Answer to Question #226133 in Differential Equations for Hasnain ali

Question

reduce into canonical form
Uxx-2sinxUxy-cos^2xUyy-cosxUy=0

Answer 1

$\Delta = a_{12}^2 - {a_{11}}{a_{22}} = {( - \sin x)^2} - 1 \cdot \left( { - {{\cos }^2}x} \right) = {\sin ^2}x + {\cos ^2}x = 1 > 0$ - we have an equation of hyperbolic type.

Characteristic equation:

$\frac{{dy}}{{dx}} = \frac{{{a_{12}} + \sqrt \Delta }}{{{a_{11}}}} = \frac{{ - \sin x + 1}}{1} \Rightarrow y = \cos x + x + {C_1}$

or

$\frac{{dy}}{{dx}} = \frac{{{a_{12}} - \sqrt \Delta }}{{{a_{11}}}} = \frac{{ - \sin x - 1}}{1} \Rightarrow y = \cos x - x + {C_2}$

Let

$\xi (x,y) = y - \cos x - x,\,\,\eta (x,y) = y - \cos x + x,\,\,u(x,y) = v(\xi ,\,\eta )$

Then

${u_x} = {v_\xi }{\xi _x} + {v_\eta }{\eta _x} = (\sin x - 1){v_\xi } + (\sin x + 1){v_\eta }$

${u_y} = {v_\xi }{\xi _y} + {v_\eta }{\eta _y} = {v_\xi } + {v_\eta }$

${u_{xx}} = \cos x{v_\xi } + (\sin x - 1)\left( {{v_{\xi \xi }}{\xi _x} + {v_{\xi \eta }}{\eta _x}} \right) + \cos x{v_\eta } + (\sin x + 1)\left( {{v_{\eta \xi }}{\xi _x} + {v_{\eta \eta }}{\eta _x}} \right) = \\= \cos x{v_\xi } + (\sin x - 1)\left( {(\sin x - 1){v_{\xi \xi }} + (\sin x + 1){v_{\xi \eta }}} \right) + \cos x{v_\eta } + (\sin x + 1)\left( {(\sin x - 1){v_{\eta \xi }} + (\sin x + 1){v_{\eta \eta }}} \right) = \\= \cos x{v_\xi } + {(\sin x - 1)^2}{v_{\xi \xi }} + 2\left( {{{\sin }^2}x - 1} \right){v_{\xi \eta }} + \cos x{v_\eta } + {(\sin x + 1)^2}{v_{\eta \eta }} = \\ = {(\sin x - 1)^2}{v_{\xi \xi }} - 2{\cos ^2}x{v_{\xi \eta }} + {(\sin x + 1)^2}{v_{\eta \eta }} + \cos x{v_\xi } + \cos x{v_\eta }$

${u_{yy}} = {v_{\xi \xi }}{\xi _y} + {v_{\xi \eta }}{\eta _y} + {v_{\eta \xi }}{\xi _y} + {v_{\eta \eta }}{\eta _y} = {v_{\xi \xi }} + 2{v_{\xi \eta }} + {v_{\eta \eta }}$

${u_{yx}} = {v_{\xi \xi }}{\xi _x} + {v_{\xi \eta }}{\eta _x} + {v_{\eta \xi }}{\xi _x} + {v_{\eta \eta }}{\eta _x} = (\sin x - 1){v_{\xi \xi }} + (\sin x + 1){v_{\xi \eta }} + (\sin x - 1){v_{\eta \xi }} + (\sin x + 1){v_{\eta \eta }} = \\ = (\sin x - 1){v_{\xi \xi }} + 2\sin x{v_{\xi \eta }} + (\sin x + 1){v_{\eta \eta }}$

Substitute the found values ​​into the original equation:

${(\sin x - 1)^2}{v_{\xi \xi }} - 2{\cos ^2}x{v_{\xi \eta }} + {(\sin x + 1)^2}{v_{\eta \eta }} + \cos x{v_\xi } + \cos x{v_\eta } -\\ - 2\sin x\left( {(\sin x - 1){v_{\xi \xi }} + 2\sin x{v_{\xi \eta }} + (\sin x + 1){v_{\eta \eta }}} \right) - \\- {\cos ^2}x\left( {{v_{\xi \xi }} + 2{v_{\xi \eta }} + {v_{\eta \eta }}} \right) - \cos x\left( {{v_\xi } + {v_\eta }} \right) = 0$

${v_{\xi \xi }}\left( {{{\sin }^2}x - 2\sin x + 1 - 2{{\sin }^2}x + 2\sin x - {{\cos }^2}x} \right) + \\ + {v_{\xi \eta }}\left( { - 2{{\cos }^2}x - 4{{\sin }^2}x - 2{{\cos }^2}x} \right) + \\+ {v_{\eta \eta }}\left( {{{\sin }^2}x + 2\sin x + 1 - 2{{\sin }^2}x - 2\sin x - {{\cos }^2}x} \right) + \\ + {v_\xi }\left( {\cos x - \cos x} \right) + {v_\eta }\left( {\cos x - \cos x} \right) = 0$

$\left( { - {{\sin }^2}x - {{\cos }^2}x + 1} \right){v_{\xi \xi }} + \left( { - 4{{\cos }^2}x - 4{{\sin }^2}x} \right){v_{\xi \eta }} + \left( { - {{\sin }^2}x - {{\cos }^2}x + 1} \right){v_{\eta \eta }} = 0$

$\left( { - 1 + 1} \right){v_{\xi \xi }} - 4{v_{\xi \eta }} + \left( { - 1 + 1} \right){v_{\eta \eta }} = 0 \Rightarrow {v_{\xi \eta }} = 0$

Answer: ${v_{\xi \eta }} = 0$ , where $\xi (x,y) = y - \cos x - x,\,\,\eta (x,y) = y - \cos x + x,\,\,u(x,y) = v(\xi ,\,\eta )$