# PDE separation of variables

Hi could someone guide me this problem

It says ,

$u_t - u_{xx}-2 u_x=0$

Use the method of separation of variables to find all possible solutions.

Could someone help me out for this problem. I'm beginner at PDE. I would be much appreciated if you able to show some partial work so that I can understand.

Thanks in advance for taking my consideration

If $u(x,t) = X(x)T(t)$, then

$$\frac{T'}{T} = \frac{X''}{X} + 2 \frac{X'}{X}$$

LHS depends on $t$ only, RHS depends on $x$ only, so both sides are equal to a constant, say $\lambda$. You can take it from here...

• should i need to make $\lambda$ into 3 cases. Example let $\lambda$ = p^2(for positive), $\lambda$ = -p^2 (for negative), and $\lambda$ =0 Commented Jan 3, 2013 at 15:46
• Those 3 "cases" are all attempts to avoid imaginary numbers. What is the equation modeling (e.g., harmonic motion with damping)? What are the initial/boundary conditions? These questions will guide you on the allowed values of $\lambda$. Commented Jan 3, 2013 at 16:01
• The question didnt mention anything. It just ask me to solve. So what should I do ? Commented Jan 3, 2013 at 16:09
• You cannot get a unique solution without some initial condition (in time) and boundary conditions (in space). If not given, then a) make them up, or b) ask whoever asked the question. Is this homework, or is it an actual problem you have encountered in, e.g., research? Commented Jan 3, 2013 at 16:11
• In any case, if you can solve the individual ODE's, then you can construct a general solution. Can you do this? Commented Jan 3, 2013 at 16:13

Case $1$: $\text{Re}(t)\geq0$

Let $u(x,t)=X(x)T(t)$ ,

Then $X(x)T'(t)-X''(x)T(t)-2X'(x)T(t)=0$

$X(x)T'(t)=X''(x)T(t)+2X'(x)T(t)$

$X(x)T'(t)=(X''(x)+2X'(x))T(t)$

$\dfrac{T'(t)}{T(t)}=\dfrac{X''(x)+2X'(x)}{X(x)}=-(f(s))^2-1$

$\begin{cases}\dfrac{T'(t)}{T(t)}=-(f(s))^2-1\\X''(x)+2X'(x)+((f(s))^2+1)X(x)=0\end{cases}$

$\begin{cases}T(t)=c_3(s)e^{-t((f(s))^2+1)}\\X(x)=\begin{cases}c_1(s)e^{-x}\sin(xf(s))+c_2(s)e^{-x}\cos(xf(s))&\text{when}~f(s)\neq0\\c_1xe^{-x}+c_2e^{-x}&\text{when}~f(s)=0\end{cases}\end{cases}$

$\therefore u(x,t)=C_1xe^{-x-t}+C_2e^{-x-t}+\int_sC_3(s)e^{-x-t((f(s))^2+1)}\sin(xf(s))~ds+\int_sC_4(s)e^{-x-t((f(s))^2+1)}\cos(xf(s))~ds~\text{or}~C_1xe^{-x-t}+C_2e^{-x-t}+\sum\limits_sC_3(s)e^{-x-t((f(s))^2+1)}\sin(xf(s))+\sum\limits_sC_4(s)e^{-x-t((f(s))^2+1)}\cos(xf(s))$

Case $2$: $\text{Re}(t)\leq0$

Let $u(x,t)=X(x)T(t)$ ,

Then $X(x)T'(t)-X''(x)T(t)-2X'(x)T(t)=0$

$X(x)T'(t)=X''(x)T(t)+2X'(x)T(t)$

$X(x)T'(t)=(X''(x)+2X'(x))T(t)$

$\dfrac{T'(t)}{T(t)}=\dfrac{X''(x)+2X'(x)}{X(x)}=(f(s))^2-1$

$\begin{cases}\dfrac{T'(t)}{T(t)}=(f(s))^2-1\\X''(x)+2X'(x)+(1-(f(s))^2)X(x)=0\end{cases}$

$\begin{cases}T(t)=c_3(s)e^{t((f(s))^2-1)}\\X(x)=\begin{cases}c_1(s)e^{-x}\sinh(xf(s))+c_2(s)e^{-x}\cosh(xf(s))&\text{when}~f(s)\neq0\\c_1xe^{-x}+c_2e^{-x}&\text{when}~f(s)=0\end{cases}\end{cases}$

$\therefore u(x,t)=C_1xe^{-x-t}+C_2e^{-x-t}+\int_sC_3(s)e^{-x+t((f(s))^2-1)}\sinh(xf(s))~ds+\int_sC_4(s)e^{-x+t((f(s))^2-1)}\cosh(xf(s))~ds~\text{or}~C_1xe^{-x-t}+C_2e^{-x-t}+\sum\limits_sC_3(s)e^{-x+t((f(s))^2-1)}\sinh(xf(s))+\sum\limits_sC_4(s)e^{-x+t((f(s))^2-1)}\cosh(xf(s))$

Hence $u(x,t)=\begin{cases}C_1xe^{-x-t}+C_2e^{-x-t}+\int_sC_3(s)e^{-x-t((f(s))^2+1)}\sin(xf(s))~ds+\int_sC_4(s)e^{-x-t((f(s))^2+1)}\cos(xf(s))~ds&\text{when}~\text{Re}(t)\geq0\\C_1xe^{-x-t}+C_2e^{-x-t}+\int_sC_3(s)e^{-x+t((f(s))^2-1)}\sinh(xf(s))~ds+\int_sC_4(s)e^{-x+t((f(s))^2-1)}\cosh(xf(s))~ds&\text{when}~\text{Re}(t)\leq0\end{cases}$

or $\begin{cases}C_1xe^{-x-t}+C_2e^{-x-t}+\sum\limits_sC_3(s)e^{-x-t((f(s))^2+1)}\sin(xf(s))+\sum\limits_sC_4(s)e^{-x-t((f(s))^2+1)}\cos(xf(s))&\text{when}~\text{Re}(t)\geq0\\C_1xe^{-x-t}+C_2e^{-x-t}+\sum\limits_sC_3(s)e^{-x+t((f(s))^2-1)}\sinh(xf(s))+\sum\limits_sC_4(s)e^{-x+t((f(s))^2-1)}\cosh(xf(s))&\text{when}~\text{Re}(t)\leq0\end{cases}$

This is already the general solution of $u_t-u_{xx}-2u_x=0$ . Note that when without any I.C.s, the form of $f(s)$ can choose arbitrary, but when I.C.s are given, the form of $f(s)$ and the choice whether using the integration kernel or using the summation kernel should choose wisely in order to accommodate the I.C.s to get the most nice form of the solution, especially the number of I.C.s is more than two.

• Thank you so much doraemonpaul!!!! You really save me. I'm glad you able to help me till the end. Thanks once again man!! Commented Jan 5, 2013 at 17:11
• @Garett I want to make sure you know that on this site you can (and should) accept the best answers to your questions by clicking a checkmark to the left of the question. This serves multiple purposes, in particular it tells other readers that the issue was resolved.
– user53153
Commented Jan 5, 2013 at 20:43