I am trying to find an expression for a change of variables (invertible and $C^{\infty}$) of a linear differential operator in $\mathbb R ^d \newcommand{\t}{\tilde} \newcommand{\p}{\partial}$.

i) case $d = 1$. A linear differential operator has the form $$ A = a_0 + a_1 \dfrac{d}{dx}.$$ Consider the change of variable $ y = f(x) $. I want to find a differential operator $\t A = \t a_0 + \t a_1 \dfrac{d}{dy}$ s.t. $$ \t A( u(y))|_{y=f(x)} = A(u(x)).$$

To obtain this identity, it must be $$ \t a_0(f(x)) + \t a_1(f(x)) \dfrac{d u(y)}{dy} \Bigg|_{y=f(x)} = a_0(x) + a_1(x) \dfrac{du(x)}{dx}.$$ But how should I take $\t a_0, \t a_1$ to obtain this equality?

ii) case $ d = 2$. A linear differential operator has the form $$ A= a_{00} + a_{10} \dfrac{\p}{\p x_1} + a_{01} \dfrac{\p}{\p x_2} + a_{11} \dfrac{\p^2}{\p x_2} + a_{20} \dfrac{\p^2}{\p x_1 ^2} + a_{02} \dfrac{\p^2}{\p x_2^2}.$$ Considering the change of variable $Y = (y_1, y_2) = (f_1(x_1, x_2), f_2(x_1, x_2)) = F(X)$, we will have that the operator $\t A$ expresses $A$ in the new coordinates if $$ \t a_{00}(F(X)) + \dots + \t a_{02}(F(X)) \dfrac{\p^2 u(y_1, y_2))}{\p y_2^2} \Bigg|_{Y= F(X)} \\ = a_{00}(X) + \dots + a_{02}(X) \dfrac{\p^2 u(x_1, x_2)}{\p x_2^2}.$$ Again, how can I find the coefficients $\t a_{ij}$?

iii) general case. Consider in $\mathbb R ^d $ the change of coordinates $$Y = (y_1, \dots, y_d) = (f_1(x_1, \dots, x_d), \dots , f_d(x_1, \dots, x_d)) = F(X).$$ I will the use multi-index notation $$ P = (p_1, \dots, p_d), |P| = p_1 + \dots + p_d,\\ \p ^P_X = \dfrac{\p ^{|P|}}{\p x_1^{p_1} \dots \p x_d^{p_d}}, \p ^P_Y = \dfrac{\p ^{|P|}}{\p y_1^{p_1} \dots \p y_d^{p_d}}.$$ A linear differential operator has the form $$ A = \sum_{|P| \leq m} a_P(X) \p ^P_X. $$ How can I find the coefficients of $$ \t A = \sum_{|P| \leq m} \t a_P(Y) \p ^P_Y $$ so that it is the expression of $A$ in the new coordinates $Y$?

  • $\begingroup$ For the first case: since $\frac{d}{dx}=\frac{dy}{dx}\frac{d}{dy}$ then $\tilde{a_0}=a_0$ and $\tilde{a_1}=a_1y'(x)$. $\endgroup$ – janmarqz Jan 8 '14 at 2:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.