Show the equation\(\left( {\frac{{{\partial ^2}u}}{{\partial {x^2}}}} \right) + \left( {\frac{{{\partial ^2}u}}{{\partial {y^2}}}} \right) = {e^{ - 2s}}\left( {\left( {\frac{{{\partial ^2}u}}{{\partial {s^2}}}} \right) + \left( {\frac{{{\partial ^2}u}}{{\partial {t^2}}}} \right)} \right)\) if\(u = f(x,y){\rm{,}}\) where\({\rm{ }}x = {e^s}\cos t\) and\(y = {e^s}\sin t.\)

The function is\({\rm{z = u(x, y)}}.\)

Obtain the partial derivative\(\frac{{\partial u}}{{\partial s}}.\)

\(\begin{aligned}{l}\frac{{\partial u}}{{\partial s}} = \frac{{\partial u}}{{\partial x}} \cdot \frac{{\partial x}}{{\partial s}} + \frac{{\partial u}}{{\partial y}} \cdot \frac{{\partial y}}{{\partial s}}\\ = \frac{{\partial u}}{{\partial x}} \cdot \frac{\partial }{{\partial s}}\left( {{e^s}\cos t} \right) + \frac{{\partial u}}{{\partial y}} \cdot \frac{\partial }{{\partial s}}\left( {{e^s}\sin t} \right)\\ = \frac{{\partial u}}{{\partial x}} \cdot {e^s}\cos t + \frac{{\partial u}}{{\partial y}} \cdot {e^s}\sin t\end{aligned}\)

Thus, the partial derivative\(\frac{{\partial u}}{{\partial s}} = \frac{{\partial u}}{{\partial x}} \cdot {e^s}\cos t + \frac{{\partial u}}{{\partial y}} \cdot {e^s}\sin t.\)

Obtain the partial derivative\(\frac{{du}}{{dt}}.\)

\(\begin{aligned}{l}\frac{{\partial u}}{{\partial t}} = \frac{{\partial u}}{{\partial x}} \cdot \frac{{\partial x}}{{\partial t}} + \frac{{\partial u}}{{\partial y}} \cdot \frac{{\partial y}}{{\partial t}}\\ = \frac{{\partial u}}{{\partial x}} \cdot \frac{\partial }{{\partial t}}\left( {{e^s}\cos t} \right) + \frac{{\partial u}}{{\partial y}} \cdot \frac{\partial }{{\partial t}}\left( {{e^s}\sin t} \right)\\ = \frac{{\partial u}}{{\partial x}}\left( { - {e^s}\sin t} \right) + \frac{{\partial u}}{{\partial y}}\left( {{e^s}\cos t} \right)\end{aligned}\)

Thus, the partial derivative\(\frac{{\partial u}}{{\partial t}} = \frac{{\partial u}}{{\partial x}}\left( { - {e^s}\sin t} \right) + \frac{{\partial u}}{{\partial y}}\left( {{e^s}\cos t} \right).\)

Obtain\(\frac{{{\partial ^2}u}}{{\partial {s^2}}}.\)

Take partial derivative of\(\frac{{\partial u}}{{\partial s}}.\)with respect to\(s.\)

\(\begin{aligned}{l}\frac{{{\partial ^2}u}}{{\partial {s^2}}} = \frac{{\partial u}}{{\partial x}} \cdot {e^s}\cos t + {e^s}\cos t \cdot \frac{\partial }{{\partial s}}\left( {\frac{{\partial u}}{{\partial x}}} \right) + \frac{{\partial u}}{{\partial y}} \cdot {e^s}\sin t + {e^s}\sin t \cdot \frac{\partial }{{\partial s}}\left( {\frac{{\partial u}}{{\partial y}}} \right)\\ = \left( {\begin{aligned}{*{20}{l}}{\frac{{\partial u}}{{\partial x}} \cdot {e^s}\cos t + {e^s}\cos t\left( {\frac{{{\partial ^2}u}}{{\partial {x^2}}} \cdot \frac{{\partial x}}{{\partial s}} + \frac{{{\partial ^2}u}}{{\partial y\partial x}} \cdot \frac{{\partial y}}{{\partial s}}} \right) + }\\{\frac{{\partial u}}{{\partial y}} \cdot {e^s}\sin t + {e^s}\sin t\left( {\frac{{{\partial ^2}u}}{{\partial {y^2}}} \cdot \frac{{\partial y}}{{\partial s}} + \frac{{{\partial ^2}u}}{{\partial x\partial y}} \cdot \frac{{\partial x}}{{\partial s}}} \right)}\end{aligned}} \right)\\ = \left( {\begin{aligned}{*{20}{c}}{\frac{{\partial u}}{{\partial x}} \cdot {e^s}\cos t + {e^s}\cos t\left( {\frac{{{\partial ^2}u}}{{\partial {x^2}}} \cdot {e^s}\cos t + \frac{{{\partial ^2}u}}{{\partial y\partial x}} \cdot {e^s}\sin t} \right) + }\\{\frac{{\partial u}}{{\partial y}} \cdot {e^s}\sin t + {e^s}\sin t\left( {\frac{{{\partial ^2}u}}{{\partial {y^2}}} \cdot {e^s}\sin t + \frac{{{\partial ^2}u}}{{\partial x\partial y}} \cdot {e^s}\cos t} \right)}\end{aligned}} \right)\end{aligned}\)

