How can I proceed with this equation?

[Metronome]

I am trying to solve [math]\begin{cases}\sin(\pi x) = h(x) + g(-x) \\ 0 = \frac{\partial}{\partial t}[h(2t + x) + g(2t - x)]_{t = 0}\end{cases}[/math]
For the derivatives, I have been confused by ambiguity in Lagrange's notation (prime symbols), so I am trying to do the whole thing very slowly using Leibniz's notation (d and [imath]\partial[/imath] symbols). Starting with the second equation, I make the substitutions [imath]y = 2t + x[/imath] and [imath]z = 2t - x[/imath] and get [imath]0 = \frac{\partial}{\partial t}[h(y) + g(z)]_{t = 0} \implies 0 = [\frac{dh(y)}{dy} * 2 + \frac{dg(z)}{dz} * 2]_{t = 0} \implies 0 = [\frac{dh(y)}{dy} + \frac{dg(z)}{dz}]_{t = 0}[/imath].

It seems too soon to plug [imath]0[/imath] in for [imath]t[/imath],since there are still derivatives which have not yet been resolved. How can I proceed from here (without using Lagrange's notation for derivatives)?

 

[bigjohn 2]

on the right track......you know that h ' ( x ) + g ' (-x ) = 0...[ when t = 0 ]..now look at the derivatives of the 1st equation

 

[Metronome]

Heya, I am not clear on the leap from my final step to the statement you made. Also, I mentioned in the problem that I am trying to completely avoid the prime derivative notation in favor of Leibniz notation, as the former has too many meanings.

 

[Cubist]

I don't know how to derive the answer step-by-step, but somehow I made a lucky guess at the final answer (or at least one possible answer). Maybe this will help you to find a complete solution, so here it is within a spoiler...

Spoiler: answer

h(x) = sin(pi*x)/2 + c
g(x) = -sin(pi*x)/2 - c

I'm not sure if the substitutions that you made will help. In my own, failed, attempt I rewrote the first equation as

[math]h(x) = \sin(\pi x) - g(-x)[/math]
and then substituted this into the second equation BEFORE doing anything with the partial derivative. But I wasn't able to make much progress after that

 

[Cubist]

I'm not sure if the substitutions that you made will help. In my own, failed, attempt I rewrote the first equation as

[math]h(x) = \sin(\pi x) - g(-x)[/math]
and then substituted this into the second equation BEFORE doing anything with the partial derivative. But I wasn't able to make much progress after that

I've got a step-by-step answer now. It turns out that my suggestion (quoted above) is a good starting point. OP, can you continue from there?

 

[Metronome]

I've got a step-by-step answer now. It turns out that my suggestion (quoted above) is a good starting point. OP, can you continue from there?

[imath]h(x) = \sin(πx) − g(−x)\newline 0 = \frac{\partial}{\partial t}[\sin(\pi(2t + x)) - g(-(2t + x)) + g(2t - x)]_{t = 0}\newline 0 = 2\pi \cos(\pi x) + \frac{\partial}{\partial t}[g(2t - x) - g(-2t - x)]_{t = 0}\newline v = 2t - x, w = -2t - x\newline 0 = 2\pi \cos(\pi x) + \frac{\partial}{\partial t}[g(v) - g(w)]_{t = 0}\newline 0 = 2\pi \cos(\pi x) + \frac{\partial}{\partial t}[2\frac{dg(v)}{dv} - (-2)\frac{dg(w)}{dw}]_{t = 0}\newline 0 = \pi \cos(\pi x) + \frac{\partial}{\partial t}[\frac{dg(v)}{dv} + \frac{dg(w)}{dw}]_{t = 0}\newline[/imath]

I think this might be worse because now I have [imath]g[/imath] as a function of two different variables appearing in the equation, and still have the original problem of how to proceed from here?

 

[Cubist]

Thanks for showing your attempt! While I was writing up a hint about my step-by-step solution I noticed a mistake that I'd made (a plus instead of a minus). So unfortunately I don't actually have a solution to guide you. I'm probably going to be busy on other things tomorrow, therefore I'll show you a work-in-progress idea that might help you to see a way forward. However it contains a couple of very unsatisfactory steps...

