- #1
PhilDSP
- 643
- 15
I'm using variable substitution to solve a problem. Finding the relationships between the first derivatives of both sets is straightforward using the chain rule, but I'm uncertain if the way I'm determining the second derivative relationships is correct.
Given a description of a problem expressed in the x and y variables, I make the substitution as follows
[itex]r = 3x + y \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ s = x + \frac{1}{5}y[/itex]
The chain rule gives
[itex]\frac{\partial}{\partial x} = 3 \frac{\partial}{\partial r} + \frac{\partial}{\partial s} \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{\partial}{\partial y} = \frac{\partial}{\partial r} + \frac{1}{5} \frac{\partial}{\partial s}[/itex]
Solving for [itex]\frac{\partial}{\partial r}[/itex] and [itex]\frac{\partial}{\partial s}[/itex] I get
[itex]\frac{\partial}{\partial r} = - \frac{1}{2} \frac{\partial}{\partial x} + \frac{5}{2} \frac{\partial}{\partial y} \ \ \ \ \ \ \ \ \frac{\partial}{\partial s} = \frac{5}{2} \frac{\partial}{\partial x} - \frac{15}{2} \frac{\partial}{\partial y}[/itex]
Are the second derivative relationships from the chain rule then ?
[itex]\frac{\partial^2}{\partial x^2} = - \frac{1}{2} \frac{\partial^2}{\partial r^2} + \frac{5}{2} \frac{\partial^2}{\partial s^2} \ \ \ \ \ \ \ \ \frac{\partial^2}{\partial y^2} = \frac{5}{2} \frac{\partial^2}{\partial r^2} - \frac{15}{2} \frac{\partial^2}{\partial s^2}[/itex]
Maybe this a strange example as the [itex]\frac{5}{2} \frac{\partial}{\partial y}[/itex] and [itex]\frac{5}{2} \frac{\partial}{\partial x}[/itex] in the separate equations seem to make the relationships symmetrical.
Given a description of a problem expressed in the x and y variables, I make the substitution as follows
[itex]r = 3x + y \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ s = x + \frac{1}{5}y[/itex]
The chain rule gives
[itex]\frac{\partial}{\partial x} = 3 \frac{\partial}{\partial r} + \frac{\partial}{\partial s} \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{\partial}{\partial y} = \frac{\partial}{\partial r} + \frac{1}{5} \frac{\partial}{\partial s}[/itex]
Solving for [itex]\frac{\partial}{\partial r}[/itex] and [itex]\frac{\partial}{\partial s}[/itex] I get
[itex]\frac{\partial}{\partial r} = - \frac{1}{2} \frac{\partial}{\partial x} + \frac{5}{2} \frac{\partial}{\partial y} \ \ \ \ \ \ \ \ \frac{\partial}{\partial s} = \frac{5}{2} \frac{\partial}{\partial x} - \frac{15}{2} \frac{\partial}{\partial y}[/itex]
Are the second derivative relationships from the chain rule then ?
[itex]\frac{\partial^2}{\partial x^2} = - \frac{1}{2} \frac{\partial^2}{\partial r^2} + \frac{5}{2} \frac{\partial^2}{\partial s^2} \ \ \ \ \ \ \ \ \frac{\partial^2}{\partial y^2} = \frac{5}{2} \frac{\partial^2}{\partial r^2} - \frac{15}{2} \frac{\partial^2}{\partial s^2}[/itex]
Maybe this a strange example as the [itex]\frac{5}{2} \frac{\partial}{\partial y}[/itex] and [itex]\frac{5}{2} \frac{\partial}{\partial x}[/itex] in the separate equations seem to make the relationships symmetrical.