show that Ux+3Uy=0 has a solution of the form u(x,y)=f(ax+cy) for a proper choice of a and ad c. find a ad c
To show that the equation Ux + 3Uy = 0 has a solution of the form u(x,y) = f(ax+cy), we can start by assuming that u(x,y) can be expressed in this form.
Let's differentiate the expression u(x,y) = f(ax+cy) with respect to x and y:
u_x = f'(ax+cy)(a)
u_y = f'(ax+cy)(c)
Now, let's substitute u_x and u_y in the equation Ux + 3Uy = 0:
Ux + 3Uy = (f'(ax+cy)(a)) + 3(f'(ax+cy)(c)) = 0
Factor out f'(ax+cy):
f'(ax+cy)(a+3c) = 0
Since f'(ax+cy) can't be equal to zero (otherwise u(x,y) = f(ax+cy) would be a constant function), we must have a+3c = 0.
Therefore, a = -3c.
So, the proper choice of a and c to satisfy the equation Ux + 3Uy = 0 and the form u(x,y) = f(ax+cy) is a = -3 and c = 1 (or a = 3 and c = -1).
To prove that the equation Ux + 3Uy = 0 has a solution of the form u(x,y) = f(ax + cy), we can follow these steps:
Step 1: Let's assume that u(x,y) has the form u(x,y) = f(ax + cy).
Step 2: We need to find the partial derivatives of u with respect to x and y. Using the chain rule, we have:
- ∂u/∂x = f'(ax + cy) * (a)
- ∂u/∂y = f'(ax + cy) * (c)
Step 3: Substitute the partial derivatives into the original equation Ux + 3Uy = 0:
- ∂u/∂x + 3(∂u/∂y) = 0
- f'(ax + cy) * a + 3 * f'(ax + cy) * c = 0
Step 4: Factor out f'(ax + cy) from the equation:
- f'(ax + cy) * (a + 3c) = 0
Step 5: Since f'(ax + cy) cannot be zero for all (x, y), we set a + 3c = 0:
- a + 3c = 0
Step 6: Solve the equation for a and c:
- a = -3c
So, we have found that a = -3c. Therefore, the proper choice for a and c such that Ux + 3Uy = 0 has a solution of the form u(x,y) = f(ax + cy) is a = -3 and c = 1.