written 8.8 years ago by
teamques10
★ 69k
|
•
modified 8.8 years ago
|
x$
+y=2e^{\theta} cos\phi \; \;
x - y \;=\; 2ie^{\theta} sin\phi
$
$ \\ $
Adding given equations,
$ \\ \; \\ \therefore
2x \;=\; 2e^{\theta} cos\phi \; \;
+ 2ie^{\theta} sin\phi
\\ \; \\
\therefore
x \;=\; e^{\theta} cos\phi \; \;
+ ie^{\theta} sin\phi \;=\; e^{\theta} \cdot e^{i\theta} \;=\; e^{\theta + i\phi}
\\ \; \\
\therefore
\dfrac{\partial x}{\partial \theta} \;=\;
e^{\theta + i\phi} (1) \;= e^{\theta + i\phi} \; = \; x
\\ \; \\
\dfrac{\partial x}{\partial \phi} \;=\;
e^{\theta + i\phi} (1) \;= ie^{\theta + i\phi} \; = \; ix
$
$ \\ $
Subtracting given equations,
$
\therefore
2y \;=\; 2e^{\theta} cos\phi \; \;
- 2ie^{\theta} sin\phi
\\ \; \\
\therefore
y \;=\; e^{\theta} cos\phi \; \;
- ie^{\theta} sin\phi \;=\; e^{\theta} \cdot e^{-i\theta} \;=\; e^{\theta - i\phi}
\\ \; \\
\therefore
\dfrac{\partial y}{\partial \theta} \;=\;
e^{\theta - i\phi} (1) \;= e^{\theta - i\phi} \; = \; y
\\ \; \\
\dfrac{\partial y}{\partial \phi} \;=\;
e^{\theta - i\phi} (-i) \;= -ie^{\theta - i\phi} \; = \; -iy
$
$ \\ $
Given u is a function of $x$ and $y$
$\$
$
\dfrac{\partial u}{\partial \theta} \;=\;
\dfrac{\partial u}{\partial x} \cdot \dfrac{\partial x}{\partial \theta} \;+\;
\dfrac{\partial u}{\partial y} \cdot
\dfrac{\partial y}{\partial \theta}
\\ \; \\
$
$
\therefore
\dfrac{\partial u}{\partial \theta} \;=\;
x\dfrac{\partial u}{\partial x} +
y\dfrac{\partial u}{\partial y}
$
.......From (i) and (ii)
$ \\ $
$
\therefore
\dfrac{\partial^2 u}{\partial \theta^2} \;=\;
\dfrac{\partial}{\partial \theta}
\Bigg[ x\dfrac{\partial u}{\partial x} +
y\dfrac{\partial u}{\partial y} \Bigg]
\\ \; \\ \; \\
= x\dfrac{\partial}{\partial \theta} \Big( \dfrac{\partial u}{\partial x} \Big) +
\dfrac{\partial u}{\partial x}
\dfrac{\partial x}{\partial \theta}
+
y \dfrac{\partial}{\partial \theta}
\Big( \dfrac{\partial u}{\partial y} \Big) +
\dfrac{\partial u}{\partial y} \cdot
\dfrac{\partial y}{\partial \theta}
\\ \; \\ \; \\
=
\dfrac{\partial x}{\partial \theta} \cdot
\dfrac{\partial u}{\partial x} +
x \Bigg[ \dfrac{\partial}{\partial x}
\Big( \dfrac{\partial u}{\partial x} \Big)
\dfrac{\partial x}{\partial \theta} +
\dfrac{\partial}{\partial y}
\Big( \dfrac{\partial u}{\partial x} \Big)
\dfrac{\partial y}{\partial \theta}
\Bigg] \\
+
\dfrac{\partial y}{\partial \theta} \cdot
\dfrac{\partial u}{\partial y}+
y \Bigg[ \dfrac{\partial}{\partial x}
\Big( \dfrac{\partial u}{\partial y} \Big)
