Mixed Partial Derivatives

**Rocky** · February 22nd 2011, 11:38 AM

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!

**Moo** · February 22nd 2011, 11:45 AM

Hello,

show your working and we'll tell you what's wrong

(although I don't think that'll be me ^^)

**Rocky** · February 22nd 2011, 11:56 AM

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!

**Opalg** · February 22nd 2011, 01:14 PM

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

**Rocky** · February 24th 2011, 06:37 AM

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!

**Opalg** · February 24th 2011, 07:51 AM

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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