Question #242986

Show that u(x,y)=excosy and v(x,y)=exsiny are harmonic for all values of (x,y).


1
Expert's answer
2021-09-28T10:37:00-0400

Let us show that u(x,y)=excosyu(x,y)=e^x\cos y and v(x,y)=exsinyv(x,y)=e^x\sin y are harmonic for all values of (x,y).(x,y). Taking into account that u(x,y)x=excosy,u2(x,y)x2=excosy,u(x,y)y=exsiny,u2(x,y)y2=excosy,\frac{\partial u(x,y)}{\partial x}=e^x\cos y, \frac{\partial u^2(x,y)}{\partial x^2}=e^x\cos y, \frac{\partial u(x,y)}{\partial y}=-e^x\sin y, \frac{\partial u^2(x,y)}{\partial y^2}=-e^x\cos y,

we conclude that u2(x,y)x2+u2(x,y)y2=excosyexcosy=0,\frac{\partial u^2(x,y)}{\partial x^2}+\frac{\partial u^2(x,y)}{\partial y^2}=e^x\cos y-e^x\cos y=0, and hence the function u(x,y)=excosyu(x,y)=e^x\cos y is harmonic.


Since v(x,y)x=exsiny,v2(x,y)x2=exsiny,v(x,y)y=excosy,v2(x,y)y2=exsiny,\frac{\partial v(x,y)}{\partial x}=e^x\sin y, \frac{\partial v^2(x,y)}{\partial x^2}=e^x\sin y, \frac{\partial v(x,y)}{\partial y}=e^x\cos y, \frac{\partial v^2(x,y)}{\partial y^2}=-e^x\sin y,

we conclude that v2(x,y)x2+v2(x,y)y2=exsinyexsiny=0,\frac{\partial v^2(x,y)}{\partial x^2}+\frac{\partial v^2(x,y)}{\partial y^2}=e^x\sin y-e^x\sin y=0, and hence the function v(x,y)=exsinyv(x,y)=e^x\sin y is harmonic.


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