Answer to Question #94198 in Mechanics | Relativity for Zwivhuya

Question

If c1 and c2 are constant vectors and @ is a constant scalar, show that h=e^(-@x)(c1sin@x+c2cos@y)

Answer 1

Let's write the condition clearly since the existing condition is incomplete and a little bit confusing:

If $\textbf{c}_1$and $\textbf{c}_2$ are constant vectors and $\lambda$ is a constant scalar, show that

$$\textbf{h}=e^{-\lambda x}[\textbf{c}_1\text{sin}(\lambda x)+\textbf{c}_2 \text{cos}(\lambda y)].$$

satisfies the partial equation

$$\frac{\partial^2\textbf{h}}{\partial x^2}+\frac{\partial^2\textbf{h}}{\partial y^2}=0.$$

Solution

Simply take the second partial derivatives of the given vector:

$$\Big(\frac{\partial^2\textbf{}}{\partial x^2}+\frac{\partial^2\textbf{}}{\partial y^2}\Big)\textbf{h}=\\ \space\\ =\Big[\frac{\partial^2\textbf{}}{\partial x^2}+\frac{\partial^2\textbf{}}{\partial y^2}\Big]e^{-\lambda x}[\textbf{c}_1\text{sin}(\lambda x)+\textbf{c}_2 \text{cos}(\lambda y)]=\\ \space\\ =\frac{\partial^2\textbf{}}{\partial x^2}e^{-\lambda x}\textbf{c}_1\text{sin}(\lambda x)+\frac{\partial^2\textbf{}}{\partial x^2}e^{-\lambda x}\textbf{c}_2 \text{cos}(\lambda y)+\\ \space\\ +\frac{\partial^2\textbf{}}{\partial y^2}e^{-\lambda x}\textbf{c}_1\text{sin}(\lambda x)+\frac{\partial^2\textbf{}}{\partial y^2}e^{-\lambda x}\textbf{c}_2 \text{cos}(\lambda y)=\\ \space\\ =-2\textbf{c}_1\lambda^2e^{-\lambda x}\text{cos}(\lambda x)+\textbf{c}_2\lambda^2e^{-\lambda x}\text{cos}(\lambda y)+\\ \space\\ +0-\textbf{c}_2\lambda^2e^{-\lambda x}\text{cos}(\lambda y)=\\ \space\\ =-2\textbf{c}_1\lambda^2e^{-\lambda x}\text{cos}(\lambda x).$$

We can't show what we were asked to show if we wrote the correct condition :(