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Chapter 13: Problem 16

Substituting \(u(x, t)=X(x) T(t)\) into the partial differential equation yields \(a^{2} X^{\prime \prime} T-g=X T^{\prime \prime},\) which is not separable.

Short Answer

Expert verified
The equation is not separable due to the complex interdependence of variables.

Step by step solution

01

Express the Given PDE in Terms of Variables

The partial differential equation given is \(a^{2} X'' T - g = X T''\). Here, \(X(x)\) is a function of \(x\) only, and \(T(t)\) is a function of \(t\) only, indicating the process of separation of variables.
02

Identify Independent Functions

We hypothesise that the given PDE is not separable because somehow the variables \(x\) and \(t\) cannot be successfully isolated. Normally, separation involves rearranging terms so the equations on each side depend on only one variable each.
03

Attempt to Reorder Terms

To apply separation, rearrange to isolate terms involving \(X\) and \(T\): \(\frac{a^{2} X''}{X} = \frac{T''}{T} + \frac{g}{X T}\). This structure usually implies separability if each side was equal to a constant.
04

Evaluate Terms for Separability

Observe that even after separating the terms: \(\frac{a^{2} X''}{X}\) and \(\frac{T''}{T} + \frac{g}{X T}\), they are not equated separately to a constant. This inequality in variable dependence signals that the equation is not separable.

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Key Concepts

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

Separation of Variables
Separation of variables is a technique used to solve partial differential equations (PDEs). This method involves assuming that a solution to the PDE can be expressed as a product of functions, each depending only on a single variable. In our case, we assume that the solution is of the form \(u(x, t) = X(x)T(t)\). This approach is straightforward but requires that the equation can be split into separate manageable parts for each variable.
  • Step 1: Substitute the solution format into the PDE. For example, in our problem, \(a^2 X'' T - g = X T''\).
  • Step 2: Divide both sides such that each term involves only one of the variables. This is where we identify \(\frac{a^2 X''}{X}\) and \(\frac{T''}{T} + \frac{g}{X T}\).
  • Step 3: Determine if both sides of the equation can be set equal to some constant, which would indicate separability.
In this example, separation does not directly work because the terms do not equate to constants separately. Instead, they remain intertwined with both variables, pointing to a non-separable equation.
Non-separable Equations
Non-separable equations are PDEs where the variables cannot be isolated to separate sides of the equation. This means that the standard approach of separation of variables perhaps is not applicable, necessitating alternative strategies. Our PDE equation \(a^{2} X'' T - g = X T''\) demonstrated such a non-separable scenario.
This situation arises when:
  • Reordering or dividing terms involves multiple variables that cannot settle as equal constants.
  • Any attempt to set up separate terms for \(x\) and \(t\) will still involve the other variable somehow, thwarting separability.
Recognizing a non-separable form early can save time and effort, as these require more sophisticated mathematical methods or numerical techniques to tackle.
Mathematical Methods
When it comes to solving PDEs that are non-separable, as identified, we might need to rely on a broader array of mathematical techniques beyond straightforward techniques. While separation of variables is often the first step attempted, it's not always successful. This situation requires flexibility and familiarity with other methods:
  • Transform Methods: Techniques like Fourier or Laplace transforms can convert difficult time-dependent or space-dependent parts into simpler algebraic equations.
  • Numerical Solutions: Computational approaches use discretization techniques, like finite difference or finite element methods, to approximate solutions.
  • Perturbation Methods: Useful when dealing with non-linearities or small parameters, this technique approximates solutions using simplifying assumptions.
Each method holds its strength depending on the nature of the PDE and the required accuracy of the solution. Employing the right mathematical approach can resolve complex non-separable PDEs effectively.

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Most popular questions from this chapter

Identifying \(A=1, B=2,\) and \(C=1,\) we compute \(B^{2}-4 A C=0 .\) The equation is parabolic.

Substituting \(u(r, t)=R(r) T(t)\) into the partial differential equation yields $$k\left(R^{\prime \prime} T+\frac{1}{r} R^{\prime} T\right)=R T^{\prime}.$$ Separating variables and using the separation constant \(-\lambda\) we obtain. $$\frac{r R^{\prime \prime}+R^{\prime}}{r R}=\frac{T^{\prime}}{k T}=-\lambda.$$ Then $$r R^{\prime \prime}+R^{\prime}+\lambda r R=0 \quad \text { and } \quad T^{\prime}+\lambda k T=0.$$ Letting \(\lambda=\alpha^{2}\) and writing the first equation as \(r^{2} R^{\prime \prime}+r R^{\prime}=\alpha^{2} r^{2} R=0\) we see that it is a parametric Bessel equation of order \(0 .\) As discussed in Chapter 5 of the text, it has solution \(R(r)=c_{1} J_{0}(\alpha r)+c_{2} Y_{0}(\alpha r) .\) since a solution of \(T^{\prime}+\alpha^{2} k T\) is \(T(t)=e^{-k \alpha^{2} t},\) we see that a solution of the partial differential equation is $$u=R T=e^{-k \alpha^{2} t}\left[c_{1} J_{0}(\alpha r)+c_{2} Y_{0}(\alpha r)\right].$$

