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

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)`.

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)=(2)/(ln(10))((x^(2) +y^(2))^(-1) -2x^(2) (x^(2) +y^(2))^(-2))`

`u_(yy)=(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
`(2)/(ln(10)) ((2)/((x^(2)+y^(2)) ) - (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
`(\partial u)/(\partial x)=(df)/(ds)(\partial s)/(\partial x) +(df)/(dw) (\partial w)/(\partial x)`

because we have
`(\partial s)/(\partial x) =1` and
`(\partial w)/(\partial x) =1`

then
`(\partial u)/(\partial x) =f'(s)+f'(w)`

⇒
`(\partial^(2) u )/(\partial x^(2) ) =(\partial)/(\partial x) (f'(s))+ (\partial)/(\partial x) (f'(w))`

⇒
`(\partial^(2) u )/( \partial x^(2) ) =(d)/(ds) (f'(s))(\partial s)/(\partial x) +(d)/(ds) (f'(w))(\partial w)/(\partial x)`

⇒
`(\partial^(2) u )/( \partial x^(2) ) =f''(s)+f''(w)`

Regarding the derivatives with respect to time

`(\partial u)/(\partial t)=(df)/(ds) (\partial s)/(\partial t)+(df)/(dw) (\partial w)/(\partial t)=-(df)/(ds) =-f'(s)`

then
`(\partial^(2) u )/(\partial t^(2) ) =(\partial)/(\partial t) (-f'(s))=-(d)/(ds) (f'(s))(\partial s)/(\partial t) =f''(s)`

we see that
`(\partial^(2) u )/( \partial x^(2) ) =f''(s)+f''(w) \\eq f''(s)=(\partial^(2) u )/(\partial t^(2) )`

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