The partial derivatives\(\frac{{{\partial ^2}u}}{{\partial y\partial x}} = \frac{{{\partial ^2}u}}{{\partial x\partial y}}\), as functions are continuous.

Simplify\(\frac{{{\partial ^2}u}}{{\partial {s^2}}}\)as follows.

\(\begin{aligned}{l}\frac{{{\partial ^2}u}}{{\partial {s^2}}} = \left( {\begin{aligned}{*{20}{l}}{\frac{{\partial u}}{{\partial x}} \cdot {e^s}\cos t + {e^s}\cos t\left( {\frac{{{\partial ^2}u}}{{\partial {x^2}}} \cdot {e^s}\cos t + \frac{{{\partial ^2}u}}{{\partial x\partial y}} \cdot {e^s}\sin t} \right) + }\\{\frac{{\partial u}}{{\partial y}} \cdot {e^s}\sin t + {e^s}\sin t\left( {\frac{{{\partial ^2}u}}{{\partial {y^2}}} \cdot {e^s}\sin t + \frac{{{\partial ^2}u}}{{\partial x\partial y}} \cdot {e^s}\cos t} \right)}\end{aligned}} \right)\\ = \left( {\begin{aligned}{*{20}{l}}{\frac{{\partial u}}{{\partial x}} \cdot {e^s}\cos t + \frac{{\partial u}}{{\partial y}} \cdot {e^s}\sin t + {e^{2s}}{{\cos }^2}t \cdot \frac{{{\partial ^2}u}}{{\partial {x^2}}} + }\\{2{e^{2s}}\cos t\sin t \cdot \frac{{{\partial ^2}u}}{{\partial x\partial y}} + {e^{2s}}{{\sin }^2}t \cdot \frac{{{\partial ^2}u}}{{\partial {y^2}}}}\end{aligned}} \right)\end{aligned}\)

Similarly, find the value of\(\frac{{{\partial ^2}u}}{{\partial {t^2}}}\) as shown.

Obtain\(\frac{{{\partial ^2}u}}{{\partial {t^2}}}\)

Take partial derivative on both sides of\(\frac{{\partial u}}{{\partial t}}\) with respect to\(t.\)

\(\begin{aligned}{l}\frac{{{\partial ^2}u}}{{\partial {t^2}}} = \frac{{\partial u}}{{\partial x}} \cdot \left( { - {e^s}\cos t} \right) - {e^s}\sin t \cdot \frac{\partial }{{\partial t}}\left( {\frac{{\partial u}}{{\partial x}}} \right) + \frac{{\partial u}}{{\partial y}} \cdot \left( { - {e^s}\sin t} \right) + {e^s}\cos t \cdot \frac{\partial }{{\partial t}}\left( {\frac{{\partial u}}{{\partial y}}} \right)\\ = \left( {\begin{aligned}{*{20}{l}}{\left. {\frac{{\partial u}}{{\partial x}} \cdot \left( { - {e^s}\cos t} \right) - {e^s}\sin t\left( {\frac{{{\partial ^2}u}}{{\partial {x^2}}} \cdot \frac{{\partial x}}{{\partial t}} + \frac{{{\partial ^2}u}}{{\partial y\partial x}} \cdot \frac{{\partial y}}{{\partial t}}} \right) - } \right)}\\{\frac{{\partial u}}{{\partial y}} \cdot {e^s}\sin t + {e^s}\cos t\left( {\frac{{{\partial ^2}u}}{{\partial {y^2}}} \cdot \frac{{\partial y}}{{\partial s}} + \frac{{{\partial ^2}u}}{{\partial x\partial y}} \cdot \frac{{\partial x}}{{\partial s}}} \right)}\end{aligned}} \right)\\ = \left( {\begin{aligned}{*{20}{l}}{\frac{{\partial u}}{{\partial x}} \cdot \left( { - {e^s}\cos t} \right) - {e^s}\sin t\left( {\frac{{{\partial ^2}u}}{{\partial {x^2}}} \cdot {e^s}\cos t + \frac{{{\partial ^2}u}}{{\partial y\partial x}} \cdot {e^s}\sin t} \right) - }\\{\frac{{\partial u}}{{\partial y}} \cdot {e^s}\sin t + {e^s}\cos t\left( {\frac{{{\partial ^2}u}}{{\partial {y^2}}} \cdot {e^s}\sin t + \frac{{{\partial ^2}u}}{{\partial x\partial y}} \cdot {e^s}\cos t} \right)}\end{aligned}} \right)\end{aligned}\)

The partial derivatives\(\frac{{{\partial ^2}u}}{{\partial y\partial x}} = \frac{{{\partial ^2}u}}{{\partial x\partial y}}\), as the functions are continuous.

