $g(\bar{s})$ at some fixed $\bar s$.

Partial attempts: Below are my attempts / incomplete solutions

  • If $g(s)= 0$ then $h(t)=\gamma(0,t)$. If $g'(s) = 0$ then $h(t)=\frac{\partial \gamma(0,t)}{\partial s}$. So $$ \frac{\partial^2 (\gamma(s,t)-h(t))}{\partial t^2}=(g(s))^2 f''(t \cdot g(s)) $$ Evaluating this at $t=0$ leads to $g(s) = \pm \sqrt{\frac{\partial^2 \gamma(s,0)-h(t)}{\partial t^2}}$. Substituting in, we solve for $f$.

  • If $g(0)=c_1 \neq 0$ and $g'(0) =c_2\neq 0$. Then, $$ \frac{\partial^3 \gamma(0,0)}{\partial t^2 \partial s}=2g'(0)g(0) f''(0 \cdot g(0)) +0\cdot g'(0)(g(0))^2 f'''(0 \cdot g(0))=g'(0)g(0) =2c_2c_1\neq 0 $$ Hence, \begin{align*} \frac{\partial \gamma(0,t)}{\partial s} &=tg'(0)f'(t \cdot g(0)) =c_2 tf'(c_1t) \end{align*} So, we can solve for $f$ up to $c_1$ by \begin{align*} \frac{\partial \gamma(0,t)}{\partial s} \Big / \frac{\partial^3 \gamma(0,0)}{\partial t^2 \partial s} &= \frac{t}{2c_1}f'(c_1t) \end{align*} Hence, integrating out we can solve for $f$ up to $c_1$. I'm not sure what to do from here.

  • Alternatively if $g(s) \neq 0$ and $g'(s) \neq 0$, then \begin{align*} \gamma(s,t)&=f(tg(s))+h(t) \\ \frac{\partial \gamma(s,t)}{\partial s}&=tg'(s)f'(tg(s))\\ \frac{\partial^2 \gamma(s,t)}{\partial s \partial t}&=g'(s)f'(tg(s))+tg(s)g'(s)f''(tg(s))\\ \frac{\partial^3 \gamma(s,t)}{\partial s \partial t^2} &=2g(s)g'(s)f''(tg(s))+t(g(s))^2g'(s)f'''(tg(s)) \end{align*} Then, $$ \frac{\partial^3 \gamma(s,0)}{\partial t^2 \partial s}=2g(s)g'(s) f''(0 \cdot g(s)) +0 \cdot (g(s))^2 g'(s) f'''(0 \cdot g(s))=2g(s)g'(s) \neq 0 $$ So, $$ f'''(tg(s))f(tg(s))=\frac{4 (\frac{\partial \gamma(s,t)}{\partial s})(t^2\frac{\partial \gamma^3(s,t)}{\partial s\partial t^2}-2t\frac{\partial \gamma^2(s,t)}{\partial s\partial t}+2\frac{\partial \gamma(s,t)}{\partial s})}{t^4(\left.\frac{\partial^2 f(tg(s))}{\partial s \partial t^2}\right|_{t=0})^2} $$ Now fix $s$. Then by the above equation and initial conditions, we can solve for $f(tg(s))$. Not sure where to go from here.

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I´m not sure, but I have the following results:

The case $g=0$ has already been considered. I will analize two case:

Case 1: $g(s)\neq 0$ for all $s\in \mathbb R$ and $g'\neq 0$. I consider $s_1\in \mathbb R$ and $g(s_1)=c_1\neq 0,$ and $g'(s_1)=c_2\neq 0$

Since $$\gamma(s,t)=h(t)+f(tg(s)) \ \ \ \ \ (1)$$ we have $$h(t)=\gamma(s_1,t)-f(c_1t)$$ moreover $$g^2(s)=\gamma_{tt}(s,0)-h''(0)=\gamma_{tt}(s,0)-\gamma(s_1,0)+c^2_1 \ \ \ \ \ (2)$$ We need find $f$ and $h$.

Deriving the equality number (1) with respecto to $s$, we have $$\gamma_s(s,t)=tf'(tg(s))g'(s) \Rightarrow \gamma_s(s_1,t)=c_2tf'(c_1t), \ \ (3)$$ since $$\gamma_s(s_1,0)=\gamma_{st}(s_1,0)=0$$ then $$\gamma_{s}(s_1,t)=t^2\Psi_1(t) \ \ \ (4)$$

From (3) and (4) $$t\Psi_1(t)=c_2f'(c_1t)$$ Integrating $$f(c_1t)=\frac{c_1}{c_2}\int_{0}^{t} x\Psi_1(x)dx \ \ \ (5),$$ so $$h(t)=\gamma(s_1,t)-f(c_1t)=\frac{c_1}{c_2}\int_{0}^{t} x\Psi_1(x)dx. \ \ \ (6)$$

From (2), (5) and (6) we get the solutions.

Case 2: If $g'=0$ then $g(s)=C\neq 0$, so we rewrite (1): $$\psi(0,t)=f(Ct)+h(t)\Rightarrow h(t)=\psi(0,t)-f(Ct).$$

In this case $h$ depend of $f$.

Regards

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  • $\begingroup$ What is $\Psi_1$ (how do you get equations (4))? Is it possible to express $c_1$ and / or $c_2$ in terms of $\gamma$? $\endgroup$ – user103828 Feb 1 '15 at 7:24
  • $\begingroup$ For (4) I assumed that $\gamma \in C^{\infty}$ then Via Taylor Series we can obtain 4. $\endgroup$ – liliane Feb 1 '15 at 12:39
  • $\begingroup$ Moreover I think that all solution are linearly dependent. $\endgroup$ – liliane Feb 3 '15 at 0:07

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