\dfrac{\partial x}{\partial \theta} +
\dfrac{\partial}{\partial y}
\Big( \dfrac{\partial u}{\partial y} \Big)
\dfrac{\partial y}{\partial \theta}
\Bigg]
\\ \; \\ \; \\ \; \\
=x \dfrac{\partial u}{\partial x}+
x \bigg[ x\dfrac{\partial^2 u}{\partial x^2}
+ y\dfrac{\partial^2 u}{\partial y \partial x}
\bigg] + \\
y \dfrac{\partial u}{\partial y}+
y \bigg[ x\dfrac{\partial^2 u}{\partial x \partial y}
+ y\dfrac{\partial^2 u}{\partial y^2}
\bigg]
\\ \; \\ \; \\
= x \dfrac{\partial u}{\partial x}+
y \dfrac{\partial u}{\partial y}+
x^2 \dfrac{\partial^2 u}{\partial x^2}
+ xy \dfrac{\partial^2 u}{\partial x \partial y} +
\\ xy \dfrac{\partial^2 u}{\partial x \partial y} +
y^2 \dfrac{\partial^2 u}{\partial y^2} \ldots
\big\{ \dfrac{\partial^2 u}{\partial y \partial x} = \dfrac{\partial^2 u}{\partial x \partial y} \big\}
\\ \; \\ \; \\ \; \\
\therefore
\dfrac{\partial^2 u}{\partial {\theta}^2} \;=\;
x \dfrac{\partial u}{\partial x}+
y \dfrac{\partial u}{\partial y}+
x^2 \dfrac{\partial^2 u}{\partial x^2}
+ 2xy \dfrac{\partial^2 u}{\partial x \partial y} +
y^2 \dfrac{\partial^2 u}{\partial y^2}
$
$\\ \; \\ \; \\$
Now,
$
\dfrac{\partial u}{\partial \phi} \;=\;
\dfrac{\partial u}{\partial x} \cdot
\dfrac{\partial x}{\partial \phi} \;+\;
\dfrac{\partial u}{\partial y} \cdot
\dfrac{\partial y}{\partial \phi}
$
$ \ \; \ $
$
\therefore
\dfrac{\partial u}{\partial \phi} \;=\;
\dfrac{\partial u}{\partial x} (ix) +
\dfrac{\partial u}{\partial y} (-iy)
$ ........From (i) and (ii)
$\$
$
\therefore \dfrac{\partial u}{\partial \phi} \;=\;
i \Bigg[
x \dfrac{\partial u}{\partial x} -y
\dfrac{\partial u}{\partial y}
\Bigg]
$
$ \\ $
$ \dfrac{\partial^2 u}{\partial {\phi}^2} \;=\;
\dfrac{\partial}{\partial {\phi}}
\Bigg\{
i \bigg[
x \dfrac{\partial u}{\partial x} -y
\dfrac{\partial u}{\partial y}
\bigg]
\Bigg\}
\\ \; \\ \; \\ \; \\
=
i \Bigg\{ x
\dfrac{\partial }{\partial \phi} \Big( \dfrac{\partial u}{\partial x} \Big) +
\Big( \dfrac{\partial x}{\partial \phi} \Big)
\Big( \dfrac{\partial u}{\partial x} \Big) \\
-y
\dfrac{\partial }{\partial \phi} \Big( \dfrac{\partial u}{\partial y} \Big) +
\Big( \dfrac{\partial y}{\partial \phi} \Big)
\Big( \dfrac{\partial u}{\partial y} \Big)
\Bigg\}
\\ \; \\ \; \\ \; \\
i \Bigg\{ x
\bigg[ \dfrac{\partial }{\partial x}
\Big( \dfrac{\partial u}{\partial x} \Big)
\dfrac{\partial x}{\partial \phi}
+
\dfrac{\partial }{\partial y}
\Big( \dfrac{\partial u}{\partial x} \Big)
\dfrac{\partial y}{\partial \phi}
\bigg] \;+\;
