u_{xx}-u=0

**dwsmith** · December 23rd 2010, 02:22 PM

Find the general solution to $u_{xx}-u=0$

The solution is $\displaystyle u(x,y)=f(y)e^x+g(y)e^{-x}$

How is this obtained?

Thanks.

I solved the first part:

$\displaystyle P(x)=-1\Rightarrow e^{\int P(x)dx}=e^{-\int dx}=e^{-x}$

$\displaystyle \int\frac{\partial}{\partial x}\left[e^{-x}u_x\right]dx=\int 0dx\Rightarrow e^{-x}u_x(x,y)=f(y)$

Should that be in the integral the partial derivative or just derivative?

$\displaystyle\Rightarrow u_x(x,y)=e^xf(y)$

**pickslides** · December 23rd 2010, 03:29 PM

I think of this problem as find the solution $u(x,y)$ such that $u(x,y) = u_{xx}$

As you only have a partial with respect to the same variable then the solution is similar to that of a 2nd order ODE. I.e $y'' = y$ has the solution $y=e^x$ and $y= e^{-x}$

The difference in your problem is that $f(y)$ and $g(y)$ need to be introduced but are constant with respect to $x$

**pickslides** · December 23rd 2010, 03:31 PM

Originally Posted by dwsmith

$\displaystyle P(x)=-1\Rightarrow e^{\int P(x)dx}=e^{-\int dx}=e^{-x}$

Why not try separation of variables?

**dwsmith** · December 23rd 2010, 03:31 PM

Originally Posted by pickslides

I think of this problem as find the solution $u(x,y)$ such that $u(x,y) = u_{xx}$

As you only have a partial with respect to the same variable then the solution is similar to that of a 2nd order ODE. I.e $y'' = y$ has the solution $y=e^x$ and $y= e^{-x}$

The difference in your problem is that $f(y)$ and $g(y)$ need to be introduced but are constant with respect to $x$

I just updated my post as you replied.

My problem is should it be partial derivatives or written like an ODE with just the derivative?

Should I use the same method for the integration again?

**dwsmith** · December 23rd 2010, 03:32 PM

Originally Posted by pickslides

Why not try separation of variables?

That was what I thought of when the light bulb turned on.

**dwsmith** · December 23rd 2010, 03:36 PM

I understand your method which is simple but I would like to finish the problem how I started it. If you wouldn't mind, how should I finish it off?

Thanks.

If I just integrate, I wouldn't obtain the e^{-x} so I am not sure how to proceed.

**pickslides** · December 23rd 2010, 03:46 PM

Originally Posted by dwsmith

I understand your method which is simple but I would like to finish the problem how I started it. If you wouldn't mind, how should I finish it off?

Thanks.

If I just integrate, I wouldn't obtain the e^{-x} so I am not sure how to proceed.

Well your attempt suggests finding an integrating factor in a 2nd order DE, will that work? Should you not be finding a characteristic equation?

;D

**dwsmith** · December 23rd 2010, 03:49 PM

I just looked it up. Integrating factors aren't restricted to first order linear equations.

**pickslides** · December 23rd 2010, 04:21 PM

I would say $\displaystyle u_{xx}-u(x,y)=0$

Gives characteristic equation in the form $\displaystyle r^2-1= (r-1)(r+1)=0 \implies r_{1,2}=\pm 1$ and $\displaystyle y=c_1e^{r_1x}+c_2e^{r_2x}$ therefore seek a solution in the form of $\displaystyle y=c_1e^x+c_2e^{-x}$

As we have a PDE $\displaystyle c_1$ and $\displaystyle c_2$ can also be separate functions of $\displaystyle y$ .

**dwsmith** · December 23rd 2010, 04:45 PM

Now, I am looking to find the sol. of $u_{xx}-u=0$ which satisfies the auxiliary conditions

$u(0,y)=\varphi (y) \ , \ u_x(0,y)=\psi (y)$

$u(0,y)=f(y)+g(y)=\varphi (y)$

$u_x(0,y)=f(y)-g(y)=\psi (y)$

How do I incorporate this into the solution?

$u(x,y)=e^xf(y)+e^{-x}g(y)$

**dwsmith** · December 23rd 2010, 09:00 PM

Is this the correct procedure?

$u(0,y)=f(y)+g(y)=\varphi (y)$

$u_x(0,y)=f(y)-g(y)=\psi (y)$

By adding the two equations and simplifying, we obtain $\displaystyle f(y)=\frac{\varphi (y)+\psi (y)}{2}$

Now, solving for $g(y)$ I get $\displaystyle g(y)=\frac{\varphi (y)-\psi (y)}{2}$

$\displaystyle u(x,y)=e^x\left[\frac{\varphi (y)+\psi (y)}{2}\right]+e^{-x}\left[\frac{\varphi (y)-\psi (y)}{2}\right]$

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