Question

$$\text{ For }u=e^x\cos y$$, (a) verify that ${\partial }^{2}u/\partial x\partial y={\partial }^{2}u/\partial y\partial x$; (b) verify that ${\partial }^{2}u/\partial x^2+{\partial }^{2}u/\partial y^2=0$.

Short answer

Both mixed partial derivatives are equal, and the sum of the second partial derivatives with respect to $x$ and $y$ is zero.

Step-by-step solution

Compute $\frac{\partial u}{\partial x}$

Given $u=e^x\cos y$, find $\frac{\partial u}{\partial x}$:$$\frac{\partial u}{\partial x}=\cos y\cdot \frac{\partial }{\partial x}(e^x)=e^x\cos y$$

Compute $\frac{\partial u}{\partial y}$

Now, find $\frac{\partial u}{\partial y}$:$$\frac{\partial u}{\partial y}=e^x\cdot \frac{\partial }{\partial y}(\cos y)=-e^x\sin y$$

Compute $\frac{{\partial }^{2}u}{\partial x\partial y}$

Find the mixed partial derivative $\frac{{\partial }^{2}u}{\partial x\partial y}$ by differentiating $\frac{\partial u}{\partial y}$ with respect to $x$:$$\frac{{\partial }^{2}u}{\partial x\partial y}=\frac{\partial }{\partial x}(-e^x\sin y)=-\sin y\cdot \frac{\partial }{\partial x}(e^x)=-e^x\sin y$$

Compute $\frac{{\partial }^{2}u}{\partial y\partial x}$

Now find $\frac{{\partial }^{2}u}{\partial y\partial x}$ by differentiating $\frac{\partial u}{\partial x}$ with respect to $y$:$$\frac{{\partial }^{2}u}{\partial y\partial x}=\frac{\partial }{\partial y}(e^x\cos y)=e^x\cdot \frac{\partial }{\partial y}(\cos y)=-e^x\sin y$$

Verify equality of mixed partial derivatives

Since both $\frac{{\partial }^{2}u}{\partial x\partial y}$ and $\frac{{\partial }^{2}u}{\partial y\partial x}$ are equal, we have verified:$$\frac{{\partial }^{2}u}{\partial x\partial y}=\frac{{\partial }^{2}u}{\partial y\partial x}$$

Compute $\frac{{\partial }^{2}u}{\partial x^2}$

Find $\frac{{\partial }^{2}u}{\partial x^2}$ by differentiating $\frac{\partial u}{\partial x}$ with respect to $x$:$$\frac{{\partial }^{2}u}{\partial x^2}=\frac{\partial }{\partial x}(e^x\cos y)=\cos y\cdot \frac{\partial }{\partial x}(e^x)=e^x\cos y$$

Compute $\frac{{\partial }^{2}u}{\partial y^2}$

Find $\frac{{\partial }^{2}u}{\partial y^2}$ by differentiating $\frac{\partial u}{\partial y}$ with respect to $y$:$$\frac{{\partial }^{2}u}{\partial y^2}=\frac{\partial }{\partial y}(-e^x\sin y)=-e^x\cdot \frac{\partial }{\partial y}(\sin y)=-e^x\cos y$$

Verify Laplace equation

Now, check if $\frac{{\partial }^{2}u}{\partial x^2}+\frac{{\partial }^{2}u}{\partial y^2}=0$:$$\frac{{\partial }^{2}u}{\partial x^2}+\frac{{\partial }^{2}u}{\partial y^2}=e^x\cos y-e^x\cos y=0$$

Key concepts

mixed partial derivatives

Mixed partial derivatives involve taking the partial derivative of a function more than once, but with respect to different variables each time. For instance, if we have a function $u(x,y)$ and we're finding the mixed partial derivative $\frac{{\text{∂}}^{2}u}{\text{∂}x\text{∂}y}$, we first take the derivative with respect to $y$ and then with respect to $x$.

In the original problem, we used the function $u=e^x\text{cos}y$ and calculated its mixed partial derivatives as follows:

• From $\frac{\text{∂}u}{\text{∂}y}=-e^x\text{sin}y$, we find $\frac{{\text{∂}}^{2}u}{\text{∂}x\text{∂}y}$ by differentiating with respect to $x$.
• From $\frac{\text{∂}u}{\text{∂}x}=e^x\text{cos}y$, we then find $\frac{{\text{∂}}^{2}u}{\text{∂}y\text{∂}x}$ by differentiating with respect to $y$.

We ultimately observed that both mixed partial derivatives are equal: $\frac{{\text{∂}}^{2}u}{\text{∂}x\text{∂}y}=\frac{{\text{∂}}^{2}u}{\text{∂}y\text{∂}x}=-e^x\text{sin}y$. This equality should hold for any sufficiently smooth functions, verifying the symmetry of mixed partial derivatives in many practical cases.

Laplace equation

The Laplace equation is a second-order partial differential equation given by $abl{a}^{2}u=0$. For functions of two variables, it takes the form:

• $\frac{{\text{∂}}^{2}u}{\text{∂}{x}^2}+\frac{{\text{∂}}^{2}u}{\text{∂}{y}^2}=0$

It is used in many fields of science and engineering, including electromagnetism, fluid dynamics, and potential theory.

For the given function $u=e^x\text{cos}y$, we computed:

• $\frac{{\text{∂}}^{2}u}{\text{∂}{x}^2}=e^x\text{cos}y$.
• $\frac{{\text{∂}}^{2}u}{\text{∂}{y}^2}=-e^x\text{cos}y$.

Summing these two second-order partial derivatives confirmed the Laplace equation: $\frac{{\text{∂}}^{2}u}{\text{∂}{x}^2}+\frac{{\text{∂}}^{2}u}{\text{∂}{y}^2}=e^x\text{cos}y-e^x\text{cos}y=0$. This shows that $u$ satisfies the Laplace equation, making it a harmonic function.

step-by-step solutions

Step-by-step solutions are helpful in solving complex problems, especially in mathematics. They break down the problem into manageable steps, making it easier to follow the logic and calculations. Here's a brief overview of how we approached the given problem:

• First, we found the partial derivatives $\frac{\text{∂}u}{\text{∂}x}$ and $\frac{\text{∂}u}{\text{∂}y}$.
• Next, we took the second partial derivatives $\frac{{\text{∂}}^{2}u}{\text{∂}x\text{∂}y}$ and $\frac{{\text{∂}}^{2}u}{\text{∂}y\text{∂}x}$.
• We verified that these mixed partial derivatives are equal, checking the symmetry property.
• Finally, we computed $\frac{{\text{∂}}^{2}u}{\text{∂}{x}^2}$ and $\frac{{\text{∂}}^{2}u}{\text{∂}{y}^2}$, summing them to show that $u$ satisfies the Laplace equation.

By breaking the task into smaller parts, each step becomes easier to understand and solve. This method can be applied to a wide range of mathematical problems, making it a powerful tool for learning and problem-solving.