Question

Find the solution of the equation div(grad z)(x,y)=(e^(-x))cosy which tends to zero as x tends to infinity.

Answer 1

Answer on Question# #47877 – Mathematics – Differential Calculus | Equations

Question:

Find the solution of the equation

$$\operatorname{div}(\operatorname{grad} z)(x, y) = e^{-x} \cos y,$$

which tends to zero as $x$ tends to infinity.

Solution:

Let us rewrite equation (1) by means of the Laplace operator:

$$\operatorname{div}(\operatorname{grad} z)(x, y) = (\nabla \cdot \nabla z(x, y)) = \nabla^2 z(x, y) = \Delta z(x, y) = \frac{\partial^2 z(x, y)}{\partial x^2} + \frac{\partial^2 z(x, y)}{\partial y^2} = e^{-x} \cos y.$$

Consider the right side of this equation. As the function $f(x, y) = e^{-x} \cos y$ is a periodic in $y$-direction, then we can search the solution in the form

$$z(x, y) = f(x) \cos y,$$

Substituting (2) in the equation (1a), we have

$$\frac{\partial^2 z(x, y)}{\partial x^2} = f'' \cos y, \quad \frac{\partial^2 z(x, y)}{\partial y^2} = -f \cos y,$$ $$f'' \cos y - f \cos y = e^{-x} \cos y,$$ $$f'' - f = e^{-x},$$

Therefore we obtain the second-order linear ordinary differential equation. The general solution of this equation is

$$f(x) = C_1 e^x + C_1 e^{-x} - x \frac{e^{-x}}{2},$$

By definition $f(y) = \cos y$ is the bounded function. So, from the problem condition it follows that

$$z(x, y)_{x \to \infty} \to 0 \Rightarrow f(x)_{x \to \infty} \to 0.$$

Now, using (5) and (4) we receive $\left(C_1 e^x + C_1 e^{-x} - x \frac{e^{-x}}{2}\right)_{x \to \infty} \to 0$

As we see, the limit condition holds, if $C_1 = 0$, $C_2 = 1$ (note that $\lim_{x \to \infty} \left(e^{-x} - x \frac{e^{-x}}{2}\right) = 0$). Thus, the solution of equation (1) is

$$z(x, y) = \left(e^{-x} - x \frac{e^{-x}}{2}\right) \cos y.$$

Answer: $z(x, y) = \left(e^{-x} - x \frac{e^{-x}}{2}\right) \cos y$.

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