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Assume a solution of the linear homogeneous partial differential equation having the "separation of variables" form given. Either demonstrate that solutions having this form exist, by deriving appropriate separation equations, or explain why the technique fails. \(u_{t}(x, t)=\left(1+t^{2}\right)\left(1+x^{2}\right) u_{x x}(x, t), \quad u(x, t)=X(x) T(t)\)

Short Answer

Expert verified
Answer: No, the given linear homogeneous PDE does not allow a solution of the form \(u(x, t) = X(x) T(t)\) using the separation of variables technique because we couldn't separate the variables and derive two separate ODEs after substituting the partial derivatives into the given PDE.

Step by step solution

01

Partial Derivative of u(x,t) with Respect to t

We're given that \(u(x,t) = X(x) T(t)\). Let's first find the partial derivative of \(u\) with respect to \(t\): \(u_t(x, t) = X(x) \frac{d}{dt} T(t)\).
02

Partial Derivative of u(x,t) With Respect to x Twice

Now, we'll find the second partial derivative of \(u(x, t)\) with respect to \(x\): \(u_x(x, t) = T(t) \frac{d}{dx} X(x)\), \(u_{xx}(x, t) = T(t) \frac{d^2}{dx^2} X(x)\).
03

Substitute the Partial Derivatives into the Given PDE

We substitute \(u_t\) and \(u_{xx}\) from steps 1 and 2 into the given PDE: \(X(x) \frac{d}{dt} T(t) = (1+t^2)(1+x^2)T(t) \frac{d^2}{dx^2} X(x)\).
04

Try to Separate the Variables

Now, we'll try to separate the variables by dividing both sides by \(X(x)T(t)\): \(\frac{1}{T(t)}\frac{d}{dt}T(t) = \frac{(1+t^2)(1+x^2)}{X(x)} \frac{d^2}{dx^2} X(x)\). At this point, it's clear that we can't separate the variables successfully because the function \((1 + t^2)(1 + x^2)\) on the right-hand side includes both \(x\) and \(t\). Therefore, we can't derive two separate ODEs for \(X(x)\) and \(T(t)\), and the separation of variables technique fails.
05

Conclusion

In conclusion, the linear homogeneous PDE given doesn't allow a solution of the form \(u(x, t) = X(x) T(t)\) using the separation of variables technique. The technique fails because we couldn't separate the variables and derive two separate ODEs after substituting the partial derivatives into the given PDE.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Separation of Variables
The separation of variables is a powerful technique used to solve partial differential equations (PDEs). The basic idea is to assume that the solution can be written as a product of functions, each depending only on a single variable. For example, for a PDE involving variables \(x\) and \(t\), we assume a form \(u(x, t) = X(x) T(t)\).
This assumption allows us to break down complex PDEs into simpler, separate ordinary differential equations (ODEs). We take each part, like \(X(x)\) and \(T(t)\), and solve them independently.
However, as with the given exercise, separation of variables may not always be possible. It fails if the original equation has terms that can’t be split into functions of only \(x\) or \(t\). As we saw, if the multiplication term involves both \(x\) and \(t\), like \((1+t^2)(1+x^2)\), separating them is not feasible. This mixed term prevents the equation from being split into two independent equations, stopping us from finding a solution using this method.
Homogeneous Equations
A homogeneous equation is one where all terms are of the same degree, or it's structured such that if you multiply any solution by a constant, it's still a solution. This concept is essential for understanding solutions to linear PDEs.
In our example, the PDE is homogeneous because there are no standalone terms. Everything on both sides of the equation depends on \(u\) or its derivatives. Homogeneous equations often allow for simplifications and are easier to analyze under the assumption of separation of variables.
When attempting to solve homogeneous equations using separation of variables, we rely on this property to look for solutions like \(u(x, t) = X(x)T(t)\). This simplifies analyzing the differential equation, but this exercise showed that the presence of mixed terms can still pose challenges, preventing successful separation.
Second-Order Derivatives
Second-order derivatives are crucial in partial differential equations, influencing how functions change. In our equation, \( u_{xx} \) represents the second derivative of \( u \) concerning \( x\). Similarly, parts of equations involving \( u_{tt} \), \( u_{xx} \), etc., are second-order terms.
These derivatives give information about the curvature and acceleration in the respective directions. In physical applications, such as in heat or wave equations, second-order derivatives describe rate changes in movement or energy distribution.
For the given PDE:
  • \(u_{xx}(x, t) = T(t) \frac{d^2}{dx^2} X(x)\)
is crucial as it captures how \(u\) changes with \(x\) squared, and it interacts with the term \((1+t^2)(1+x^2)\), influencing the differential relationship. These interactions often define the nature of solutions and determine the complexity of achieving a general solution. Recognizing and properly interpreting second-order derivatives is essential for solving such equations.

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Most popular questions from this chapter

Assume a solution of the linear homogeneous partial differential equation having the "separation of variables" form given. Either demonstrate that solutions having this form exist, by deriving appropriate separation equations, or explain why the technique fails. \(u_{t}(x, t)=u_{x x}(x, t)+x^{2} u(x, t), \quad u(x, t)=X(x) T(t)\)

D'Alembert's Solution of the Wave Equation Given the partial differential equation \(u_{n}(x, t)-c^{2} u_{x x}(x, t)=0\), define new independent variables \(\xi=x-c t, \eta=x+c t\). (a) Find constants \(a_{1}, a_{2}, b_{1}\), and \(b_{2}\) such that \(x=a_{1} \eta+a_{2} \xi\) and \(t=b_{1} \eta+b_{2} \xi\). Show that the determinant of this transformation, \(a_{1} b_{2}-a_{2} b_{1}\), is nonzero [establishing that there is a unique correspondence between points in the \(x t\)-plane and points in the \(\xi \eta\)-plane]. (b) In terms of the new variables, show that the wave equation transforms into \(u_{\xi \eta}=0\). You will need to use the chain rule-for example, $$ \frac{\partial u}{\partial x}=\frac{\partial u}{\partial \xi} \frac{\partial \xi}{\partial x}+\frac{\partial u}{\partial \eta} \frac{\partial \eta}{\partial x}, \quad \frac{\partial u}{\partial t}=\frac{\partial u}{\partial \xi} \frac{\partial \xi}{\partial t}+\frac{\partial u}{\partial \eta} \frac{\partial \eta}{\partial t} $$ (c) Show that the general solution of \(u_{\xi \eta}=0\) is \(u=p(\xi)+q(\eta)\), where \(p\) and \(q\) are arbitrary, twice continuously differentiable functions. Since \(\xi=x-c t, \eta=x+c t\), equation (17) follows. (d) Establish the formula in equation (18) for the solution \(u(x, t)\).

In each exercise, use the separation of variables representation developed in Exercise 16 to determine the membrane displacement \(u(x, y, t)\) for the specified initial displacement and velocity. $$f(x, y)=\sin \left(\frac{\pi x}{a}\right) \sin \left(\frac{\pi y}{b}\right), g(x, y)=0$$

(a) As in Example 2, use (14) and (15) to solve the initial value problem $$ \begin{aligned} &u_{t}(x, t)=u_{x x}(x, t), \quad 0

In each exercise, (a) Show by direct substitution that the linear combination of functions is a solution of the given homogeneous linear partial differential equation. (b) Determine values of the constants so that the linear combination satisfies the given supplementary condition. \(u(x, t)=c_{1} e^{-t} \sin x+c_{2} e^{-4 t} \sin 2 x, \quad u_{x x}-u_{t}=0\) \(u(x, 0)=3 \sin 2 x-\sin x\)

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