Show that :(I) u = e^xsiny is a solution of Laplace's equation.
(ii) u = x^2+t^2 is a solution of the wave equation.
To show that a given function is a solution of a certain differential equation, we need to substitute the function into the differential equation and verify that it satisfies the equation.
I. To show that u = e^(xsin(y)) is a solution of Laplace's equation, we need to substitute it into the Laplace's equation and check whether it holds.
Laplace's equation is given by:
∇^2u = ∂^2u/∂x^2 + ∂^2u/∂y^2 = 0,
where ∇^2 denotes the Laplacian operator.
Let's evaluate the second derivatives with respect to x and y:
∂^2u/∂x^2 = ∂/∂x(e^(xsin(y)) * sin(y))
= e^(xsin(y)) * cos(y) * sin(y)
∂^2u/∂y^2 = ∂/∂y(e^(xsin(y)) * sin(y))
= e^(xsin(y)) * x * cos(y) * sin(y)
Now, substitute these results into the Laplace's equation:
∂^2u/∂x^2 + ∂^2u/∂y^2 = e^(xsin(y)) * cos(y) * sin(y) + e^(xsin(y)) * x * cos(y) * sin(y)
Since we are looking for a solution where ∇^2u = 0, the above expression should simplify to 0.
Notice that both terms have a common factor of e^(xsin(y)) * cos(y) * sin(y). By factoring out this common factor, we get:
e^(xsin(y)) * cos(y) * sin(y) * (1 + x) = 0
For this equation to hold, either e^(xsin(y)) * cos(y) * sin(y) = 0 or (1 + x) = 0.
Since (1 + x) = 0 is not true for all x, we focus on the first case:
e^(xsin(y)) * cos(y) * sin(y) = 0
To satisfy this equation, we must have e^(xsin(y)) = 0 or cos(y) * sin(y) = 0.
However, e^(xsin(y)) cannot be zero for any values of x and y because the exponential function e^z is always positive for any real value of z. Therefore, e^(xsin(y)) = 0 does not hold.
We are left with cos(y) * sin(y) = 0.
This equation holds for y = nπ, where n is any integer. However, we are interested in solutions that are not identically zero.
Hence, we conclude that u = e^(xsin(y)) is not a solution of Laplace's equation.
II. To show that u = x^2 + t^2 is a solution of the wave equation, we need to substitute it into the wave equation and check whether it holds.
The wave equation is given by:
∂^2u/∂t^2 - c^2∇^2u = 0,
where c is the wave speed, and ∇^2 denotes the Laplacian operator.
First, let's evaluate the second derivative with respect to t:
∂^2u/∂t^2 = ∂^2/∂t^2(x^2 + t^2)
= 2
Now, evaluate the Laplacian of u:
∇^2u = ∂^2u/∂x^2 + ∂^2u/∂y^2 + ∂^2u/∂z^2
Since u is given as a function of x and t only, there are no y- or z- derivatives involved. Therefore, this simplifies to:
∇^2u = ∂^2u/∂x^2
Now, evaluate the second derivative with respect to x:
∂^2u/∂x^2 = ∂^2/∂x^2(x^2 + t^2)
= 2
Now, substitute the results into the wave equation:
∂^2u/∂t^2 - c^2∇^2u = 2 - c^2 * 2
This expression can be simplified to:
2 - 2c^2
For this equation to hold, 2 - 2c^2 = 0.
Simplifying the equation further, we find:
c^2 = 1
Therefore, the wave equation holds for any value of c^2 = 1.
Hence, we can conclude that u = x^2 + t^2 is indeed a solution of the wave equation.