ix \dfrac{\partial u}{\partial x} \\
-y
\bigg[ \dfrac{\partial }{\partial x}
\Big( \dfrac{\partial u}{\partial y} \Big)
\dfrac{\partial y}{\partial \phi}
+
\dfrac{\partial }{\partial y}
\Big( \dfrac{\partial u}{\partial x} \Big)
\dfrac{\partial y}{\partial \phi}
\bigg] \;+\;
ix \dfrac{\partial u}{\partial x}
- (-iy \dfrac{\partial u}{\partial y} )
\Bigg\}
\\ \; \\ \; \\ \; \\
i \Bigg\{
x \bigg[ (ix) \dfrac{\partial^2 u}{\partial x^2}
+ (-iy) \dfrac{\partial^2 u}{\partial y \partial x}
\bigg]
+ ix\dfrac{\partial u}{\partial x} \\
-
y \bigg[ (ix) \dfrac{\partial^2 u}{\partial x \partial y}
+ (-iy) \dfrac{\partial^2 u}{\partial y^2}
\bigg]
+ iy\dfrac{\partial u}{\partial y}
\Bigg\}
\\ \; \\ \; \\ \; \\
= i^2
\Bigg \{ x^2
\dfrac{\partial^2 u}{\partial x^2} -xy \dfrac{\partial^2 u}{\partial y \partial x} +
x \dfrac{\partial u}{\partial x} \\ -xy \dfrac{\partial^2 u}{\partial x \partial y}
+ y^2\dfrac{\partial^2 u}{\partial y^2}
+ y \dfrac{\partial u}{\partial y}
\Bigg\}
$
$ \\ \; \\ \; \\ $
$ =
-1
\Bigg \{ x^2
\dfrac{\partial^2 u}{\partial x^2} -2xy \dfrac{\partial^2 u}{\partial y \partial x} +
x \dfrac{\partial u}{\partial x}
+ y^2\dfrac{\partial^2 u}{\partial y^2}
+ y \dfrac{\partial u}{\partial y}
\Bigg\}
$
$ \\ \; \; i^2 =-1 \\
\& \dfrac{\partial^2 u}{\partial y \partial x} \;=\;
\dfrac{\partial^2 u}{\partial x \partial y}
\\ \; \\ \; \\ \; \\
\therefore
\dfrac{\partial^2 u}{\partial \phi^2} \;=\;
-x^2 \dfrac{\partial^2 u}{\partial x^2} +
\\ 2xy \dfrac{\partial^2 u}{\partial y \partial x}
-y^2 \dfrac{\partial^2 u}{\partial y^2}-
x \dfrac{\partial u}{\partial x}
-
y \dfrac{\partial u}{\partial y} \; \; \ldots (iv)
\\ \; \\ \; \\ \; \\
\therefore
\dfrac{\partial^2 u}{\partial \theta^2} +
\dfrac{\partial^2 u}{\partial \phi^2}
\; = \;
x^2 \dfrac{\partial^2 u}{\partial x^2} +
2xy \dfrac{\partial^2 u}{\partial y \partial x}
+y^2 \dfrac{\partial^2 u}{\partial y^2}+
x \dfrac{\partial u}{\partial x}
+ \\
y \dfrac{\partial u}{\partial y}
-x^2 \dfrac{\partial^2 u}{\partial x^2} +
\\ 2xy \dfrac{\partial^2 u}{\partial y \partial x}
-y^2 \dfrac{\partial^2 u}{\partial y^2}-
x \dfrac{\partial u}{\partial x}
-
y \dfrac{\partial u}{\partial y}
\\ \; \\ \; \\ \; \\
= 4xy \dfrac{\partial^2 u}{\partial x \partial y}
\\ \; \\ \; \\ \; \\
$
$
\dfrac{\partial^2 u}{\partial \theta^2} +
\dfrac{\partial^2 u}{\partial \phi^2} \;=\;
4xy \dfrac{\partial^2 u}{\partial x \partial y}
$ ............Hence Proved
and 3 others joined a min ago.
teamques10
★ 69k