Question pde question. please help me to PROVE THE PERIODIC CASE! Thanks!:-) Corollary 4.11 Suppose that f. g are even functions, and for every t > 0 the func- tion F(,1) is even too. Then for every 1 > 0 the solution u(·,t) of the Cauchy problem (4.13)H4.14) is also even. Similarly, the solution is an odd function or a periodic function with a period L (as a function of x) if the data are odd functions or periodic functions with a period L. Proof We prove the first part of the corollary. The other parts can be shown simi- larly. Let u be the solution of the problem and define the function v(x, t) = u(-x,1). Clearly, vx(x,1)= -ux(-x,t), v(x, t) = u(-x,1) and Vxx(x, 1) = Uxx(-x,1), V., (X, 1) = U/(-X, 1). Therefore, Vr(x, t)-c-vxx(x, t) = 4,(-x, t)-c?uxx(-x,1) = F(-X,1)=F(x,1) - 00. Thus, v is a solution of the nonhomogeneous wave equation (4.13). Furthermore, v(x,0) = u(-x,0) = f(-x) = f(x), v(x,0) = u(-x,0) = g(-x) = g(x). It means that v is also a solution of the initial value problem (4.13)-(4.14). Since the solution of this problem is unique, we have v(x, 1) = u(x, t), which implies u(-X, 1) = u(x, t).
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pde question. please help me to PROVE THE PERIODIC CASE! Thanks!:-)
Transcribed Image Text: Corollary 4.11 Suppose that f. g are even functions, and for every t > 0 the func- tion F(,1) is even too. Then for every 1 > 0 the solution u(·,t) of the Cauchy problem (4.13)H4.14) is also even. Similarly, the solution is an odd function or a periodic function with a period L (as a function of x) if the data are odd functions or periodic functions with a period L. Proof We prove the first part of the corollary. The other parts can be shown simi- larly. Let u be the solution of the problem and define the function v(x, t) = u(-x,1). Clearly, vx(x,1)= -ux(-x,t), v(x, t) = u(-x,1) and Vxx(x, 1) = Uxx(-x,1), V., (X, 1) = U/(-X, 1). Therefore, Vr(x, t)-c-vxx(x, t) = 4,(-x, t)-c?uxx(-x,1) = F(-X,1)=F(x,1) - 00. Thus, v is a solution of the nonhomogeneous wave equation (4.13). Furthermore, v(x,0) = u(-x,0) = f(-x) = f(x), v(x,0) = u(-x,0) = g(-x) = g(x). It means that v is also a solution of the initial value problem (4.13)-(4.14). Since the solution of this problem is unique, we have v(x, 1) = u(x, t), which implies u(-X, 1) = u(x, t).
More Transcribed Image Text: Corollary 4.11 Suppose that f. g are even functions, and for every t > 0 the func- tion F(,1) is even too. Then for every 1 > 0 the solution u(·,t) of the Cauchy problem (4.13)H4.14) is also even. Similarly, the solution is an odd function or a periodic function with a period L (as a function of x) if the data are odd functions or periodic functions with a period L. Proof We prove the first part of the corollary. The other parts can be shown simi- larly. Let u be the solution of the problem and define the function v(x, t) = u(-x,1). Clearly, vx(x,1)= -ux(-x,t), v(x, t) = u(-x,1) and Vxx(x, 1) = Uxx(-x,1), V., (X, 1) = U/(-X, 1). Therefore, Vr(x, t)-c-vxx(x, t) = 4,(-x, t)-c?uxx(-x,1) = F(-X,1)=F(x,1) - 00. Thus, v is a solution of the nonhomogeneous wave equation (4.13). Furthermore, v(x,0) = u(-x,0) = f(-x) = f(x), v(x,0) = u(-x,0) = g(-x) = g(x). It means that v is also a solution of the initial value problem (4.13)-(4.14). Since the solution of this problem is unique, we have v(x, 1) = u(x, t), which implies u(-X, 1) = u(x, t).
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Given PDE: u_(tt)-c^(2)u_(xx)=F(x,t)cdots(4.13)" IC: "{:[u(x","0)=f(x)quad dots(4.14)],[u_(t)(x","0)=g(x)quad cdots(u_(1):}]:}Also, it is given that F,f,g are periodic functions with period 1 (as a function of x ).So,{:[F(x+L","t)=F(x","t)],[f(x+L)=f(x)],[g(x+L)=g(x)]}quad cdots(4.15)Ret U(x,z) be the solution of the cauchy problem (4.13)-(4. 14) To show u(x,t) is perisdic i.e, u(x+L,t)=u(x,t):Define the function v(x,t)=u(x+L,t)So,{:[v_(x)(x","t)=u_(x)(x+L","t)",",v_(t)(x","t)=u_(t)(x+L","t)],[v_(x_(x))(x","t)=u_(xx)(x+L","t)",",v_(tt)(x","t)=u_(tt)(x+L","t)]:}Therefore,{:[v_(tt) ... See the full answer