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Chapter 6: Problem 31

Determine the solvability conditions for the nonhomogeneous boundralue problem: \(u^{\prime}(\pi / 4)=\beta\).

Short Answer

Expert verified
The solvability conditions for the given boundary value problem can't be determined without additional information.

Step by step solution

01

Understanding the Problem

The given problem \(u^{\prime}(\pi / 4) = \beta\) is a boundary value problem as it specifies the value of the derivative of the function at a particular point, here at \(x=\pi / 4\). The solvability of this problem, in simplest terms, would mean a set of conditions under which a solution to the equation exists.
02

Identifying Solvability Conditions

This is a first order differential equation, and it's known an initial value problem of the form \(u^{\prime} = f(x, u)\) has a unique solution provided \(f\) and \(\partial f / \partial u\) are continuous within a region containing the initial point. However, we don't have a function f(x,u) here, the derivative at a single point \(x = pi / 4\) is given, not an interval. This limits our ability to reach a conclusion about the solvability directly.
03

Conclusion

Without additional information, no general conclusion can be given about the solvability of the given boundary problem since only the derivative value at a single point is given

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Most popular questions from this chapter

Consider the boundary value problem: \(y^{\prime \prime}-y=x, x \in(0,1)\), with Idary conditions \(y(0)=y(1)=0\). a. Find a closed form solution without using Green's functions. b. Determine the closed form Green's function using the properties of Green's functions. Use this Green's function to obtain a solution of the boundary value problem. c. Determine a series representation of the Green's function. Use this Green's function to obtain a solution of the boundary value problem. d. Confirm that all of the solutions obtained give the same results.

The Hermite polynomials, \(H_{n}(x)\), satisfy the following: i. \(=\int_{-\infty}^{\infty} e^{-x^{2}} H_{n}(x) H_{m}(x) d x=\sqrt{\pi} 2^{n} n ! \delta_{n, m}\) ii. \(H_{n}^{\prime}(x)=2 n H_{n-1}(x)\). iii. \(H_{n+1}(x)=2 x H_{n}(x)-2 n H_{n-1}(x)\) iv. \(H_{n}(x)=(-1)^{n} e^{x^{2}} \frac{d^{n}}{d x^{n}}\left(e^{-x^{2}}\right) .\) ng these, show that a. \(H_{n}^{\prime \prime}-2 x H_{n}^{\prime}+2 n H_{n}=0\). [Use properties ii. and iii.] b. \(\int_{-\infty}^{\infty} x e^{-x^{2}} H_{n}(x) H_{m}(x) d x=\sqrt{\pi} 2^{n-1} n !\left[\delta_{m, n-1}+2(n+1) \delta_{m, n+1}\right] .\) [Use properties i. and iii.]

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Consider the problem: $$ \frac{\partial^{2} G}{\partial x^{2}}=\delta\left(x-x_{0}\right), \quad \frac{\partial G}{\partial x}\left(0, x_{0}\right)=0, \quad G\left(\pi, x_{0}\right)=0 $$ a. Solve by direct integration. b. Compare this result to the Green's function in part b of Problem \(31 .\) c. Verify that \(G\) is symmetric in its arguments.

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