Spoiler: my attempt

Sorry, I didn't have time to LaTeX

d/dt [ sin(pi*(2*t+x)) - g(-(2*t+x)) + g(2*t-x) ] = 0

Calculate the partial derivative, but delay putting t=0

 pi*cos(pi*(2*t + x)) + dg(-2*t-x)/dt + dg(2*t-x)/dt = 0

Let q(u) = g(u - x)

 pi*cos(pi*(2*t + x)) + dq(-2*t)/dt + dq(2*t)/dt = 0

Make an unsatisfactory GUESS that function q is odd. It certainly can't be even otherwise q'(-x) = -q'(x)
Since odd q'(-x) = q'(x). Thus...

 pi*cos(pi*(2*t + x)) + 2*dq(-2*t)/dt = 0
                             |
For some reason it doesn't work if we put "+" instead of "-" in above. Another unsatisfactory thing :-(

   dq(-2*t)/dt = -pi*cos(pi*(2*t + x))/2

Integrate wrt t, and add a constant of integration

-q(-2*t)/2 = - sin(pi*(2*t + x))/4 + c/2

Put t=0

q(0) = sin(pi*x)/2 + c = g(-x)

Therefore

g(x) = - sin(pi*x)/2 - c

 

[Metronome]

Suppose I invent a nonstandard notation to clarify this. For the purposes of this question, I will stipulate an expansion of Leibniz's notation and denote it in [imath]\color{blue}{blue}[/imath]. The expansion is as follows: Expressions can appear in the "numerator" and/or "denominator" of ordinary and partial derivatives (i.e., if [imath]b = a^4[/imath], then [imath]\frac{dc}{db}[/imath] could be written as [imath]\color{blue}{\frac{dc}{d(a^4)}}[/imath]; additionally, a value in parentheses to be plugged in for a variable (or expression) after differentiation also includes the variable, in the form "variable [imath]\rightarrow[/imath] value" (i.e., [imath]\color{blue}{\frac{dc}{d(a^4)}(a^4 \rightarrow 2)}[/imath] more clearly states what might be meant by the more standard but ambiguous [imath]c'(2)[/imath]). Now onto the problem.



[math]\begin{cases}\sin(\pi x) = h(x) + g(-x) \\ 0 = \frac{\partial}{\partial t}[h(2t + x) + g(2t - x)]_{t = 0}\end{cases}[/math]


By the chain rule, the second equation becomes [imath]0 = 2\color{blue}{\frac{dh}{d(2t + x)}(t \rightarrow 0)} + 2\color{blue}{\frac{dg}{d(2t - x)}(t \rightarrow 0)}[/imath], easily rewritten as [imath]\color{blue}{\frac{dh}{d(2t + x)}(t \rightarrow 0)} + \color{blue}{\frac{dg}{d(2t - x)}(t \rightarrow 0)} = 0[/imath], but because the functions [imath]h(x,\ t)[/imath] and [imath]g(x,\ t)[/imath] are such that they only contain [imath]t[/imath] in the pattern of [imath]2t + x[/imath] and [imath]2t - x[/imath], respectively, this is the same as [imath]\color{blue}{\frac{dh}{d(2t + x)}((2t + x) \rightarrow x)} + \color{blue}{\frac{dg}{d(2t - x)}((2t - x) \rightarrow -x)} = 0[/imath]. This is ostensibly what bigjohn 2 means by the notation [imath]h'(x) + g'(-x) = 0[/imath].



The substitution [imath]\tilde g(x) = g(-x)[/imath] leads to the system



[math]\begin{cases}h(x) + \tilde g(x) = \sin(πx) \\ \color{blue}{\frac{dh}{d(2t + x)}((2t + x) \rightarrow x)} - \color{blue}{\frac{d\tilde g}{d(2t - x)}((2t - x) \rightarrow x)} = 0\end{cases}[/math]


The second equation now needs to be antidifferentiated, but I don't know with respect to what, so I have [imath]\int \color{blue}{\frac{dh}{d(2t + x)}((2t + x) \rightarrow x)} - \color{blue}{\frac{d\tilde g}{d(2t - x)}((2t - x) \rightarrow x)}\ d? = \int 0\ d?[/imath]. My first hypothesis was that I should antidifferentiate both sides with respect to [imath]t[/imath], producing an arbitrary function [imath]f(x)[/imath] on the RHS, because the equation was partially differentiated with respect to [imath]t[/imath] *initially*. I also considered antidifferentiating with respect to [imath]2t + x[/imath] or [imath]2t - x[/imath], because these are the expressions [imath]h[/imath] and [imath]\tilde g[/imath] are respectively differentiated with respect to *currently*, but the problem with such an approach is that I cannot antidifferentiate different terms in the equation with respect to different expressions. Finally, I could antidifferentiate with respect to [imath]x[/imath], producing an arbitrary function [imath]f(t)[/imath] on the RHS.



How should this equation be antidifferentiated? With respect to what variable or expression is the antiderivative, what is the rational behind this choice, and what is the solution to this antidifferentiated equation? Please do not rely on any understanding being conveyed by Lagrange's notation.