# wave equation pde problem

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Hi everyone,

I'm having a bit of trouble with this pde problem:

i get the answer to be u(x,t)=0 but i am guessing thats not right.

is the general solution to this problem: u(x,t) = f(x+ct) + g(x-ct) ??

thanks

sarah

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I don´t have the time thinking about that atm, but since I suppose others might have the same question: Are the small subscripts at the u derivatives with respect to that parameter?

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That's how I understand it; $u_t = \frac{\partial u}{\partial t}, u_{tt} = \frac{\partial^2 u}{\partial t^2}$ etc.

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well, u(x,t)=0 cannot be right, since the equation in a becomes 0=x*t

Have you tried separation of variables? That method is usually the first bullet out of my gun when given problems like that.

Let u(x,t)=X(x)*T(t), then you should be able to write 2 equations, one for X(x) (and only a function of x) and T(t) for only a function of t. Then solve the two ODEs.

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ok, i tried again and this is what i get...

i included a non-homogenous term...

for (a)

for (b)

what do you guys think?

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I don´t have the time thinking about that atm, but since I suppose others might have the same question: Are the small subscripts at the u derivatives with respect to that parameter?

yes they are

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well, u(x,t)=0 cannot be right, since the equation in a becomes 0=x*t

Have you tried separation of variables? That method is usually the first bullet out of my gun when given problems like that.

Let u(x,t)=X(x)*T(t), then you should be able to write 2 equations, one for X(x) (and only a function of x) and T(t) for only a function of t. Then solve the two ODEs.

i can't see how to apply seperation of variables (at least for part (a)).... i get this after plugging u(x,t) = X(x)T(t) into the orginal PDE:

XT'' = (c^2)X''T + xt

but you can't seperate the x's and t's ...? lol i am probably just completely wrong!

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i can't see how to apply seperation of variables (at least for part (a)).... i get this after plugging u(x,t) = X(x)T(t) into the orginal PDE:

XT'' = (c^2)X''T + xt

but you can't seperate the x's and t's ...? lol i am probably just completely wrong!

Separation of variables doesn't always work, and I think since the extra term is added, not multiplied, that you can't use it here. Your answers look like they should work for the extra term, but you also have the basic part, which should be an exponential of some sort.

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You need to write down the most general expression which could give the last term, and then work out the coefficients. For example, for (a) write down a polynomial in x and t up to four powers (in x or t) with arbitrary coefficients. Stick it in and solve for the coefficients. That gives you one solution, but then you can add anything to that solution which satisfies $u_{xx}=c^2u_{tt}$ and it will still be a solution. So you also need to find the solution to that inhomogenous equation and add it in.

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You need to write down the most general expression which could give the last term, and then work out the coefficients. For example, for (a) write down a polynomial in x and t up to four powers (in x or t) with arbitrary coefficients. Stick it in and solve for the coefficients. That gives you one solution, but then you can add anything to that solution which satisfies $u_{xx}=c^2u_{tt}$ and it will still be a solution. So you also need to find the solution to that inhomogenous equation and add it in.

ok, i think i did something like that...i said that part (a) has a solution of:

$u(x,t) = f(x+ct)+g(x-ct)-\frac{x^3t}{6c^2}$

then using the initial conditions i solved for f and g... and came up with my answer of $u(x,t) = \frac{1}{6}xt^3$

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its ok i think its correct, its some other questions that i am having problems with now!

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You can add ex+ct to that and still satisfy the solution, which means that it must be included in the answer.

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You can add ex+ct to that and still satisfy the solution, which means that it must be included in the answer.

how do you know to add that? i see that it works, but at what point do you figure out to add it? that is it doesnt come out in any of the steps i take to solve the problem...

does this mean for this problem: that i have to do a similar thing? (i.e. add $e^{x+ct}$ ?

i get for that problem that u(x,t) = 1 for x > ct and u(x,t) = 0 for 0 < x < ct but that seems to simple, am i forgetting something completely obvious swansont? :S

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I would have solved the second-order part of the problem first, and then add in the solution to the extra function. A second deriviative equal to itself is an exponential of some form (real or imaginary). You really have to look at whether something that looks like e-x fits as well; the general solution will be a linear combination of the two, as Severian already stated. Since part of solving differential equations can be to guess the answer, you should try and remember the general solution form to some of the common equations.

You also need to consider (depending on the problem) whether other terms that are linear in the variables will fit, since they disappear when you differentiate twice. Again, as Severian had suggested, you can write out a polynomial and find the solution.

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but don't we also need to satify the initial conditions?

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but don't we also need to satify the initial conditions?

Yes; I was still looking at the general solution. The constraints would seem to eliminate the exponential (real or imaginary) solution.

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Yes; I was still looking at the general solution. The constraints would seem to eliminate the exponential (real or imaginary) solution.

ah ok, yeah that all makes sense now.

since you people have been so helpful with this PDE question, if any of you guys have time, i would really love some help with some of my other questions

thanks guys!

Sarah

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