# PDE arbitrary function F(x+t)

1. Oct 4, 2006

### Dragonfall

Solve $$u_{xx}-3u_{xt}-4u_{tt}=0$$ with initial conditions $$u(x,0)=x^2, u_t(x,0)=e^x$$.

I got that u is an arbitrary function F(x+t), which makes no sense. I factored the operator into $$(\partial/\partial x+\partial/\partial t)(\partial/\partial x-4\partial/\partial t)u=0$$, but I can't get anywhere.

2. Oct 4, 2006

### jpr0

Say that

$$(\partial/\partial x-4\partial/\partial t)u=g$$

Then your equation becomes,

$$(\partial/\partial x+\partial/\partial t)g=0$$

Solve for g, then solve your first equation.

3. Oct 4, 2006

### Dragonfall

Let $$x'=x-4t, t'=-4x-t$$. At some point I have $$u_{x'}=f(\frac{3x'+5t'}{-17})/17$$. Am I on the right track?

4. Oct 4, 2006

### HallsofIvy

Staff Emeritus
Why does that make no sense? You have a pde that is of second order in both x and t but you give only initial conditions when t= 0. Without boundary conditions on x, you will not have a specific solution.

5. Oct 4, 2006

### Dragonfall

Well I'm copying down exactly what's written in the textbook, Partial Differential Equations: An Introduction by W. A. Strauss, S2.2 Problem 9.

6. Oct 4, 2006

### arildno

And why should your textbook make no sense?

7. Oct 4, 2006

### Dragonfall

Ok this second level questioning confuses me.

8. Oct 4, 2006

### Dragonfall

Because there exists a unique solution at the back of the book:

$$\frac{4}{5}(e^{x+t/4}-e^{x-t})+x^2+\frac{t^2}{4}$$.

Last edited: Oct 4, 2006
9. Oct 4, 2006

### jpr0

$$\left(\partial_{x}-a^{-1}\partial_{t}\right)\left(\partial_{x}+\partial_{t}\right)u\left(x,t\right)=0$$

Hence the solution is the sum of two arbitrary functions with arguments $x-t$ and $x+at$,

$$u\left(x,t\right)=F\left(x-t\right)+G\left(x+at\right)$$

We have the conditions

\begin{align*} u\left(x,0\right)& =x^{2}\\ u_{t}\left(x,0\right)& =e^{x} \end{align*}

Using $u\left(x,0\right)$

\begin{align*} x^{2}&=F\left(x\right)+G\left(x\right) \\ G\left(x\right)&=x^{2}-F\left(x\right) \end{align*}

Therefore,

$$u\left(x,t\right)=F\left(x-t\right)+\left(x+at\right)^{2}-F\left(x+at\right)$$

Next, using $u_{t}\left(x,0\right)$

\begin{align*} e^{x}& =-F^{\prime}\left(x\right)+2ax-aF^{\prime}\left(x\right) \\ F^{\prime}\left(x\right)& =\frac{2ax-e^{x}}{\left(1+a\right) }\\ F\left(x\right)& =\frac{ax^{2}-e^{x}}{\left(1+a\right)} + c \end{align*}

So

\begin{align*} u\left(x,t\right)& =\frac{a\left(x-t\right)^{2}-e^{\left(x-t\right)}}{\left(1+a\right)}+c+\left(x+at\right)^{2} -\frac{a\left(x+at\right)^{2}-e^{\left(x+at\right)}}{\left(1+a\right)} - c\\ &=\frac{e^{\left(x+at\right)}-e^{\left(x-t\right)}}{\left(1+a\right)}+\frac{\left(x+at\right)^{2}\left(1+a\right)+a\left(x-t\right)^{2}-a\left(x+at\right)^{2}}{\left(1+a\right)}\\ &=\frac{e^{\left(x+at\right)}-e^{\left(x-t\right)}}{\left(1+a\right)}+\frac{x^{2}\left(1+a\right)+\left(1+a\right)at^{2}}{\left(1+a\right)}\\ &=\frac{e^{\left(x+at\right)}-e^{\left(x-t\right)}}{\left(1+a\right)}+x^{2}+at^{2} \end{align*}

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