Solving uxx = utt for 0<x<π using separation of variables

[Jncik]

Homework Statement

solve u_xx = u_tt

for 0<x<π when u(0,t) = u(π,t) = 0 and u(x,0) = sin(2x) and u_t(π/4,0) = 0

The Attempt at a Solution

i used the method of separation of variables and have found out that

[URL]http://latex.codecogs.com/gif.latex?u%28x,t%29%20=%20%28c_{1}e^{x\sqrt{\lambda}}%20+%20c_{2}e^{-x\sqrt{\lambda}}%29%28C_{1}e^{t\sqrt{\lambda}}%20+%20C_{2}e^{-t\sqrt{\lambda}}%29[/URL]the problem is that i don't know how to use the given equations to find the final result

for example

[URL]http://latex.codecogs.com/gif.latex?u%280,t%29%20=%200%20=%3E%20%28c_{1}%20+%20c_{2}%29%28C_{1}e^{t\sqrt{\lambda}}%20+%20C_{2}e^{-t\sqrt{\lambda}}%29%20=%200%20=%3E%20?[/URL]

my professor says that we assume that there is a t for which

[URL]http://latex.codecogs.com/gif.latex?%28C_{1}e^{t\sqrt{\lambda}}%20+%20C_{2}e^{-t\sqrt{\lambda}}%29%20\neq%200[/URL]

and he finds that c₁ = -c₂

for

u(π,t) he says

[URL]http://latex.codecogs.com/gif.latex?u%28%CF%80,t%29%20=%20%28c_{1}e^{\pi%20\sqrt{\lambda}}%20-%20c_{1}e^{-\pi\sqrt{\lambda}}%29%28C_{1}e^{t\sqrt{\lambda}}%20+%20C_{2}e^{-t\sqrt{\lambda}}%29%20=%200%20=%3E%20%28c_{1}e^{\pi%20\sqrt{\lambda}}%20-%20c_{1}e^{-\pi\sqrt{\lambda}}%29%20=%200%20=%3E%20\lambda%20=%20-n^{2},%20n%20=%200,1,2,3...[/URL]

so we have

[URL]http://latex.codecogs.com/gif.latex?%28e^{\pi%20\sqrt{\lambda}}%20-%20e^{-\pi\sqrt{\lambda}}%29%20=%200%20=%3E%20e^{in\pi}%20-%20e^{-in\pi]}%20=%200%20=%3E%202sin%28n\pi%29%20=%200[/URL]

and from this he says the following which i can't understand

from u(x,0) = sin(2x) we get

λ = -4 and c₂ = 1/2

how did we get this?

Thanks in advance

 

[HallsofIvy]

Homework Statement

solve u_xx = u_tt

for 0<x<π when u(0,t) = u(π,t) = 0 and u(x,0) = sin(2x) and u_t(π/4,0) = 0

The Attempt at a Solution

i used the method of separation of variables and have found out that

[URL]http://latex.codecogs.com/gif.latex?u%28x,t%29%20=%20%28c_{1}e^{x\sqrt{\lambda}}%20+%20c_{2}e^{-x\sqrt{\lambda}}%29%28C_{1}e^{t\sqrt{\lambda}}%20+%20C_{2}e^{-t\sqrt{\lambda}}%29

[/URL]
This is wrong. I presume that, separating variables so that u(x,t)= X(x)T(t), you found that $X''/X= \lambda= T''/T$ so that you got the two ordinary differential equations $X''= \lambda X$ and $T''= \lambda T$. What you have here is the solution if $\lambda> 0$ and you have no reason to believe that.

But you can also say that $u(0,t)= X(0)T(t)= 0$ means either X(0)= 0 or T(t)= 0. Since the latter would make the solution 0 for all t, it cannot satisfy the initial conditions so we must have X(0)= 0. Similarly, $u(\pi, t)= X(\pi)T(t)= 0$ gives $X(\pi)= 0$.

So we are solving $X''= \lambda X$ with boundary conditions $X(0)= X(\pi)= 0$.

Consider three cases: $\lambda= 0$, $\lambda> 0$, $\lambda< 0$

If $\lambda= 0$, the equation is just X''= 0 and, integrating twice, $X(x)= C_1x+ C_2$. Then X(0)= C_2= 0 so $X(\pi)= C_1\pi= 0$ which implies that X is identically 0. But then we cannot satisfy the intial condition that u(x, 0)= sin(2x) so that is impossible,

If $\lambda> 0$, we can let $\lambda= \alpha^2$ for $\alpha$ non-zero. The differential equation is $X''= \alpha^2X$ and the general solution is $X(x)= C_1e^{\alpha x}+ C_2e^{-\alpha x}$ (this is what you had). Now we must have $X(0)= C_1+ C_2= 0$ and $X(\pi)= C_1e^{\alpha\pi}+ C_2e^{-\alpha\pi}$. From the first equation, $C_2= -C_1$ and, putting that into the second, $C_1e^{\alpha\pi}- C_1e^{-\alpha\pi}= C_1(e^{\alpha\pi}- e^{-\alpha\pi})= 0$.

Now, since $\alpha \ne 0$, $e^{\alpha\pi}$ and $e^{-\alpha\p}$ are not equal (one is larger than 1 and the other less than 1) and their difference cannot be 0, so we must have $C_1= 0$ and then $C_2= 0$. That is, X(x) is identically 0 so it cannot give a u that satisifies the initial condition.

Finally, if $\lambda< 0$, we can let $\lambda= -\alpha^2$ for $\alpha$ non-zer. The differential equation is $X''= -\alpha^2X$ and the general solution to that is $X(x)= C_1cos(\alpha x)+ C_2 sin(\alpha x)$. Now we must have $X(0)= C_1= 0$ and $X(\pi)= C_2 sin(\alpha\pi)= 0$. This would give $C_2= 0$ and so again a "trivial" solution unless $sin(\alpha\pi)= 0$ which does happen if $\alpha\pi$ is a multiple of $\pi$- that is, if $\alpha= n$ and $\lambda= -n^2$ with n and integer.

That is, in order to have a non-trivial solution, we must have $\lambda= -n^2$ for n some integer, in that case, $X(x)= Csin(nx)$ for unknown constant C and unknown integer n.

Now, go back to your equation for T: T''= \lambda T= -n²T. What is the general solution to that?

( Your condition "[itex]u_t(\pi/4, 0)= 0" looks very peculiar- usually it is something like $u(x, 0)= 0$ for all x, but I think $u_t(\pi/4, 0)= 0$ will give the same thing. $u_t(\pi/4, 0)= X(\pi/4)T'(0)= 0$ and $X(\pi/4)\ne 0$ so you must have $T'(\pi/4)= 0$.<br /> <br /> <br /> <blockquote data-attributes="" data-quote="" data-source="" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> the problem is that i don't know how to use the given equations to find the final result<br /> <br /> for example<br /> <br /> [URL]http://latex.codecogs.com/gif.latex?u%280,t%29%20=%200%20=%3E%20%28c_{1}%20+%20c_{2}%29%28C_{1}e^{t\sqrt{\lambda}}%20+%20C_{2}e^{-t\sqrt{\lambda}}%29%20=%200%20=%3E%20?[/URL]<br /> <br /> my professor says that we assume that there is a t for which <br /> <br /> [URL]http://latex.codecogs.com/gif.latex?%28C_{1}e^{t\sqrt{\lambda}}%20+%20C_{2}e^{-t\sqrt{\lambda}}%29%20\neq%200[/URL]<br /> <br /> and he finds that c₁ = -c₂<br /> <br /> for<br /> <br /> u(π,t) he says<br /> <br /> [URL]http://latex.codecogs.com/gif.latex?u%28%CF%80,t%29%20=%20%28c_{1}e^{\pi%20\sqrt{\lambda}}%20-%20c_{1}e^{-\pi\sqrt{\lambda}}%29%28C_{1}e^{t\sqrt{\lambda}}%20+%20C_{2}e^{-t\sqrt{\lambda}}%29%20=%200%20=%3E%20%28c_{1}e^{\pi%20\sqrt{\lambda}}%20-%20c_{1}e^{-\pi\sqrt{\lambda}}%29%20=%200%20=%3E%20\lambda%20=%20-n^{2},%20n%20=%200,1,2,3...[/URL]<br /> <br /> so we have<br /> <br /> [URL]http://latex.codecogs.com/gif.latex?%28e^{\pi%20\sqrt{\lambda}}%20-%20e^{-\pi\sqrt{\lambda}}%29%20=%200%20=%3E%20e^{in\pi}%20-%20e^{-in\pi]}%20=%200%20=%3E%202sin%28n\pi%29%20=%200[/URL]<br /> <br /> and from this he says the following which i can't understand<br /> <br /> from u(x,0) = sin(2x) we get<br /> <br /> λ = -4 and c₂ = 1/2<br /> <br /> how did we get this?<br /> <br /> Thanks in advance </div> </div> </blockquote>[/itex]

 

[Jncik]

it took me 3 hours to understand your reply but it was definitely worth it since my teacher himself would never explain anything for me to understand these basic principles

I have some stuff that I don't really get 100% though, they will be in red, any further explanations would be appreciated

if I understand correctly

generally we don't want solutions that are 0, for example in the first two parts for λ>0 and λ=0 we get two constants to be 0, which definitely gives a solution which is 0 as a whole

note that I don't know why we don't want this, you said that it doesn't satisfy the initial conditions, if you could tell me which conditions you're referring to i would be really grateful (i)

now for λ = -b^2<0 after reading your reply I tried to solve it on my own and figured out that we haveX(x) = A*cos(bx) + B*sin(bx) where A,B are just constants

from the first boundaries, by taking X(0) = 0 we get that A is 0

but when we take X(π) = 0 we have B*sin(b*π) = 0 and if I understand correctly we don't want B to be 0 since it would give a total solution for u(x,t) = 0 (now again I am not sure why 0 is not desirable since it's a solution)

so we have to get sin(b*π) to be 0, which gives us b*π = n*π where n is an integer but not 0?? (ii)

hence b = n.

from this we get that λ = -n^2.

now taking the equation T'' + n^2*T = 0 i get that

T(t) = D*cos(nt) + E*sin(nt) where D and E are just constants again

so for λ = -n^2 we have

u(x,t) = B*sin(nx) [ D*cos(nt) + E*sin(nt) ] (1)

now if I understand correctly the next step is to figure out the constants but before this

I need to find out λ and n from the equation u(x,0) = sin(2x)

Ok that's easy

We get from (1) the following equation:

B*D*sin(nx) = sin(2x)

Now from here I guess that n should be 2, but what about B and D? We just say that they can be anything? doesn't this change the solution?
iii
supposing that B and D are just meaningless n will be 2 and λ = - 4

hence we got

u(x,t) = B*sin(tx) [ D*cos(2t) + E*sin(2t) ]

Now from this we take the equation

u_t(Pi/4,0) = 0

and I have B* 1 * [ 2*E ] = 0

now, seeing the equation u(x,t) and what I have now, obviously we don't want to have B = 0 since it would give us a 0 as a result

hence E shall be 0.

so

u(x,t) = B*sin(2x) * D*cos(2t) = C sin(2x) * cos(2t) where C is a constantAgain, thank you very much for your explanation