Thus, it can be simplified as follows.

\(\begin{aligned}{l}\frac{{{\partial ^2}u}}{{\partial {t^2}}} = \left( {\begin{aligned}{*{20}{l}}{\frac{{\partial u}}{{\partial x}} \cdot \left( { - {e^s}\cos t} \right) - {e^s}\sin t\left( {\frac{{{\partial ^2}u}}{{\partial {x^2}}} \cdot {e^s}\cos t + \frac{{{\partial ^2}u}}{{\partial y\partial x}} \cdot {e^s}\sin t} \right) - }\\{\frac{{\partial u}}{{\partial y}} \cdot {e^s}\sin t + {e^s}\cos t\left( {\frac{{{\partial ^2}u}}{{\partial {y^2}}} \cdot {e^s}\sin t + \frac{{{\partial ^2}u}}{{\partial x\partial y}} \cdot {e^s}\cos t} \right)}\end{aligned}} \right)\\ = \left( {\begin{aligned}{*{20}{l}}{\frac{{\partial u}}{{\partial x}} \cdot \left( { - {e^s}\cos t} \right) - \frac{{\partial u}}{{\partial y}} \cdot {e^s}\sin t + {e^{2s}}{{\sin }^2}t\frac{{{\partial ^2}u}}{{\partial {x^2}}} - }\\{2{e^{2s}}\cos t\sin t\frac{{{\partial ^2}u}}{{\partial x\partial y}} + {e^{2s}}{{\cos }^2}t\frac{{{\partial ^2}u}}{{\partial {y^2}}}}\end{aligned}} \right)\end{aligned}\)

Find the value of the Right-Hand side of the equation, \({e^{ - 2s}}\left( {\frac{{{\partial ^2}u}}{{\partial {s^2}}} + \frac{{{\partial ^2}u}}{{\partial {t^2}}}} \right).\)

\(\begin{aligned}{l}\frac{{{\partial ^2}u}}{{\partial {s^2}}} + \frac{{{\partial ^2}u}}{{\partial {t^2}}} = {e^{2s}}{\cos ^2}t\frac{{{\partial ^2}u}}{{\partial {x^2}}} + {e^{2s}}{\sin ^2}t\frac{{{\partial ^2}u}}{{\partial {x^2}}} + {e^{2s}}{\sin ^2}t\frac{{{\partial ^2}u}}{{\partial {y^2}}} + {e^{2s}}{\cos ^2}t\frac{{{\partial ^2}u}}{{\partial {y^2}}}{e^{ - 2s}}\left( {\frac{{{\partial ^2}u}}{{\partial {s^2}}} + \frac{{{\partial ^2}u}}{{\partial {t^2}}}} \right)\\ = {e^{ - 2s}}\left\{ {{e^{2s}}{{\cos }^2}t\frac{{{\partial ^2}u}}{{\partial {x^2}}} + {e^{2s}}{{\sin }^2}t\frac{{{\partial ^2}u}}{{\partial {x^2}}} + {e^{2s}}{{\sin }^2}t\frac{{{\partial ^2}u}}{{\partial {y^2}}} + {e^{2s}}{{\cos }^2}t\frac{{{\partial ^2}u}}{{\partial {y^2}}}} \right\}\\ = \left( {{{\cos }^2}t + {{\sin }^2}t} \right)\left( {\frac{{{\partial ^2}u}}{{\partial {x^2}}} + \frac{{{\partial ^2}u}}{{\partial {y^2}}}} \right)\\ = \frac{{{\partial ^2}u}}{{\partial {x^2}}} + \frac{{{\partial ^2}u}}{{\partial {y^2}}}\end{aligned}\)

Thus, the Left-Hand side is obtained.

Hence, \(\left( {\frac{{{\partial ^2}u}}{{\partial {x^2}}}} \right) + \left( {\frac{{{\partial _u}u}}{{\partial {y^2}}}} \right) = {e^{ - 2s}}\left( {\left( {\frac{{{\partial ^2}u}}{{\partial {s^2}}}} \right) + \left( {\frac{{{\partial ^2}u}}{{\partial {t^2}}}} \right)} \right)\) is proved.