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Let $v(t)$, $t \in [0,1]$, be an unknown parameterized function into $\mathbb{R}^2$ and let $v(t)$ be defined by the differential equation
$$ \left\langle\frac{dv(t)}{dt},u(t)\right\rangle=0, $$
$\langle,\rangle$ denotes inner product with $v(0) =v_0$, $v(1)=v_1$ and $u(t)$ known. Is there a simple way to solve for $v(t)$ in terms of $u(t)$?

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You have $v(t)=(v_1(t),v_2(t))$, $u(t)=(u_1(t),u_2(t))$, $v_0=(a,b)$, $v_1=(c,d)$. Then your equation is $$ v_1'(t)u_1(t)+v_2'(t)u_2(t)=0, $$ with $v_1(0)=a$, $v_1(1)=c$, $v_2(0)=b$, $v_2(1)=d$.

If you only have that equation, you have some freedom. You can choose, say, $v_2$ to be any function satisfying its boundary conditions, and then solve for $v_1$.

For instance, you can make $v_2(t)=b+t(d-b)$, and then solve for $v_1$ the problem $$ v_1'(t)=-\frac{(d-b)u_2(t)}{u_1(t)},\ \ \ v_1(0)=a,\ \ v_1(1)=c. $$

A different choice of $v_2$ might lead a better Neumann problem, but I don't see an immediate way to determine that.

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