Answer:
The function [tex]u(x,y,z)=log ( x^{2} +y^{2})[/tex] is indeed a solution of the two dimensional Laplace equation [tex]u_{xx} +u_{yy} =0[/tex].
The wave equation [tex]u_{tt} =u_{xx}[/tex] is satisfied by the function [tex]u(x,t)=cos(4x)cos(4t)[/tex] but not by the function [tex]u(x,t)=f(x-t)+f(x+1)[/tex].
Step-by-step explanation:
To verify that the function [tex]u(x,y,z)=log ( x^{2} +y^{2})[/tex] is a solution of the 2D Laplace equation we calculate the second partial derivative with respect to x and then with respect to t.
[tex]u_{xx}=\frac{2}{ln(10)}((x^{2} +y^{2})^{-1} -2x^{2} (x^{2} +y^{2})^{-2})[/tex]
[tex]u_{yy}=\frac{2}{ln(10)}((x^{2} +y^{2})^{-1} -2y^{2} (x^{2} +y^{2})^{-2})[/tex]
then we introduce it in the equation [tex]u_{xx} +u_{yy} =0[/tex]
we get that [tex]\frac{2}{ln(10)} (\frac{2}{(x^{2}+y^{2}) } - \frac{2}{(x^{2}+y^{2} ) } )=0[/tex]
To see if the functions 1) [tex]u(x,t)=cos(4x)cos(4t)[/tex] and 2) [tex]u(x,t)=f(x-t)+f(x+1)[/tex] solve the wave equation we have to calculate the second partial derivative with respect to x and the with respect to t for each function. Then we see if they are equal.
1) [tex]u_{xx}=-16 cos (4x) cos (4t)[/tex]
[tex]u_{tt}=-16cos(4x)cos(4t)[/tex]
we see for the above expressions that [tex]u_{tt} =u_{xx}[/tex]
2) with this function we will have to use the chain rule
If we call [tex]s=x-t[/tex] and [tex]w=x+1[/tex] then we have that
[tex]u(x,t)=f(x-t)+f(x+1)=f(s)+f(w)[/tex]
So [tex]\frac{\partial u}{\partial x}=\frac{df}{ds}\frac{\partial s}{\partial x} +\frac{df}{dw} \frac{\partial w}{\partial x}[/tex]
because we have [tex]\frac{\partial s}{\partial x} =1[/tex] and [tex]\frac{\partial w}{\partial x} =1[/tex]
then [tex]\frac{\partial u}{\partial x} =f'(s)+f'(w)[/tex]
⇒ [tex]\frac{\partial^{2} u }{\partial x^{2} } =\frac{\partial}{\partial x} (f'(s))+ \frac{\partial}{\partial x} (f'(w))[/tex]
⇒[tex]\frac{\partial^{2} u }{ \partial x^{2} } =\frac{d}{ds} (f'(s))\frac{\partial s}{\partial x} +\frac{d}{ds} (f'(w))\frac{\partial w}{\partial x}[/tex]
⇒ [tex]\frac{\partial^{2} u }{ \partial x^{2} } =f''(s)+f''(w)[/tex]
Regarding the derivatives with respect to time
[tex]\frac{\partial u}{\partial t}=\frac{df}{ds} \frac{\partial s}{\partial t}+\frac{df}{dw} \frac{\partial w}{\partial t}=-\frac{df}{ds} =-f'(s)[/tex]
then [tex]\frac{\partial^{2} u }{\partial t^{2} } =\frac{\partial}{\partial t} (-f'(s))=-\frac{d}{ds} (f'(s))\frac{\partial s}{\partial t} =f''(s)[/tex]
we see that [tex]\frac{\partial^{2} u }{ \partial x^{2} } =f''(s)+f''(w) \neq f''(s)=\frac{\partial^{2} u }{\partial t^{2} }[/tex]
[tex]u(x,t)=f(x-t)+f(x+1)[/tex] doesn´t satisfy the wave equation.
