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Assume that $\Omega$ is open and connected (if $\Omega$ is not connected, then the desired conclusion is clearly false).

The case $N=1$ is easy, and the case of any natural $N$ reduces to the case of $N=1$.

Indeed, if $v_j$ is the restriction of $w_j$ to any open interval $I$ in $\Omega$ (for $j=1,2$), then $\dfrac{dv_1}{v_1}=\dfrac{dv_2}{v_2}$, $d\ln v_1=d\ln v_2$, and hence $v_1=c_Iv_2$ for some real $c_I>0$, depending only on $I$. Since $\Omega$ is connected, it is linearly connected (in the sense that any two points of $\Omega$ can be connected by finitely many straight line segments). So, $c_I$ does not depend on $I$. $\quad\Box$

Assume that $\Omega$ is open and connected (if $\Omega$ is not connected, then the desired conclusion is clearly false). By convolution with a mollifier and approximation, we may assume that $w_1$ and $w_2$ are smooth.

The case $N=1$ is easy, and the case of any natural $N$ reduces to the case of $N=1$.

Indeed, if $v_j$ is the restriction of $w_j$ to any open interval $I$ in $\Omega$ (for $j=1,2$), then $\dfrac{dv_1}{v_1}=\dfrac{dv_2}{v_2}$, $d\ln v_1=d\ln v_2$, and hence $v_1=c_Iv_2$ for some real $c_I>0$, depending only on $I$. Since $\Omega$ is connected, it is linearly connected (in the sense that any two points of $\Omega$ can be connected by finitely many straight line segments). So, $c_I$ does not depend on $I$. $\quad\Box$

Assume that $\Omega$ is open and connected (if $\Omega$ is not connected, then the desired conclusion is clearly false).

The case $N=1$ is easy, and the case of any natural $N$ reduces to the case of $N=1$.

Indeed, if $v_j$ is the restriction of $w_j$ to any open interval $I$ in $\Omega$ (for $j=1,2$), then $\dfrac{dv_1}{v_1}=\dfrac{dv_2}{v_2}$ and hence $v_1=c_Iv_2$ for some real $c_I>0$, depending only on $I$. Since $\Omega$ is connected, it is linearly connected (in the sense that any two points of $\Omega$ can be connected by finitely many straight line segments). So, $c_I$ does not depend on $I$. $\quad\Box$

Assume that $\Omega$ is open and connected (if $\Omega$ is not connected, then the desired conclusion is clearly false).

The case $N=1$ is easy, and the case of any natural $N$ reduces to the case of $N=1$.

Indeed, if $v_j$ is the restriction of $w_j$ to any open interval $I$ in $\Omega$ (for $j=1,2$), then $\dfrac{dv_1}{v_1}=\dfrac{dv_2}{v_2}$, $d\ln v_1=d\ln v_2$, and hence $v_1=c_Iv_2$ for some real $c_I>0$, depending only on $I$. Since $\Omega$ is connected, it is linearly connected (in the sense that any two points of $\Omega$ can be connected by finitely many straight line segments). So, $c_I$ does not depend on $I$. $\quad\Box$

Assume that $\Omega$ is open and connected (if $\Omega$ is not connected, then the desired conclusion is clearly false).

The case $N=1$ is easy, and the case of any natural $N$ reduces to the case of $N=1$.

Indeed, if $v_j$ is the restriction of $w_j$ to any open interval $I$ in $\Omega$ (for $j=1,2$), then $\dfrac{dv_1}{v_1}=\dfrac{dv_2}{v_2}$ and hence $v_1=c_Iv_2$ for some real $c_I>0$, depending only on $I$. Since $\Omega$ is connected, it is linearly connected. So, $c_I$ does not depend on $I$. $\quad\Box$

Assume that $\Omega$ is open and connected (if $\Omega$ is not connected, then the desired conclusion is clearly false).

The case $N=1$ is easy, and the case of any natural $N$ reduces to the case of $N=1$.

Indeed, if $v_j$ is the restriction of $w_j$ to any open interval $I$ in $\Omega$ (for $j=1,2$), then $\dfrac{dv_1}{v_1}=\dfrac{dv_2}{v_2}$ and hence $v_1=c_Iv_2$ for some real $c_I>0$, depending only on $I$. Since $\Omega$ is connected, it is linearly connected (in the sense that any two points of $\Omega$ can be connected by finitely many straight line segments). So, $c_I$ does not depend on $I$. $\quad\Box$