- #1
yungman
- 5,723
- 242
[tex]\hbox { Let }\; u(x,y)=v(x^2-y^2,2xy) \;\hbox { and let }\; t=x^2-y^2,\;s=2xy[/tex]
[tex]u_x = 2xv_t \;+\; 2yv_s[/tex]
[tex]u_{xx} = 2v_t + 4x^2 v_{tt} + 8xyv_{ts} + 4y^2 v_{ss}[/tex]
The [itex] u_{yy}[/itex] can be done the same way and is not shown here.
According to Chain Rule:
[tex]u_x = \frac{\partial v}{\partial x} \;=\; \frac{\partial v}{\partial t}\frac{\partial t}{\partial x} \;+\; \frac{\partial v}{\partial s}\frac{\partial s}{\partial x} \;=\; 2x\frac{\partial v}{\partial t} \;+\; 2y\frac{\partial v}{\partial s} [/tex]
[tex] u_{xx} = \frac{\partial^2 v}{\partial x^2} = \frac{\partial}{\partial x}[2x\frac{\partial v}{\partial t} \;+\; 2y\frac{\partial v}{\partial s}] = 4x^2\frac{\partial^2 v}{\partial t^2} \;+\; 2\frac{\partial v}{\partial t} \;+\; 4y^2 \frac{\partial^2 v}{\partial s^2} [/tex]
Where
[tex] \frac{\partial y}{\partial x} = 0 [/tex]
Please tell me what I am doing wrong? How do I miss [itex]8xyv_{ts}[/itex] term?
[tex]u_x = 2xv_t \;+\; 2yv_s[/tex]
[tex]u_{xx} = 2v_t + 4x^2 v_{tt} + 8xyv_{ts} + 4y^2 v_{ss}[/tex]
The [itex] u_{yy}[/itex] can be done the same way and is not shown here.
According to Chain Rule:
[tex]u_x = \frac{\partial v}{\partial x} \;=\; \frac{\partial v}{\partial t}\frac{\partial t}{\partial x} \;+\; \frac{\partial v}{\partial s}\frac{\partial s}{\partial x} \;=\; 2x\frac{\partial v}{\partial t} \;+\; 2y\frac{\partial v}{\partial s} [/tex]
[tex] u_{xx} = \frac{\partial^2 v}{\partial x^2} = \frac{\partial}{\partial x}[2x\frac{\partial v}{\partial t} \;+\; 2y\frac{\partial v}{\partial s}] = 4x^2\frac{\partial^2 v}{\partial t^2} \;+\; 2\frac{\partial v}{\partial t} \;+\; 4y^2 \frac{\partial^2 v}{\partial s^2} [/tex]
Where
[tex] \frac{\partial y}{\partial x} = 0 [/tex]
Please tell me what I am doing wrong? How do I miss [itex]8xyv_{ts}[/itex] term?
Last edited: