Advanced Applied Differenital Equation u tt - c 2 u xx - au xxt = 0 , for -infin

Question

Advanced Applied Differenital Equation

utt - c2uxx - auxxt = 0 , for -infinity < x, t < infinity

1- show that the total energy is a positive and negative function of time.

assume u(x,t) = 0

2- Show that u(x,t) = 0 is the only possible solution to the equation satisfying the initial conditions: u(x,0) = 0, and ut(x,0) = 0.

3- Show that there exists at most one solution to eqaution (1) that satisfy the initial condition u(x,0) = f(x) and ut(x,0) = g(x), where -inifinity < x < infinity

Explanation / Answer

Given differential equation is

utt-c2uxx-auxxt=0

T"x-c2X"t-aX"t=0

T"x=X"t(c2-a).   

Now,variable separation method,PDE become

X"+px=0 and T"+(c2-a)(pt)=0. Let us assume p=m2.

General solution become

u(x,t)=(c1cosmx+c2sinmx)(c3cos(m(c2-a)t)+c4sin(m(c2-a)t)).

1.Given energy formula is

E(t)=|-&-->+&(du/dt)2+c2(du/dx)2 dx. And given assumption is u(x,t)=0.

c1cosmx=-c2sinmx and c3cosm((c2-a)t)=-c4sinm((c2-a)t.

In the energy equation,integration is taking with respect to x.so,resulting total energy is a positive and negative function of time.

E(t) can also be written as

E(t)=|-&---.&(du/dt)2+c2(du/dx)u dx.

After substituting above equations

=(some constant +sine and cosine function interms of t).

So,it is a function of t.

(2). Given inital conditions are

u(x,0)=0 and ut(x,0)=0.

After substituting these conditions

u(x,0)=c3=0.

ut(x,0)=c4=0.

So,Finally it is proved that

u(x,t)=(0+0)=0. If u(x,0)=0 and ut(x,0)=0.

(3).If u(x,0)=f(x) and ut(x,0)=g(x)

Then by substituting these values

u(x,0)=c3(c1cosmx+c2sinmx)=f(x).

ut(x,0)=c4(c1cosmx+c2sinmx)=g(x).

Here there exists at most one solution depending on functions f(x) and g(x).

If these functions are in the form of below

f(x)=g(x)=cosx+sinx.then all coefficients are 1.

There exists at most one solution that is

u(x,t)=(cosx+sinx)(cos((c2-a)t+sin((c2-a)t.

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