Pde Verification E825F4

1. **Problem Statement:** Verify by substitution that the given functions satisfy their respective PDEs. The problems with arrows are: - (3) $u = \cos 4t \sin 2x$ for the wave equation (1): $\frac{\partial^2 u}{\partial t^2} = c^2 \frac{\partial^2 u}{\partial x^2}$ - (8) $u = e^{-9t} \sin \omega x$ for the heat equation (2): $\frac{\partial u}{\partial t} = c^2 \frac{\partial^2 u}{\partial x^2}$ - (9) $u = e^{-\pi^2 t} \cos 25x$ for the heat equation (2) - (11) $u = \arctan \frac{y}{x}$ for Laplace equation (3): $\frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = 0$ --- 2. **Verification for (3) $u = \cos 4t \sin 2x$ (Wave Equation):** - Given PDE: $$\frac{\partial^2 u}{\partial t^2} = c^2 \frac{\partial^2 u}{\partial x^2}$$ - Compute derivatives: $$\frac{\partial u}{\partial t} = -4 \sin 4t \sin 2x$$ $$\frac{\partial^2 u}{\partial t^2} = -16 \cos 4t \sin 2x$$ $$\frac{\partial u}{\partial x} = 2 \cos 4t \cos 2x$$ $$\frac{\partial^2 u}{\partial x^2} = -4 \cos 4t \sin 2x$$ - Substitute into PDE: $$-16 \cos 4t \sin 2x = c^2 (-4 \cos 4t \sin 2x)$$ - Simplify: $$-16 = -4 c^2 \implies c^2 = 4 \implies c = 2$$ - **Conclusion:** The function satisfies the wave equation if $c=2$. --- 3. **Verification for (8) $u = e^{-9t} \sin \omega x$ (Heat Equation):** - PDE: $$\frac{\partial u}{\partial t} = c^2 \frac{\partial^2 u}{\partial x^2}$$ - Compute derivatives: $$\frac{\partial u}{\partial t} = -9 e^{-9t} \sin \omega x$$ $$\frac{\partial u}{\partial x} = e^{-9t} \omega \cos \omega x$$ $$\frac{\partial^2 u}{\partial x^2} = -\omega^2 e^{-9t} \sin \omega x$$ - Substitute: $$-9 e^{-9t} \sin \omega x = c^2 (-\omega^2 e^{-9t} \sin \omega x)$$ - Simplify: $$-9 = -c^2 \omega^2 \implies c^2 \omega^2 = 9$$ - **Conclusion:** The function satisfies the heat equation if $c^2 \omega^2 = 9$. --- 4. **Verification for (9) $u = e^{-\pi^2 t} \cos 25x$ (Heat Equation):** - PDE: $$\frac{\partial u}{\partial t} = c^2 \frac{\partial^2 u}{\partial x^2}$$ - Compute derivatives: $$\frac{\partial u}{\partial t} = -\pi^2 e^{-\pi^2 t} \cos 25x$$ $$\frac{\partial u}{\partial x} = -25 e^{-\pi^2 t} \sin 25x$$ $$\frac{\partial^2 u}{\partial x^2} = -625 e^{-\pi^2 t} \cos 25x$$ - Substitute: $$-\pi^2 e^{-\pi^2 t} \cos 25x = c^2 (-625 e^{-\pi^2 t} \cos 25x)$$ - Simplify: $$-\pi^2 = -625 c^2 \implies c^2 = \frac{\pi^2}{625}$$ - **Conclusion:** The function satisfies the heat equation if $c^2 = \frac{\pi^2}{625}$. --- 5. **Verification for (11) $u = \arctan \frac{y}{x}$ (Laplace Equation):** - PDE: $$\frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = 0$$ - Let $z = \frac{y}{x}$, then $u = \arctan z$. - Compute first derivatives: $$u_x = \frac{d}{dx} \arctan z = \frac{1}{1+z^2} \cdot \frac{\partial z}{\partial x}$$ $$\frac{\partial z}{\partial x} = \frac{\partial}{\partial x} \left( \frac{y}{x} \right) = -\frac{y}{x^2} = -\frac{z}{x}$$ So, $$u_x = \frac{1}{1+z^2} \left(-\frac{z}{x}\right) = -\frac{z}{x(1+z^2)}$$ - Compute second derivative $u_{xx}$: $$u_{xx} = \frac{\partial}{\partial x} u_x = \frac{\partial}{\partial x} \left(-\frac{z}{x(1+z^2)}\right)$$ This is a bit lengthy but after calculation, $$u_{xx} = \frac{2 y x}{(x^2 + y^2)^2}$$ - Similarly, compute $u_y$ and $u_{yy}$: $$u_y = \frac{1}{1+z^2} \cdot \frac{\partial z}{\partial y} = \frac{1}{1+z^2} \cdot \frac{1}{x} = \frac{1}{x(1+z^2)}$$ $$u_{yy} = \frac{\partial}{\partial y} u_y = -\frac{2 y x}{(x^2 + y^2)^2}$$ - Add second derivatives: $$u_{xx} + u_{yy} = \frac{2 y x}{(x^2 + y^2)^2} - \frac{2 y x}{(x^2 + y^2)^2} = 0$$ - **Conclusion:** $u = \arctan \frac{y}{x}$ satisfies Laplace's equation. --- **Final answers:** - (3) satisfies wave equation if $c=2$. - (8) satisfies heat equation if $c^2 \omega^2 = 9$. - (9) satisfies heat equation if $c^2 = \frac{\pi^2}{625}$. - (11) satisfies Laplace equation unconditionally.