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Alexey Ustinov
Alexey Ustinov
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I completely disagree with the author of the remark. The Helmholtz equations and the wave equation are equivalent and the solution of the Helmholtz equation $v(\omega, \vec r)$ is associated with the solution of the wave equation $u(t,\vec r)$ $$u(t,\vec r)=\int_{-\infty}^{\infty}exp(i\omega t)v(\omega, \vec r)d\omega$$.$$u(t,\vec r)=\int_{-\infty}^{\infty}exp(i\omega t)v(\omega, \vec r)d\omega.$$ Using $t=t_0$ you get the initial conditions for the wave equation. In addition, it cannot be said that the Navier-Stokes equation is parabolic due to the presence of nonlinearity. The solution of the Navier-Stokes equation with nonlinearity differs significantly from the solution without nonlinearity.

I completely disagree with the author of the remark. The Helmholtz equations and the wave equation are equivalent and the solution of the Helmholtz equation $v(\omega, \vec r)$ is associated with the solution of the wave equation $u(t,\vec r)$ $$u(t,\vec r)=\int_{-\infty}^{\infty}exp(i\omega t)v(\omega, \vec r)d\omega$$. Using $t=t_0$ you get the initial conditions for the wave equation. In addition, it cannot be said that the Navier-Stokes equation is parabolic due to the presence of nonlinearity. The solution of the Navier-Stokes equation with nonlinearity differs significantly from the solution without nonlinearity.

I completely disagree with the author of the remark. The Helmholtz equations and the wave equation are equivalent and the solution of the Helmholtz equation $v(\omega, \vec r)$ is associated with the solution of the wave equation $u(t,\vec r)$ $$u(t,\vec r)=\int_{-\infty}^{\infty}exp(i\omega t)v(\omega, \vec r)d\omega.$$ Using $t=t_0$ you get the initial conditions for the wave equation. In addition, it cannot be said that the Navier-Stokes equation is parabolic due to the presence of nonlinearity. The solution of the Navier-Stokes equation with nonlinearity differs significantly from the solution without nonlinearity.

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Evgeniy Yakubovski
Evgeniy Yakubovski
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I completely disagree with the author of the remark. The Helmholtz equations and the wave equation are equivalent and the solution of the Helmholtz equation $v(\omega, \vec r)$ is associated with the solution of the wave equation $u(t,\vec r)$ $$u(t,\vec r)=\int_{-\infty}^{\infty}exp(i\omega t)v(\omega, \vec r)d\omega$$. Using $t=t_0$ you get the initial conditions for the wave equation. In addition, it cannot be said that the Navier-Stokes equation is parabolic due to the presence of nonlinearity. The solution of the Navier-Stokes equation with nonlinearity differs significantly from the solution without nonlinearity.

I completely disagree with the author of the remark. The Helmholtz equations and the wave equation are equivalent and the solution of the Helmholtz equation $v(\omega, \vec r)$ is associated with the solution of the wave equation $u(t,\vec r)$ $$u(t,\vec r)=\int_{-\infty}^{\infty}exp(i\omega t)v(\omega, \vec r)d\omega$$. In addition, it cannot be said that the Navier-Stokes equation is parabolic due to the presence of nonlinearity.

I completely disagree with the author of the remark. The Helmholtz equations and the wave equation are equivalent and the solution of the Helmholtz equation $v(\omega, \vec r)$ is associated with the solution of the wave equation $u(t,\vec r)$ $$u(t,\vec r)=\int_{-\infty}^{\infty}exp(i\omega t)v(\omega, \vec r)d\omega$$. Using $t=t_0$ you get the initial conditions for the wave equation. In addition, it cannot be said that the Navier-Stokes equation is parabolic due to the presence of nonlinearity. The solution of the Navier-Stokes equation with nonlinearity differs significantly from the solution without nonlinearity.

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Evgeniy Yakubovski
Evgeniy Yakubovski
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I completely disagree with the author of the remark. The Helmholtz equations and the wave equation are equivalent and the solution of the Helmholtz equation $v(\omega, \vec r)$ is associated with the solution of the wave equation $u(t,\vec r)$ $$u(t,\vec r)=\int_{-\infty}^{\infty}exp(i\omega t)v(\omega, \vec r)d\omega$$. In addition, it cannot be said that the Navier-Stokes equation is parabolic due to the presence of nonlinearity.