Problem: 2
Verify that \(u(x,t)=\frac{1}{\sqrt{4\pi kt}}e^{\frac{-x^{2}}{4 kt}}\) is a solution of \[u_{t}=ku_{xx} \]
Step 1: Restate the Problem
We need to verify that \[ u(x,t)=\frac{1}{\sqrt{4\pi kt}}e^{-\frac{x^2}{4kt}} \] satisfies the heat equation \[ u_t=ku_{xx}. \]
Step-2
Write
\[
u(x,t)=A(t)B(x,t), \quad \text{with} \quad A(t)=\frac{1}{\sqrt{4\pi kt}}, \quad B(x,t)=e^{-\frac{x^2}{4kt}}.
\]
Differentiate \(A(t)\):
\begin{align}
A(t) &= (4\pi kt)^{-1/2}, \notag\\[6pt]
A'(t) &= -\frac{1}{2}(4\pi kt)^{-3/2}\cdot(4\pi k)
= -\frac{1}{2t}\frac{1}{\sqrt{4\pi kt}} = -\frac{1}{2t}A(t). \tag{1}
\end{align}
Differentiate \(B(x,t)\) with respect to \(t\):
\begin{align}
B_t(x,t) &= e^{-\frac{x^2}{4kt}}\cdot \frac{d}{dt}\left(-\frac{x^2}{4kt}\right)
= e^{-\frac{x^2}{4kt}}\cdot \frac{x^2}{4kt^2}. \tag{2}
\end{align}
Apply the product rule:
\begin{align}
u_t(x,t) &= A'(t)B(x,t) + A(t)B_t(x,t) \notag\\[6pt]
&= A(t)e^{-\frac{x^2}{4kt}}\left(-\frac{1}{2t} + \frac{x^2}{4kt^2}\right) \notag\\[6pt]
&= u(x,t)\left(-\frac{1}{2t} + \frac{x^2}{4kt^2}\right). \tag{3}
\end{align}
Step-3
First derivative with respect to \(x\):
\begin{align}
u_x(x,t) &= A(t)\frac{d}{dx}\left(e^{-\frac{x^2}{4kt}}\right)
= A(t)e^{-\frac{x^2}{4kt}}\left(-\frac{2x}{4kt}\right) \notag\\[6pt]
&= -\frac{x}{2kt} u(x,t). \tag{4}
\end{align}
Second derivative with respect to \(x\):
\begin{align}
u_{xx}(x,t) &= \frac{d}{dx}\left(-\frac{x}{2kt} u(x,t)\right) \notag\\[6pt]
&= -\frac{1}{2kt} u(x,t) – \frac{x}{2kt} u_x(x,t) \notag\\[6pt]
&= -\frac{1}{2kt} u(x,t) – \frac{x}{2kt}\left(-\frac{x}{2kt} u(x,t)\right) \notag\\[6pt]
&= -\frac{1}{2kt} u(x,t) + \frac{x^2}{4k^2t^2} u(x,t) \notag\\[6pt]
&= u(x,t)\left(-\frac{1}{2kt} + \frac{x^2}{4k^2t^2}\right). \tag{5}
\end{align}
Multiply by \(k\):
\begin{align}
k u_{xx}(x,t) = u(x,t)\left(-\frac{1}{2t} + \frac{x^2}{4kt^2}\right). \tag{6}
\end{align}
Step-4: Conclusion
Comparing equations (3) and (6): \[ u_t(x,t)= u(x,t)\left(-\frac{1}{2t} + \frac{x^2}{4kt^2}\right) \] and \[ k u_{xx}(x,t)= u(x,t)\left(-\frac{1}{2t} + \frac{x^2}{4kt^2}\right), \] we conclude that \[ u_t(x,t)= k u_{xx}(x,t). \] Thus, the given function \(u(x,t)\) is indeed a solution of the heat equation.
Summary
We verified that \[ u(x,t)=\frac{1}{\sqrt{4\pi kt}}e^{-\frac{x^2}{4kt}} \] satisfies \(u_t=ku_{xx}\) by calculating \(u_t\) and \(u_{xx}\) step by step and showing that they match via the relation \[ u_t(x,t)= k u_{xx}(x,t). \]
Related Docs
FAQs
Partial Differential Equations
- What is a partial differential equation (PDE)?
A PDE is an equation that involves unknown multivariable functions and their partial derivatives. It describes how the function changes with respect to multiple independent variables.
- How do PDEs differ from ordinary differential equations (ODEs)?
Unlike ODEs, which involve derivatives with respect to a single variable, PDEs involve partial derivatives with respect to two or more independent variables.
- What are the common types of PDEs?
PDEs are generally classified into three types based on their characteristics:
- Elliptic: e.g., Laplace’s equation
- Parabolic: e.g., the heat equation
- Hyperbolic: e.g., the wave equation
- What role do boundary and initial conditions play?
- Boundary conditions specify the behavior of the solution along the edges of the domain.
- Initial conditions are used in time-dependent problems to define the state of the system at the start.
- What methods are commonly used to solve PDEs?
There are several techniques, including:
- Analytical methods like separation of variables, Fourier and Laplace transforms, and the method of characteristics
- Numerical methods such as finite difference, finite element, and spectral methods
- What is the method of separation of variables?
This method assumes that the solution can be written as a product of functions, each depending on only one of the independent variables. This assumption reduces the PDE to a set of simpler ODEs that can be solved individually.
- In which fields are PDEs applied?
PDEs model a wide range of phenomena across various fields including physics (heat transfer, fluid dynamics), engineering (stress analysis, electromagnetics), finance (option pricing models), and more.
- What distinguishes linear from nonlinear PDEs?
- Linear PDEs have terms that are linear with respect to the unknown function and its derivatives, making them more tractable analytically.
- Nonlinear PDEs include terms that are nonlinear, often leading to complex behaviors and requiring specialized methods for solution.
- How do you determine the order of a PDE?
The order of a PDE is defined by the highest derivative (partial derivative) present in the equation. For example, if the highest derivative is a second derivative, the PDE is second order.
- What are some common challenges in solving PDEs?
Challenges include finding closed-form analytical solutions, handling complex geometries and boundary conditions, and the significant computational effort required for accurate numerical solutions.
