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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

Ali Caglayan
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Garett
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  • You know how to accept answer in http://math.stackexchange.com/questions/271053, but you don't know how to accept answer in this question? – doraemonpaul Jan 06 '13 at 10:23

2 Answers2

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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...

Ron Gordon
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    should i need to make $\lambda$ into 3 cases. Example let $\lambda$ = p^2(for positive), $\lambda$ = -p^2 (for negative), and $\lambda$ =0 – Garett Jan 03 '13 at 15:46
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    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 $\lambda$. – Ron Gordon Jan 03 '13 at 16:01
  • The question didnt mention anything. It just ask me to solve. So what should I do ? – Garett Jan 03 '13 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? – Ron Gordon Jan 03 '13 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? – Ron Gordon Jan 03 '13 at 16:13
  • For X ODE part i'm stuck. X''(x)+2X'(x)-$\lambda$X(x)=0 – Garett Jan 03 '13 at 16:17
  • Write $X(t) = A e^{r x}$, plug into equation, solve resulting quadratic ($r^2 + 2 r - \lambda = 0$) for $r$. Get 2 solutions $r_1$ and $r_2$, general solution is $X(x) = A_1 e^{r_1 x} + A_2 e^{r_2 x}$. $A_1$ and $A_2$ are constants determined by boundary conditions. – Ron Gordon Jan 03 '13 at 16:21
  • ok I understand but where is your lambda missing??? – Garett Jan 03 '13 at 16:32
  • It's contained in the roots $r_1$ and $r_2$. – Ron Gordon Jan 03 '13 at 16:35
  • Ok thank you very much rlgordonma for your help. You really helped me alot in this question. Once again thanks – Garett Jan 03 '13 at 16:38
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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.

doraemonpaul
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  • Thank you so much doraemonpaul!!!! You really save me. I'm glad you able to help me till the end. Thanks once again man!! – Garett Jan 05 '13 at 17:11
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    @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. –  Jan 05 '13 at 20:43