Solution Found!
Consider u t = k 2u x2 + Q(x, t) subject to u(0, t) = 0, u x (L, t) = 0, and u(x, 0) =
Chapter 9, Problem 9.3.4(choose chapter or problem)
Consider
\(\frac{\partial u}{\partial t}=k \frac{\partial^{2} u}{\partial x^{2}}+Q(x, t)\)
subject to \(u(0, t)=0, \frac{\partial u}{\partial x}(L, t)=0\), and u(x, 0) = g(x).
(a) Derive (9.3.29) from (9.3.28). [Hint: Let f(x) = 1.]
(b) Show that (9.3.33) satisfies (9.3.31).
(c) Derive (9.3.30). [Hint: Show for any continuous f(x) that
\(\int_{-\infty}^{\infty} f\left(x_{0}\right) \delta\left(x-x_{0}\right) d x_{0}=\int_{-\infty}^{\infty} f\left(x_{0}\right) \delta\left(x_{0}-x\right) d x_{0}\)
by letting \(x_{0}-x=s\) in the integral on the right. ]
(d) Derive (9.3.34). [Hint: Evaluate \(\int_{-\infty}^{\infty} f(x) \delta\left[c\left(x-x_{0}\right)\right.\)] dx by making the change of variables \(y=c\left(x-x_{0}\right)\).]
Questions & Answers
QUESTION:
Consider
\(\frac{\partial u}{\partial t}=k \frac{\partial^{2} u}{\partial x^{2}}+Q(x, t)\)
subject to \(u(0, t)=0, \frac{\partial u}{\partial x}(L, t)=0\), and u(x, 0) = g(x).
(a) Derive (9.3.29) from (9.3.28). [Hint: Let f(x) = 1.]
(b) Show that (9.3.33) satisfies (9.3.31).
(c) Derive (9.3.30). [Hint: Show for any continuous f(x) that
\(\int_{-\infty}^{\infty} f\left(x_{0}\right) \delta\left(x-x_{0}\right) d x_{0}=\int_{-\infty}^{\infty} f\left(x_{0}\right) \delta\left(x_{0}-x\right) d x_{0}\)
by letting \(x_{0}-x=s\) in the integral on the right. ]
(d) Derive (9.3.34). [Hint: Evaluate \(\int_{-\infty}^{\infty} f(x) \delta\left[c\left(x-x_{0}\right)\right.\)] dx by making the change of variables \(y=c\left(x-x_{0}\right)\).]
ANSWER:Step 1 of 4
(a)
Consider the following equation
\(f\left( x \right) = \int\limits_{ - \infty }^\infty {f\left( {{x_i}} \right)\delta \left( {x - {x_i}} \right)d{x_i}}\) ……(9.3.28)
Since \(f\left( x \right)\) can be any continuous function, let it be 1.
\(1 = \int\limits_{ - \infty }^\infty {\delta \left( {x - {x_i}} \right)d{x_i}}\) ……(9.3.29)
Hence, the equation (9.3.29) is derived from (9.3.28).