# Solving a partial DE

1. May 26, 2004

### jimbo007

hi all
i have been trying to solve to following problem,

$$\frac{\partial^2 u}{\partial x^2} - \frac{\partial^2 u}{\partial y^2} + 2\frac{\partial u}{\partial x} + u = 0$$

$$u=u(x,y)$$

after a bit of work using the change of variables

$$\zeta=\zeta(x,y)=y-x$$

and

$$\eta=\eta(x,y)=y+x$$

i obtain

$$- 4\frac{\partial u}{\partial \zeta \partial \eta} = 2\frac{\partial u}{\partial\zeta} - 2\frac{\partial u}{\partial\eta}-u$$

but i am unsure how to solve this, i used maple to solve this problem and it gave out a fairly harmless answer so i am pretty sure there would be any easy way to solve the above equation.

could someone kindly show me how to obtain a solution to this problem

Last edited: May 26, 2004
2. May 26, 2004

### arildno

I would suggest a separation of variables solution,
u(x,y)=f(x)g(y)

3. May 28, 2004

### Max0526

Hi;
The first substitution to use is "running wave" substitution:
$$z=x-ct$$
and you will have:
$$\frac{\partial^2 u}{\partial x^2} - \frac{\partial^2 u}{\partial y^2} + 2\frac{\partial u}{\partial x} + u = 0$$
$$u_{xx}-u_{yy}+2u_x+u=0$$
$$z=x-ct$$
$$u_{zz}-c^2u_{zz}+2u_z+u=0$$
$$(1-c^2)u_{zz}+2u_z+u=0$$
The last equation is a linear ODE of the 2nd order while c<>+1 and c<>-1. Otherwise we have the first order ODE
$$2u_z+u=0$$
Anyways both equations are simple and can be solved easily.
Still, this is not the only way of reduction of the given PDE (there could be another substitutions v=v(u,x,y), z=z(u,x,y) that will give us ordinary DEs). If you need some other forms for v and z, let me know, I can find some for you.
Best of luck,
Max.

4. May 28, 2004

### arildno

Here's a rather general solution using separation of variables:

1.
$$u(x,y)=f(x)g(y)\to\frac{f''+2f'+f}{f}=\frac{g''}{g}$$

2. Since LHS is a function of x, wheras RHS is a function of y, equality is only possible if they are equal to some constant k.
I set a subscript k on the functions at this moment:
$$\frac{f_{k}''+2f_{k}'+f_{k}}{f_{k}}=k\to{f}_{k}(x)=A_{k}e^{-x-\sqrt{k}x}+B_{k}e^{-x+\sqrt{k}x}$$
$$\frac{g_{k}''}{g_{k}}=k\to{g}_{k}(x)=C_{k}e^{-\sqrt{k}y}+D_{k}e^{\sqrt{k}y}$$

3. Since the original differential equation is linear, a sum of two solutions (for example, for different k-values) is also a solution, and we may write:

$$u(x,y)=$$
$$e^{-x}\int_{-\infty}^{\infty}\mathcal{C}_{(k,+)}Cosh(\sqrt{k}(x+y))+\mathcal{C}_{(k,-)}Cosh(\sqrt{k}(x-y))dk+$$
$$e^{-x}\int_{-\infty}^{\infty}\mathcal{S}_{(k,+)}Sinh(\sqrt{k}(x+y))+\mathcal{S}_{(k,-)}Sinh(\sqrt{k}(x-y))dk$$

4. Negative values of k will yield trigonometric functions; the unknown functions of k,
$$\mathcal{C}_{(k,+)},\mathcal{C}_{(k,-)},\mathcal{S}_{(k,+)},\mathcal{S}_{(k,-)}$$
may be suited to satisfy boundary conditions.

Last edited: May 28, 2004
5. May 28, 2004