# Thread:

1. ## Mixed Partial Derivatives

Solve the equation u_xx - 3u_xt - 4u_tt = 0 subject to the initial conditions u(x,0) = x^2 and u_t(x,t) = e^x.

Hint: consider a change of coordinates.

Having spent a large amount of time getting wrong solutions to the above question, any help that can be offered would be very much appreciated!

Thanks!

2. Hello,

show your working and we'll tell you what's wrong (although I don't think that'll be me ^^)

3. Ok no problem!

I have seen something similar where you factor out the operator: (d_xx - 3d_xt - 4d_tt) = 0

This gives us: (d_t + d_x)(4d_t + d_x)u = 0

I then have attempted to define new coordinates. I was led to believe that I should try, for example: a = x +t and b = -4x + t (though this does not appear to work so perhaps this is where I have gone wrong?).

We then want to find a v(a,b) such that it solves the equation.

We then get;

u_x = v_a - 4v_b
u_t = v_a + v_b

I then continue...but will avoid writing out the rest of it now in case my coordinates are wrong! If anybody can advise me if they are or are not correct then I will post the rest of my solution on that basis!

Many thanks!

4. Originally Posted by Rocky
I have seen something similar where you factor out the operator: (d_xx - 3d_xt - 4d_tt) = 0

This gives us: (d_t + d_x)(4d_t + d_x)u = 0 I think you mean (d_x + d_t)(d_x – 4d_t)u = 0.

I then have attempted to define new coordinates. I was led to believe that I should try, for example: a = x +t and b = -4x + t (though this does not appear to work so perhaps this is where I have gone wrong?).
I get a = x – t and b = 4x + t.

We then want to find a v(a,b) such that it solves the equation.

We then get;

u_x = v_a - 4v_b
u_t = v_a + v_b

I then continue...
Keep going, you're on the right lines. Now use the chain rule to find $u_{xx},\; u_{xt}$ and $u_{tt}$ in terms of $v_{aa},\ v_{ab}$ and $v_{bb}$. For example, $u_{xx} = \frac{\partial}{\partial a}(u_x)\frac{\partial a}{\partial x} + \frac{\partial}{\partial b}(u_x)\frac{\partial b}{\partial x}.$

You should then find that $u_{xx}-3u_{xt}-4u_{tt}$ is some constant times $v_{ab}$ (the terms in $v_{aa}$ and $v_{bb}$ all cancel out). So your equation is equivalent to $v_{ab} = 0$, whose general solution is $v(a,b) = f(a)+g(b)$ for arbitrary (twice-differentiable) functions f, g. So the solution for u(x,t) is $u(x,t) = f(x-t) + g(4x+t)$.

5. That is a brilliant help thanks very much! I have now got to this stage...but am now a little confused as to how I take into the account the initial conditions.

I have attempted to somehow adapt d'Alambert's solution to the wave equation (or something along these lines) but the solution I attain does not seem to satisfy the original equation!

In particular I obtained a solution along the lines of:

(1/2) (x-t)^2 + (1/2) (4x+t)^2 + the integral of e^y dy (between the limits (x-t) and (4x+t))

Again any help in finishing this question would be much appreciated! Thanks!

6. Originally Posted by Rocky
Solve the equation u_xx - 3u_xt - 4u_tt = 0 subject to the initial conditions u(x,0) = x^2 and u_t(x,t) = e^x.
Is that last equation correct? I would expect it to be u_t(x,0) = e^x. (if it is an initial condition, then it ought to tell you what happens at time 0.)
In that case, if the general solution is $u(x,t) = f(x-t) + g(4x+t)$ then the initial conditions are

$f(x) + g(4x) = x^2,\quad -f'(x) + g'(4x) = e^x.$

Integrate the second one to get $-f(x) + \frac14g(4x) = e^x$. (There should be a constant of integration somewhere there, but it will cancel out in the eventual solution, so let's ignore it.)

Solve those two simultaneous equations for f(x) and g(4x) to get

$f(x) = \frac15(x^2-4e^x),\quad g(4x) = \frac45(x^2+e^x).$

Therefore

$f(x-t) = \frac15\bigl((x-t)^2-4e^{x-t}\bigr),\quad g(4x+t) = g\bigl(4(x+\frac t4)\bigr) = \frac45\bigl((x+\frac t4)^2+e^{x+(t/4)}).$

Add those together to get the solution for u(x,t).
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