You signed in with another tab or window. Reload to refresh your session.You signed out in another tab or window. Reload to refresh your session.You switched accounts on another tab or window. Reload to refresh your session.Dismiss alert
if $(x(t),v(t))$ is a solution of this system, then $E(t)=v^2/2+V(x)$ is constant in time with a value $E_0$ determined by the initial condition. To solve this with deepXDE I can (1) just enforce the system (*) in the loss function or (2) I can add an additional term to the loss function $(v^2/2+V(x)-E_0)^2$.
Of course, more complicated systems may have multiple constraints and/or will satisfy additional equations. For example, if in the above system (*) I add a linear "friction" term $-v$ to the rhs of the second equation then the solution satisfies
reacted with thumbs up emoji reacted with thumbs down emoji reacted with laugh emoji reacted with hooray emoji reacted with confused emoji reacted with heart emoji reacted with rocket emoji reacted with eyes emoji
Uh oh!
There was an error while loading. Please reload this page.
-
ODEs and PDEs usually have quantities that are conserved by the dynamics: For example consider the simple system
if$(x(t),v(t))$ is a solution of this system, then $E(t)=v^2/2+V(x)$ is constant in time with a value $E_0$ determined by the initial condition. To solve this with deepXDE I can (1) just enforce the system (*) in the loss function or (2) I can add an additional term to the loss function $(v^2/2+V(x)-E_0)^2$ .
Of course, more complicated systems may have multiple constraints and/or will satisfy additional equations. For example, if in the above system (*) I add a linear "friction" term$-v$ to the rhs of the second equation then the solution satisfies
My question is this: is it better to just solve the "barebone" system, or is it better to add as many additional constrains are known?
Beta Was this translation helpful? Give feedback.
All reactions