MHB Verifying Solution: U(x, y) = cos(x) cosh(y) for PDE U_xx + U_yy = 0

[karush]

Verify that the given function is a solution to the partial differential equation

$U_{xx} + U_{yy} = 0$

$U_1 (x, y) =cos(x) cosh(y) $

Sorry I really didn't know how to to get U(x, y) into this
$$\frac{dU^2}{d^2 x}+\frac{dU^2}{d^2 y}=0$$

 

[evinda]

Verify that the given function is a solution to the partial differential equation

$U_{xx} + U_{yy} = 0$

$U_1 (x, y) =cos(x) cosh(y) $

Sorry I really didn't know how to to get U(x, y) into this
$$\frac{dU^2}{d^2 x}+\frac{dU^2}{d^2 y}=0$$

Hello! I wish you a happy new year!$U_{1x}= -\sin x cosh y $

$U_{1xx}=-\cos x \cosh y$

$U_{1y}=\cos x \sinh y$

$U_{1yy}=\cos x \cosh y$

So what is $U_{1xx}+ U_{1yy}$ equal to?

 

[karush]

OK I see that they just cancel each other So the double derivative you just deal with whatever x or y is?

 

[evinda]

OK I see that they just cancel each other So the double derivative you just deal with whatever x or y is?

If tou want to find $U_{1xx}$ you derive twice in respect to x and you consider $\cosh y$ as a constant.

 

[karush]

OK now we have $U_2 (x, y) =ln(x^2 +y^2)$
So
$$\frac{dU_2 ^2 } {dx^2} =\frac{-2\left({x}^{2}-{y}^{2 }\right)}{({x}^{2}+{y}^{2})}$$
And
$$\frac{dU_2 ^2 } {dy^2} =\frac{2\left({x}^{2}-{y}^{2 }\right)}{({x}^{2}{y}^{2})}$$
Then
U_xx + U_yy = 0

 

[MarkFL]

OK now we have $U_2 (x, y) =ln(x^2 +y^2)$
So
$$\frac{dU_2 ^2 } {dx^2} =\frac{-2\left({x}^{2}-{y}^{2 }\right)}{({x}^{2}+{y}^{2})}$$
And
$$\frac{dU_2 ^2 } {dy^2} =\frac{2\left({x}^{2}-{y}^{2 }\right)}{({x}^{2}{y}^{2})}$$
Then
U_xx + U_yy = 0

If $$U_2(x,y)=\ln\left(x^2+y^2\right)$$

then:

$${U_2}_{xx}=-\frac{2\left(x^2-y^2\right)}{\left(x^2+y^2\right)^2}$$

$${U_2}_{yy}=\frac{2\left(x^2-y^2\right)}{\left(x^2+y^2\right)^2}$$

And so yes, $U_2$ is a solution to the PDE:

$$U_{xx}+U_{yy}=0$$