Question

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 1

Answer:

The function  $u(x,y,z)=log ( x^{2} +y^{2})$ is indeed a solution of the two dimensional Laplace equation  $u_{xx} +u_{yy} =0$.

The wave equation  $u_{tt} =u_{xx}$ is satisfied by the function $u(x,t)=cos(4x)cos(4t)$ but not by the function $u(x,t)=f(x-t)+f(x+1)$.

Step-by-step explanation:

To verify that the function  $u(x,y,z)=log ( x^{2} +y^{2})$ is a solution of the 2D Laplace equation we calculate the second partial derivative with respect to x and then with respect to t.

$u_{xx}=\frac{2}{ln(10)}((x^{2} +y^{2})^{-1} -2x^{2} (x^{2} +y^{2})^{-2})$

$u_{yy}=\frac{2}{ln(10)}((x^{2} +y^{2})^{-1} -2y^{2} (x^{2} +y^{2})^{-2})$

then we introduce it in the equation  $u_{xx} +u_{yy} =0$

we get that  $\frac{2}{ln(10)} (\frac{2}{(x^{2}+y^{2}) } - \frac{2}{(x^{2}+y^{2} ) } )=0$

To see if the functions 1) $u(x,t)=cos(4x)cos(4t)$ and 2)    $u(x,t)=f(x-t)+f(x+1)$ 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)  $u_{xx}=-16 cos (4x) cos (4t)$

   $u_{tt}=-16cos(4x)cos(4t)$

we see for the above expressions that  $u_{tt} =u_{xx}$

2) with this function we will have to use the chain rule

 If we call  $s=x-t$ and  $w=x+1$  then we have that

 $u(x,t)=f(x-t)+f(x+1)=f(s)+f(w)$

So  $\frac{\partial u}{\partial x}=\frac{df}{ds}\frac{\partial s}{\partial x} +\frac{df}{dw} \frac{\partial w}{\partial x}$

because we have  $\frac{\partial s}{\partial x} =1$ and   $\frac{\partial w}{\partial x} =1$

then  $\frac{\partial u}{\partial x} =f'(s)+f'(w)$

⇒ $\frac{\partial^{2} u }{\partial x^{2} } =\frac{\partial}{\partial x} (f'(s))+ \frac{\partial}{\partial x} (f'(w))$

⇒$\frac{\partial^{2} u }{ \partial x^{2} } =\frac{d}{ds} (f'(s))\frac{\partial s}{\partial x} +\frac{d}{ds} (f'(w))\frac{\partial w}{\partial x}$

⇒ $\frac{\partial^{2} u }{ \partial x^{2} } =f''(s)+f''(w)$

Regarding the derivatives with respect to time

$\frac{\partial u}{\partial t}=\frac{df}{ds} \frac{\partial s}{\partial t}+\frac{df}{dw} \frac{\partial w}{\partial t}=-\frac{df}{ds} =-f'(s)$

then  $\frac{\partial^{2} u }{\partial t^{2} } =\frac{\partial}{\partial t} (-f'(s))=-\frac{d}{ds} (f'(s))\frac{\partial s}{\partial t} =f''(s)$

we see that  $\frac{\partial^{2} u }{ \partial x^{2} } =f''(s)+f''(w) \neq f''(s)=\frac{\partial^{2} u }{\partial t^{2} }$

$u(x,t)=f(x-t)+f(x+1)$  doesn´t satisfy the wave equation.