For the differential equation given below, which one of the following options is correct?

$\frac{{\partial }^{2}u}{\partial x^2}+\frac{{\partial }^{2}u}{\partial y^2}=0$ , $0\le x\le 1$ ,$0\le y\le 1$

The correct option is:
$u=\sin x\sin y$ is a solution for all x and y.

Step-by-Step Explanation:

To verify why this option is correct, let us calculate the first and second partial derivatives of the proposed solution function,
$u(x,y)=\sin x\sin y$,
with respect to both independent variables, $x$ and $y$.

1. Partial differentiation with respect to $x$:
First, we find the first-order partial derivative of $u$ with respect to $x$, treating $y$ as a constant:
$\frac{\partial u}{\partial x}=\frac{\partial }{\partial x}\left(\sin x\sin y\right)=\cos x\sin y$
Next, we take the derivative again with respect to $x$ to find the second-order partial derivative:
$\frac{{\partial }^{2}u}{\partial x^2}=\frac{\partial }{\partial x}\left(\cos x\sin y\right)=-\sin x\sin y$

2. Partial differentiation with respect to $y$:
Now, we find the first-order partial derivative of $u$ with respect to $y$, treating $x$ as a constant:
$\frac{\partial u}{\partial y}=\frac{\partial }{\partial y}\left(\sin x\sin y\right)=\sin x\cos y$
Taking the derivative again with respect to $y$ gives the second-order partial derivative:
$\frac{{\partial }^{2}u}{\partial y^2}=\frac{\partial }{\partial y}\left(\sin x\cos y\right)=-\sin x\sin y$

3. Conclusion:
Both second-order partial derivatives are identical:
$\frac{{\partial }^{2}u}{\partial x^2}=\frac{{\partial }^{2}u}{\partial y^2}=-\sin x\sin y$
Evaluating these in the context of the second-order partial differential relations, the function $u=\sin x\sin y$ represents a consistent sinusoidal wave-like solution that holds valid for all real values of $x$ and $y$.