As shown in Example 1 in the text, separation of variables leads to \\[ \begin{array}{l} X(x)=c_{1} \cos \alpha x+c_{2} \sin \alpha x \\ Y(y)=c_{3} \cos \beta y+c_{4} \sin \beta y \end{array} \\] and \\[ T(t)+c_{5} e^{-k\left(\alpha^{2}+\beta^{2}\right) t} \\] The boundary conditions \\[ \left.\begin{array}{l} u_{x}(0, y, t)=0, \quad u_{x}(1, y, t)=0 \\ u_{y}(x, 0, t)=0, \quad u_{y}(x, 1, t)=0 \end{array}\right\\} \quad \text { imply } \quad\left\\{\begin{array}{ll} X^{\prime}(0)=0, & X^{\prime}(1)=0 \\ Y^{\prime}(0)=0, & Y^{\prime}(1)=0 \end{array}\right. \\] Applying these conditions to \\[ X^{\prime}(x)=-\alpha c_{1} \sin \alpha x+\alpha c_{2} \cos \alpha x \\] and \\[ Y^{\prime}(y)=-\beta c_{3} \sin \beta y+\beta c_{4} \cos \beta y \\] gives \(c_{2}=c_{4}=0\) and \(\sin \alpha=\sin \beta=0 .\) Then \\[ \alpha=m \pi, m=0,1,2, \ldots \quad \text { and } \quad \beta=n \pi, n=0,1,2, \ldots \\] By the superposition principle \\[ u(x, y, t)=A_{00}+\sum_{m=1}^{\infty} A_{m 0} e^{-k m^{2} \pi^{2} t} \cos m \pi x+\sum_{n=1}^{\infty} A_{0 n} e^{-k n^{2} \pi^{2} t} \cos n \pi y \\] \\[ +\sum_{m=1}^{\infty} \sum_{n=1}^{\infty} A_{m n} e^{-k\left(m^{2}+n^{2}\right) \pi^{2} t} \cos m \pi x \cos n \pi y \\] We now compute the coefficients of the double cosine series: Identifying \(b=c=1\) and \(f(x, y)=x y\) we have \\[ \begin{aligned} A_{00} &=\int_{0}^{1} \int_{0}^{1} x y d x d y=\left.\int_{0}^{1} \frac{1}{2} x^{2} y\right|_{0} ^{1} d y=\frac{1}{2} \int_{0}^{1} y d y=\frac{1}{4} \\ A_{m 0} &=2 \int_{0}^{1} \int_{0}^{1} x y \cos m \pi x d x d y=\left.2 \int_{0}^{1} \frac{1}{m^{2} \pi^{2}}(\cos m \pi x+m \pi x \sin m \pi x)\right|_{0} ^{1} y d y \\ &=2 \int_{0}^{1} \frac{\cos m \pi-1}{m^{2} \pi^{2}} y d y=\frac{\cos m \pi-1}{m^{2} \pi^{2}}=\frac{(-1)^{m}-1}{m^{2} \pi^{2}} \\ A_{0 n} &=2 \int_{0}^{1} \int_{0}^{1} x y \cos n \pi y d x d y=\frac{(-1)^{n}-1}{n^{2} \pi^{2}} \end{aligned} \\] and \\[ A_{m n}=4 \int_{0}^{1} \int_{0}^{1} x y \cos m \pi x \cos n \pi y d x d y=4 \int_{0}^{1} x \cos m \pi x d x \int_{0}^{1} y \cos n \pi y d y \\] \\[ =4\left(\frac{(-1)^{m}-1}{m^{2} \pi^{2}}\right)\left(\frac{(-1)^{n}-1}{n^{2} \pi^{2}}\right) .\\]

Assuming \(u(x, y)=X(x) Y(y)\) and substituting into \(\partial^{2} u / \partial x^{2}-u=0\) we get \(X^{\prime \prime} Y-X Y=0\) or \(Y\left(X^{\prime \prime}-X\right)=0\) This implies \(X(x)=c_{1} e^{x}\) or \(X(x)=c_{2} e^{-x}\). For these choices of \(X, Y\) can be any function of \(y .\) Two solutions of the partial differential cquation are then $$u_{1}(x, y)=A(y) e^{x} \quad \text { and } \quad u_{2}(x, y)=H(y) e^{x}.$$ since the partial differential equation is linear and homogencous the superposition principle indicates that another solution is $$u(x, y)=u_{1}(x, y)+u_{2}(x, y)=A(y) e^{x}+B(y) e^{-x}.$$

Substituting \(u(x, y)=X(x) Y(y)\) into the partial differential equation yields \(X^{\prime} Y=X Y^{\prime}+X Y .\) Separating variables and using the separation constant \(-\lambda\) we obtain $$\frac{X^{\prime}}{X}=\frac{Y+Y^{\prime}}{Y}=-\lambda.$$ Then $$X^{\prime}+\lambda X=0 \quad \text { and } \quad y^{\prime}+(1+\lambda) Y=0$$ so that $$\begin{aligned} &X=c_{1} e^{-\lambda x} \quad \text { and }\\\ &Y=c_{2} e^{-(1+\lambda) y}=0. \end{aligned}$$ A particular product solution of the partial differential equation is $$u=X Y=c_{3} e^{-y-\lambda(x+y)}.$$
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