Verify that the function u(x, y, z) = log x^2 + y^2 is a solution of the two dimensional Laplace equation u_xx + u_yy = 0 everywhere, except of course at the origin where f is not defined. Verify that the following functions solve the wave equation, u_tt = u_xx u(x, t) = cos(4x) cos(4t) u(x, t) = f(x - t) + f(x + 1), where f is any differentiable function of one variable.

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.