Pde Verification 189028

1. **Problem Statement:** Verify by substitution that the given functions are solutions of the PDEs indicated. --- ### Problem 3: Verify $u = \cos 4t \sin 2x$ is a solution of the wave equation $$u_{tt} = c^2 u_{xx}$$ with suitable $c$. **Step 1:** Write the wave equation: $$u_{tt} = c^2 u_{xx}$$ **Step 2:** Compute partial derivatives: $$u = \cos 4t \sin 2x$$ $$u_t = -4 \sin 4t \sin 2x$$ $$u_{tt} = -4 \cdot 4 \cos 4t \sin 2x = -16 \cos 4t \sin 2x$$ $$u_x = \cos 4t \cdot 2 \cos 2x = 2 \cos 4t \cos 2x$$ $$u_{xx} = 2 \cos 4t \cdot (-2) \sin 2x = -4 \cos 4t \sin 2x$$ **Step 3:** Substitute into the wave equation: $$u_{tt} = -16 \cos 4t \sin 2x$$ $$c^2 u_{xx} = c^2 (-4 \cos 4t \sin 2x) = -4 c^2 \cos 4t \sin 2x$$ For equality: $$-16 \cos 4t \sin 2x = -4 c^2 \cos 4t \sin 2x$$ Divide both sides by $-4 \cos 4t \sin 2x$ (nonzero for general $t,x$): $$\cancel{-16} / \cancel{-4} = c^2$$ $$4 = c^2$$ **Step 4:** Therefore, $$c = 2$$ **Answer:** $u = \cos 4t \sin 2x$ satisfies the wave equation with $c=2$. --- ### Problem 8: Verify $u = e^{-9t} \sin \omega x$ is a solution of the heat equation $$u_t = c^2 u_{xx}$$ with suitable $c$. **Step 1:** Write the heat equation: $$u_t = c^2 u_{xx}$$ **Step 2:** Compute partial derivatives: $$u = e^{-9t} \sin \omega x$$ $$u_t = -9 e^{-9t} \sin \omega x$$ $$u_x = e^{-9t} \omega \cos \omega x$$ $$u_{xx} = e^{-9t} \omega^2 (-\sin \omega x) = -\omega^2 e^{-9t} \sin \omega x$$ **Step 3:** Substitute into heat equation: $$u_t = -9 e^{-9t} \sin \omega x$$ $$c^2 u_{xx} = c^2 (-\omega^2 e^{-9t} \sin \omega x) = -c^2 \omega^2 e^{-9t} \sin \omega x$$ For equality: $$-9 e^{-9t} \sin \omega x = -c^2 \omega^2 e^{-9t} \sin \omega x$$ Divide both sides by $- e^{-9t} \sin \omega x$ (nonzero for general $t,x$): $$9 = c^2 \omega^2$$ **Step 4:** Therefore, $$c = \frac{3}{\omega}$$ **Answer:** $u = e^{-9t} \sin \omega x$ satisfies the heat equation with $c = \frac{3}{\omega}$. --- ### Problem 11: Verify $u = \arctan \left( \frac{y}{x} \right)$ is a solution of Laplace's equation $$u_{xx} + u_{yy} = 0$$. **Step 1:** Write Laplace's equation: $$u_{xx} + u_{yy} = 0$$ **Step 2:** Compute partial derivatives. Let $u = \arctan \left( \frac{y}{x} \right)$. Using chain rule: $$u_x = \frac{1}{1 + (y/x)^2} \cdot \frac{d}{dx} \left( \frac{y}{x} \right) = \frac{1}{1 + \frac{y^2}{x^2}} \cdot \left( -\frac{y}{x^2} \right) = \frac{1}{\frac{x^2 + y^2}{x^2}} \cdot \left( -\frac{y}{x^2} \right) = \frac{x^2}{x^2 + y^2} \cdot \left( -\frac{y}{x^2} \right) = -\frac{y}{x^2 + y^2}$$ Similarly, $$u_y = \frac{1}{1 + (y/x)^2} \cdot \frac{d}{dy} \left( \frac{y}{x} \right) = \frac{1}{1 + \frac{y^2}{x^2}} \cdot \frac{1}{x} = \frac{x^2}{x^2 + y^2} \cdot \frac{1}{x} = \frac{x}{x^2 + y^2}$$ **Step 3:** Compute second derivatives: $$u_{xx} = \frac{\partial}{\partial x} \left( -\frac{y}{x^2 + y^2} \right) = -y \frac{\partial}{\partial x} \left( \frac{1}{x^2 + y^2} \right) = -y \left( -\frac{2x}{(x^2 + y^2)^2} \right) = \frac{2xy}{(x^2 + y^2)^2}$$ $$u_{yy} = \frac{\partial}{\partial y} \left( \frac{x}{x^2 + y^2} \right) = x \frac{\partial}{\partial y} \left( \frac{1}{x^2 + y^2} \right) = x \left( -\frac{2y}{(x^2 + y^2)^2} \right) = -\frac{2xy}{(x^2 + y^2)^2}$$ **Step 4:** Sum second derivatives: $$u_{xx} + u_{yy} = \frac{2xy}{(x^2 + y^2)^2} - \frac{2xy}{(x^2 + y^2)^2} = 0$$ **Answer:** $u = \arctan \left( \frac{y}{x} \right)$ satisfies Laplace's equation. --- **Summary:** - Problem 3: $c=2$ - Problem 8: $c=\frac{3}{\omega}$ - Problem 11: $u=\arctan(y/x)$ satisfies Laplace's equation