Check Homework on Partial Differential Wave Equation

[roldy]

Homework Statement

Consider the partial differential equation

$$u_{xx}-3u_{xt}-4u_{tt}=0$$

(a) Find the general solution of the partial differential equation in the xt-plane, if possible.
(b) Find the solution of the partial differential equation that satisfies
$$u(x,0)=x^3$$ and $$u_t(x,0)=-3x^2$$ for $$-\infty<x<\infty$$

Homework Equations

The Attempt at a Solution

(a)

$$\left(\frac{\partial^2}{\partial x^2}-\frac{3\partial^2}{\partial x\partial t}-\frac{4\partial^2}{\partial t^2}\right)u=0$$

*
$$\left(\frac{\partial}{\partial x}+\frac{\partial}{\partial t}\right)\left(\frac{\partial}{\partial x}-\frac{4\partial}{\partial t}\right)u=0$$

Set the coefficients equal to a variable...

$$\alpha=1, \beta=1, \gamma=1, \delta=-4$$

$$\zeta=\beta{x}-\alpha{t}=x-t$$
$$\eta=\delta{x}-\gamma{t}=-4x-t$$

$$\frac{\partial v}{\partial x}=\frac{\partial v}{\partial \zeta}\frac{\partial \zeta}{\partial x}+\frac{\partial v}{\partial \eta}\frac{\partial \eta}{\partial x}=\frac{\partial v}{\partial \zeta}-\frac{4\partial v}{\partial \eta}$$

$$\frac{\partial v}{\partial t}=\frac{\partial v}{\partial \zeta}\frac{\partial \zeta}{\partial t}+\frac{\partial v}{\partial \eta}\frac{\partial \eta}{\partial t}=\frac{-\partial v}{\partial \zeta}-\frac{\partial v}{\partial \eta}$$

So now substitute these two differentials into *
$$\left[\left(\frac{\partial}{\partial \zeta}-\frac{4\partial}{\partial \eta}\right)+\left(\frac{-\partial}{\partial \zeta}-\frac{\partial}{\partial \eta}\right)\right]\left[\left(\frac{\partial}{\partial \zeta}-\frac{4\partial}{\partial \eta}\right)-\left(\frac{-4\partial}{\partial \zeta}-\frac{4\partial}{\partial \eta}\right)\right]u=0$$

$$\left(\frac{-4\partial}{\partial \eta}-\frac{\partial}{\partial \eta}\right)\left(\frac{\partial}{\partial \zeta}+\frac{4\partial}{\partial \zeta}\right)u=0$$

$$\left(\frac{-5\partial}{\partial \eta}\right)\left(\frac{5\partial}{\partial \zeta}\right)u=0$$

$$\frac{\partial}{\partial \eta}\left(\frac{\partial u}{\partial \zeta}\right)=0$$

Let $$h(\zeta)=\frac{\partial u}{\partial \zeta}$$

So now integrate with respect to $$\zeta$$ holding $$\eta$$ fixed.

$$u=\int h(\zeta)d\zeta + g(\eta)=f(\zeta)+g(\eta)$$

Substitute back in for $$\zeta$$ and $$\eta$$...

$$u(x,t)=f(x-t)+g(-4x-t)$$

Now this should be the general solution. Is this right, did I do my differentiation and arithmetic correctly?

(b)

So for the functions f and g I just picked some arbitrary functions.

Let $$f(x)=e^x$$ and $$g(x)=e^{-x}$$

So now,

$$x^3=e^x+e^{4x}$$

And also,

$$-3x^2=-e^x-e^{4x}$$

And this is where I stall. I'm not sure what to do after this point. My book doesn't have an example of solving these types of problems and the notes our professor gave us doesn't show an example either.

 

[hunt_mat]

Your new equation is fine along with your general solution. however you did not apply the boundary conditions correctly, but you need to work on your boundary conditions. You have $$u=f(x-t)+g(-4x-t)$$ Applying you boundary conditions:
$$u(x,0)=f(x)+g(-4x)=x^{3}\quad\partial_{t}u(x,0)=-f'(x)-g(-4x)=-3x^{2}$$
Now solve those equations for f and g to get the solution.

 

[roldy]

So I'm not suppose to pick arbitrary functions for f and g then right?. I guess this makes sense now, two equations two unknowns.

 

[hunt_mat]

Your general solution will give you the functions as you've done now you have to fit those functions to your initial conditions. I have done it so I can say if you have the correct equation or not.

 

[roldy]

I'm not that I entirely understand what you mean in your last post.

 

[hunt_mat]

Now you have a set of ODE's to solve:
$$\begin{array}{ccc}<br /> f(x)+g(-4x) & = & x^{3} \\<br /> f'(x)+g'(-4x) & = & 3x^{2}<br /> \end{array}$$
You should be able to find what f and g are by solving the above system of ODE.

 

[roldy]

Yeah, that's what I figured I needed to do but what method involved is what I'm not sure about.

 

[hunt_mat]

In the first equation set x=0 to find that f(0)+g(0)=0, integrate the second equation to obtain:
$$f(x)-\frac{1}{4}g(x)=x^{3}+C$$
Then take the first equation away from the second one to obtain:
$$-\frac{5}{4}g(x)=C$$
This shows that g is constant. which in turn shows that:
$$f(x)=x^{3}+D$$
So the full solution becomes...

Simple integration is all that is required, what can you say about D in light of the initial conditions?

 

[roldy]

I don't understand why you are setting x=0. There's no initial condition for x=0. Also, how to you know that when you integrate the second equation your not going to get a x time a function. Meaning, how do you for example that f(x) is not (1+x)?

 

[hunt_mat]

The setting x=0 was unimportant, so ignore it.

I get a constant because f is a function of one variable only.

 

[roldy]

Could you maybe explain a little bit more on how that 1/4 got out there?

 

[roldy]

Still confused on this one.

 

[hunt_mat]

it comes into the integral in the following way:
$$\int g'(-4x)dx =-\frac{1}{4}\int g'(u)du=-\frac{1}{4}g(u)=-\frac{1}{4}g(-4x)$$
using the substitution u=-4x