<Dd> v (x 1, x 2) = u (e (x 1, x 2), f (x 1, x 2)) (\ displaystyle v (x_ (1), x_ (2)) = u (e (x_ (1), x_ (2)), f (x_ (1), x_ (2)))) </Dd> <P> and we operate as follows to go from L u (x, t) = 0 (\ displaystyle (\ mathcal (L)) u (x, t) = 0) to L v (x 1, x 2) = 0: (\ displaystyle (\ mathcal (L)) v (x_ (1), x_ (2)) = 0:) </P> <Ul> <Li> Apply the chain rule to L v (x 1 (x, t), x 2 (x, t)) = 0 (\ displaystyle (\ mathcal (L)) v (x_ (1) (x, t), x_ (2) (x, t)) = 0) and expand out giving equation e 1 (\ displaystyle e_ (1)). </Li> <Li> Substitute a (x, t) (\ displaystyle a (x, t)) for x 1 (x, t) (\ displaystyle x_ (1) (x, t)) and b (x, t) (\ displaystyle b (x, t)) for x 2 (x, t) (\ displaystyle x_ (2) (x, t)) in e 1 (\ displaystyle e_ (1)) and expand out giving equation e 2 (\ displaystyle e_ (2)). </Li> <Li> Replace occurrences of x (\ displaystyle x) by e (x 1, x 2) (\ displaystyle e (x_ (1), x_ (2))) and t (\ displaystyle t) by f (x 1, x 2) (\ displaystyle f (x_ (1), x_ (2))) to yield L v (x 1, x 2) = 0 (\ displaystyle (\ mathcal (L)) v (x_ (1), x_ (2)) = 0), which will be free of x (\ displaystyle x) and t (\ displaystyle t). </Li> </Ul> <Li> Apply the chain rule to L v (x 1 (x, t), x 2 (x, t)) = 0 (\ displaystyle (\ mathcal (L)) v (x_ (1) (x, t), x_ (2) (x, t)) = 0) and expand out giving equation e 1 (\ displaystyle e_ (1)). </Li>

Solving partial differential equations by change of variables