Table of Contents

    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 \[ u_t(x,t)= k u_{xx}(x,t). \]

    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. 

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