$u(x)$ satisfies the integral equation $u(x) = \int_0^x \sin(u(t))u(t)^p dt$ on $0 \leq x \leq 1$. show that $u(x) = 0$ on this interval Suppose that $u(x)$ is continuous and satisfies the integral equation
\begin{equation}\label{1.4.1}
        u(x) = \int_0^x \sin(u(t))u(t)^p dt
    \end{equation}
on the interval $0 \leq x \leq 1$. show that $u(x) = 0$ on this interval if $p \geq 0$.
This is what I have:
Since $\sin(u(t))u(t)^p$ is continuous, it follows from the integral definition that $u(x)$ is differentiable. Let us differentiate both sides of equation above with respect to $x$. This yields:
\begin{equation}
    u'(x) = \sin(u(x))u(x)^p
    \end{equation}
This ODE is separabale and becomes:
\begin{equation}
        \frac{1}{\sin(u(x))u(x)^p}du(x) = dx
    \end{equation}
However, this doesn't seem easily solvable so I'm not sure how to show that $u(x) = 0$ from this.
 A: Assume there exists $t\in(0,1)$ such that $u(t)>0$. Because $u$ is continuous and $u(0)=0$, we can identify $a,b\in [0,1]$ such that $u(a)=0,a<t<b,$ and $u>0$ on $(a,b)$. Since $\sin\big(u(t)\big)$ is also continuous, we can find $c\in (a,b)$ such that $\sin \big(u(t)\big)>0$ on $(a,c)$. Now consider the IVP: $$\frac{dy}{dx}=y^p\sin(y):x\in(a,c)$$ $$y(a)=0$$ We know $u$ is a solution to this IVP, so for $x_1,x_2\in (a,c)$ we have $$\int_{u(x_1)}^{u(x_2)}\frac{dy}{y^p\sin(y)}=\int_{x_1}^{x_2}dx=x_2-x_1$$ Take $x_1 \longrightarrow a^{+}$ to get $$\lim_{x_1 \longrightarrow a^{+}}\Bigg[\int_{u(x_1)}^{u(x_2)}\frac{dy}{y^p\sin(y)}\Bigg]=x_2-a$$ This reduces to $$\lim_{h \longrightarrow 0^{+}}\Bigg[\int_{h}^{u(x_2)}\frac{dy}{y^p\sin(y)}\Bigg]=x_2-a$$ However, if $p\geq 0$ the limit on the left side doesn't exist. This is by a limit comparison test to $\lim_{h\longrightarrow 0^{+}}\Big[\int_h^{u(x_2)}\frac{dy}{y^{p+1}}\Big]$. You will arrive at a similar contradiction if we assume $u(t)<0$.
A: WARNING: As pointed out by Matthew in the comments, this isn't as sound as I wanted it to be and has a few holes - namely continuity for $p<1$ and the interval for which uniqueness is guaranteed. Check out his answer, if he posts one!
I'm going to write the differential equation
$$u'=\sin(u)u^p$$
with the initial value condition $u(0)=0$. Notice that $\frac{\partial}{\partial u}\left(\sin(u)u^p\right)$ and $\sin(u)u^p$ are both continuous on the rectangle $[0,1]\times[a,b]$ for any arbitrary $a,b$. Thus, we can guarantee the uniqueness and existence of our solutions.
$u(x)=0$ is a solution to the differential equation $u'=\sin(u)u^p$. So we can argue that if $u(x)=\int_0^x\sin(u(t))u(t)^pdt$, it must satisfy the differential equation $u'=\sin(u)u^p$ with the initial condition $u(0)=0$. Therefore, $u(x)=0$.
(I hope this is correct; it's been a while since I touched a differential equation)
A: Use @Matthew Pilling's setting. Assume there exists $t\in(0,1)$ such that $u(t)>0$. Because $u$ is continuous and $u(0)=0$, we can identify $a,b\in [0,1]$ such that $u(a)=0,a<t<b,$ and $u>0$ on $(a,b)$. Since $\sin\big(u(t)\big)$ is also continuous, we can find $c\in (a,b)$ such that $\sin \big(u(t)\big)>0$ on $(a,c)$. Now consider the IVP:
$$\frac{dy}{dx}=y^p\sin(y),x\in(a,c).$$
So
$$ y'=y^p\sin(y)\le y^{p+1}, x\in(a,c)$$
and hence
$$ y^{-p-1}y'\le 1. \tag1$$
Integrating (1) from $x_0$ to $x$ ($x_0,x\in(a,c)$, $x_0<x$), one has
$$ -\frac1p\bigg[y^{-p}(x)-y^{-p}(x_0)\bigg]\le x-x_0\le1 $$
from which
$$ y^{-p}(x)\ge y^{-p}(x_0)-p. \tag2$$
Choose $x_0$ close to $a$ such that
$$ y^{-p}(x_0)> p $$
and then from (2), one has
$$ 0< y^p(x)\le \frac{y^p(x_0)}{1-py^p(x_0)}. \tag3$$
Letting $x_0\to a^+$ in (3) gives
$$ 0<y^p(x)\le0, x\in(a,c)$$
which is absurd for $x\in(a,c)$.
