# Math Help - PDE-Check

1. ## PDE-Check

Find all separable exponential solutions of:

a) u_tt= u_xx - 4 u_x + 2u

I have
u(x,t) = X(x)T(t)
X(x) = Ce^ax

X'(x) = aCe^ax
X''(x) = a^2Ce^ax

X''T-4X'+2X=0
= T(X''-4X'+2X)=0
= Ce^ax(a^2-4a+2)

b) u_xx + u_yy=0

u(x.y) = X(x)Y(y)

X''/X = -Y''/Y

X(x) = Ce^ax

not sure from here.

Thanks

2. Originally Posted by taypez
Find all separable exponential solutions of:

a) u_tt= u_xx - 4 u_x + 2u

I have
u(x,t) = X(x)T(t)
X(x) = Ce^ax

X'(x) = aCe^ax
X''(x) = a^2Ce^ax

X''T-4X'+2X=0
= T(X''-4X'+2X)=0
= Ce^ax(a^2-4a+2)

[snip]
Let u = X(x) T(t).

Then $X \frac{d^2 T}{dt^2} = T\left( \frac{d^2 X}{dx^2} - 4 \frac{d X}{dx} + 2X \right)$

$\Rightarrow \frac{1}{T} \frac{d^2 T}{dt^2} = \frac{1}{X} \left( \frac{d^2 X}{dx^2} - 4 \frac{d X}{dx} + 2X \right)$

Therefore

$\frac{1}{T} \frac{d^2 T}{dt^2} = k^2 \,$ and $\, \frac{1}{X} \left( \frac{d^2 X}{dx^2} - 4 \frac{d X}{dx} + 2X \right) = k^2$

where $k^2$ is the seperation constant and is necessarily positive for exponential solutions.

Therefore:

$\frac{d^2 T}{dt^2} - k^2 T = 0\,$ and $\, \frac{d^2 X}{dx^2} - 4 \frac{d X}{dx} + (2 - k^2) X = 0$.

Now assume exponential solutions of the form $T = A e^{\lambda t} \,$ and $\, X = B e^{\mu x} \,$ .......

3. Originally Posted by taypez
Find all separable exponential solutions of:

[snip]
b) u_xx + u_yy=0

u(x.y) = X(x)Y(y)

X''/X = -Y''/Y Mr F says: Therefore either:

1. X''/X = k^2 and -Y''/Y = k^2 => X'' - k^2 X = 0 and Y'' + k^2 Y = 0 and the exponential solution will be in X = X(x), or

2. X''/X = -k^2 and -Y''/Y = -k^2 => X'' + k^2 X = 0 and Y'' - k^2 Y = 0 and the exponential solution will be in the Y = Y(y).

X(x) = Ce^ax

not sure from here.

Thanks
..