- #1
yungman
- 5,741
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\hbox { Let }\; u(x,y)=v(x^2-y^2,2xy) \;\hbox { and let }\; t=x^2-y^2,\;s=2xy
u_x = 2xv_t \;+\; 2yv_s
u_{xx} = 2v_t + 4x^2 v_{tt} + 8xyv_{ts} + 4y^2 v_{ss}
The u_{yy} can be done the same way and is not shown here.
According to Chain Rule:
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}
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}
Where
\frac{\partial y}{\partial x} = 0
Please tell me what I am doing wrong? How do I miss 8xyv_{ts} term?
u_x = 2xv_t \;+\; 2yv_s
u_{xx} = 2v_t + 4x^2 v_{tt} + 8xyv_{ts} + 4y^2 v_{ss}
The u_{yy} can be done the same way and is not shown here.
According to Chain Rule:
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}
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}
Where
\frac{\partial y}{\partial x} = 0
Please tell me what I am doing wrong? How do I miss 8xyv_{ts} term?
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