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Question 1. Consider the second order partial differential equation (1) for an unknown real-valued function u = u(t, x), where t represents time, r represents a point in space, and c> 0 is a constant. 1. For any twice differentiable functions F = satisfies 1). F(x) and G = G(x), show that u(t, x) = F(x + ct) + G(x – ct) Partial differential equations such as 1) the unknown function and its time derivative are provided at each point in space. Along these lines, suppose we are given that u(0, x) = 9(x) and (0, x) = h(x), for some given functions g and h. are often solved as initial value problems, where the initial description of 2. Assume that u(t, x) = F(x + ct) + G(x – ct) for some functions F and G, as described in problem 1.1. If u = u(t, x) solves the initial value problem described above, show that g(x) = F(x) + G(x) and h(x) = cF' (x) – cG' (x). 3. By integrating the last equation for h(x), show that for any constant a E R, | h(s) ds = cF(x) – cG(x) – cF(a) + cG(a), and from here solve a linear system to show that ; (o(e) +[ he) de + F(a) – G(a), F(x) = and G(#) =; (9(=) – h(») ds – F(a) + G(a) - cl). 4. Lastly, given that u(t, a) value problem for 1): F(x + ct) + G(x – ct), arrive at an explicit formula for the solution to the initial 1 1 cr+ct u(t, a) = ; [g(x + ct) + g(x – ct)] + h(s)ds. 2c r-ct

Question 1. Consider the second order partial differential equation (1) for an unknown real-valued function u = u(t, x), where t represents time, r represents a point in space, and c> 0 is a constant. 1. For any twice differentiable functions F = satisfies 1). F(x) and G = G(x), show that u(t, x) = F(x + ct) + G(x – ct) Partial differential equations such as 1) the unknown function and its time derivative are provided at each point in space. Along these lines, suppose we are given that u(0, x) = 9(x) and (0, x) = h(x), for some given functions g and h. are often solved as initial value problems, where the initial description of 2. Assume that u(t, x) = F(x + ct) + G(x – ct) for some functions F and G, as described in problem 1.1. If u = u(t, x) solves the initial value problem described above, show that g(x) = F(x) + G(x) and h(x) = cF' (x) – cG' (x). 3. By integrating the last equation for h(x), show that for any constant a E R, | h(s) ds = cF(x) – cG(x) – cF(a) + cG(a), and from here solve a linear system to show that ; (o(e) +[ he) de + F(a) – G(a), F(x) = and G(#) =; (9(=) – h(») ds – F(a) + G(a) - cl). 4. Lastly, given that u(t, a) value problem for 1): F(x + ct) + G(x – ct), arrive at an explicit formula for the solution to the initial 1 1 cr+ct u(t, a) = ; [g(x + ct) + g(x – ct)] + h(s)ds. 2c r-ct

Calculus For The Life Sciences
Calculus For The Life Sciences
2nd Edition
ISBN:9780321964038
Author:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.
Publisher:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.
Chapter11: Differential Equations
Section11.1: Solutions Of Elementary And Separable Differential Equations
Problem 54E: Plant Growth Researchers have found that the probability P that a plant will grow to radius R can be...
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Question 1. Consider the second order partial differential equation (1) for an unknown real-valued function u = u(t, x), where t represents time, r represents a point in space, and c> 0 is a constant. 1. For any twice differentiable functions F = satisfies 1). F(x) and G = G(x), show that u(t, x) = F(x + ct) + G(x – ct) Partial differential equations such as 1) the unknown function and its time derivative are provided at each point in space. Along these lines, suppose we are given that u(0, x) = g(x) and (0, x) = h(x), for some given functions g and h. are often solved as initial value problems, where the initial description of 2. Assume that u(t, x) = F(x + ct) + G(x – ct) for some functions F and G, as described in problem 1.1. If u = u(t, x) solves the initial value problem described above, show that g(x) = F(x) + G(x) and h(x) = cF' (x) – cG' (x). 3. By integrating the last equation for h(x), show that for any constant a E R, | h(s) ds = cF(x) – cG(x) – cF(a) + cG(a), and from here solve a linear system to show that F(2) = (o(2) +[ h(0) ds + F(a) – G(a). F(x) = and G(x) = ; (9(=) – h(s) ds – F(a) + G(a) 4. Lastly, given that u(t, x) = F(x + ct) + G(x – ct), arrive at an explicit formula for the solution to the initial value problem for 1): 1 1 (r+ct u(t, æ) = ; [g(x + et) + g(x – ct)] + h(s)ds. 2c r-ct xye-(2²+v°+z*). Find all points (x, y, z) such that Vf(x, y, z) = Question 2. Consider the function f(x, y, z) (0,0,0).
Transcribed Image Text:Question 1. Consider the second order partial differential equation (1) for an unknown real-valued function u = u(t, x), where t represents time, r represents a point in space, and c> 0 is a constant. 1. For any twice differentiable functions F = satisfies 1). F(x) and G = G(x), show that u(t, x) = F(x + ct) + G(x – ct) Partial differential equations such as 1) the unknown function and its time derivative are provided at each point in space. Along these lines, suppose we are given that u(0, x) = g(x) and (0, x) = h(x), for some given functions g and h. are often solved as initial value problems, where the initial description of 2. Assume that u(t, x) = F(x + ct) + G(x – ct) for some functions F and G, as described in problem 1.1. If u = u(t, x) solves the initial value problem described above, show that g(x) = F(x) + G(x) and h(x) = cF' (x) – cG' (x). 3. By integrating the last equation for h(x), show that for any constant a E R, | h(s) ds = cF(x) – cG(x) – cF(a) + cG(a), and from here solve a linear system to show that F(2) = (o(2) +[ h(0) ds + F(a) – G(a). F(x) = and G(x) = ; (9(=) – h(s) ds – F(a) + G(a) 4. Lastly, given that u(t, x) = F(x + ct) + G(x – ct), arrive at an explicit formula for the solution to the initial value problem for 1): 1 1 (r+ct u(t, æ) = ; [g(x + et) + g(x – ct)] + h(s)ds. 2c r-ct xye-(2²+v°+z*). Find all points (x, y, z) such that Vf(x, y, z) = Question 2. Consider the function f(x, y, z) (0,0,0).
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9780321964038
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