Let \( u(x, t) \) be the solution of the following initial-boundary value problem: \[ \frac{\partial u}{\partial t} - \frac{\partial^2 u}{\partial x^2} = 0, \quad x \in (0, \pi), \quad t>0, \] with the boundary conditions: \[ u(0, t) = u(\pi, t) = 0, \quad u(x, 0) = \sin 4x \cos 3x. \] Then, for each \( t>0 \), the value of \( u\left( \frac{\pi}{4}, t \right) \) is
To find the solution to this problem, we are dealing with a heat equation (a type of partial differential equation). Here, the equation is given by:
\[\frac{\partial u}{\partial t} - \frac{\partial^2 u}{\partial x^2} = 0,\] with boundary conditions u(0, t) = u(\pi, t) = 0 for x \in (0, \pi) and t > 0.
The initial condition is u(x, 0) = \sin(4x) \cos(3x), which is given.
The general solution for this type of equation (heat equation on a bounded domain) can be expressed using the method of separation of variables. Assume a solution of the form:
u(x, t) = X(x)T(t).
Plugging this into the original equation, we get:
X(x)\frac{dT}{dt} = T(t)\frac{d^2X}{dx^2}.
Separating variables, we have:
\[\frac{1}{T}\frac{dT}{dt} = \frac{1}{X}\frac{d^2X}{dx^2} = -\lambda\] (a separation constant).
We solve these equations individually:
The spatial part: \(\frac{d^2X}{dx^2} + \lambda X = 0\)
Boundary conditions: X(0) = X(\pi) = 0.
These are sine functions; hence, \(\lambda_n = n^2\) and \(X_n(x) = \sin(nx)\).
The temporal part: \(\frac{dT}{dt} + \lambda T = 0\) which gives:
T(t) = e^{-\lambda t}.
Thus, the solution is a sum of terms: u(x, t) = \sum_{n=1}^{\infty} C_n e^{-n^2 t} \sin(nx).
The coefficients C_n are obtained by the initial condition:
u(x, 0) = \sin(4x) \cos(3x).
Use trigonometric identity: \(\sin(4x) \cos(3x) = \frac{1}{2} (\sin(7x) + \sin(x))\).
Comparing with u(x,0) = \sum_{n=1}^{\infty} C_n \sin(nx), the coefficients are C_7 = \frac{1}{2} and C_1 = \frac{1}{2}.
The solution becomes:
\[ u(x, t) = \frac{1}{2} e^{-49t} \sin(7x) + \frac{1}{2} e^{-t} \sin(x).\]
Now, evaluate u\left( \frac{\pi}{4}, t \right):
u\left( \frac{\pi}{4}, t \right) = \frac{1}{2} e^{-49t} \sin\left(7 \cdot \frac{\pi}{4}\right) + \frac{1}{2} e^{-t} \sin\left(\frac{\pi}{4}\right).
We calculate:
\[\sin\left(\frac{7\pi}{4}\right) = -\frac{\sqrt{2}}{2}, \quad \sin\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}.\]
Thus:
\[ u\left( \frac{\pi}{4}, t \right) = \frac{1}{2} e^{-49t} \left(-\frac{\sqrt{2}}{2}\right) + \frac{1}{2} e^{-t} \left(\frac{\sqrt{2}}{2}\right) \Rightarrow \frac{\sqrt{2}}{2}\left(\frac{1}{2}\right)\left(e^{-t} - e^{-49t}\right) \]
Finally, we can simplify this result:
\[ \frac{\sqrt{2}}{2}\left(\frac{1}{2}\right)\left(e^{-t} - e^{-49t}\right) = \frac{1}{2\sqrt{2}}\left(e^{-49t}- e^{-t}\right) = \frac{e^{-49t}}{2\sqrt{2}} (e^{48t} - 1) \]
Therefore, the correct answer is:
Option: \(\frac{e^{-49t}}{2\sqrt{2}} (e^{48t} - 1)\)
Pick the CORRECT solution for the following differential equation:
\[ \frac{dy}{dx} = e^{x - y} \]
In the differential equation $\dfrac{dy}{dx} + \alpha x y = 0$, $\alpha$ is a positive constant. If $y=1.0$ at $x=0.0$, and $y=0.8$ at $x=1.0$, the value of $\alpha$ is (rounded off to three decimal places).