PDE help

renegade05

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ut=uxxuxet,0<x<1,t>0\displaystyle u_t\, =\, u_{xx}\, -\, u\, -\, xe^{-t},\, 0\, <\, x\, <\, 1,\, t\, >\, 0

\(\displaystyle \mbox{BC: }\, u(0,\, t)\, =\, 0,\, u(1,\, t)\, =\, 0\)

\(\displaystyle \mbox{IC: }\, u(x,\, 0)\, =\, x\)

I am trying to solve this initial boundary value problem.

I am having a hard time starting out. I want to guess w(x,t) = Ax + B but that is not going to work for the Dirichlet BC.

How should I start this one?

Thanks!
 
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Great! Since there may be others who are wondering, here is how I would attempt this problem:
As we do in ordinary differential equations with a "linear non-homogeneous equation", look for the general solution to the associated "homogeneous equation, ut=uxxu\displaystyle u_t= u_{xx}- u. Try separating variables- that is, look for a solution of the form u(x,t)=X(x)T(t)\displaystyle u(x, t)= X(x)T(t) where X is a function of X only and T is a function of t only. The differential equation becomes XT=XTXTxet\displaystyle XT'= X''T- XT- xe^{-t}. If divide by u= XT, that becomes TT=XX+1\displaystyle \frac{T'}{T}= \frac{X''}{X}+ 1. The left side of the equation depends on "x" only while the right side depends on "t" only. Since they must be equal for all x and t, each side must be equal to a constant.

We can separate the equation into TT=α\displaystyle \frac{T'}{T}= \alpha and XX+1=α\displaystyle \frac{X''}{X}+ 1= \alpha. The boundary conditions are u(0,t)=X(0)T(t)=0\displaystyle u(0, t)= X(0)T(t)= 0 for all t, so we must have X(0)= 0, and u(1,t)=X(1)T(t)=0\displaystyle u(1, t)= X(1)T(t)= 0 for all t so we must have X(1)= 0.

So we start by solving the ordinary differential equation problem, X+(1α)X=0\displaystyle X''+ (1- \alpha)X= 0 with boundary conditions X(0)= X(1)= 0. But we don't know what α\displaystyle \alpha is!

So we look at "cases". If α=1\displaystyle \alpha= 1, the equation is X=0\displaystyle X''= 0 and we have, by integrating twice, X(x)= Ax+ B. But then X(0)= B= 0 and X(1)= A+ B= 0 so the we must have X(x)= 0x+ 0= 0 for all x. That is the "trivial solution". There is no "non-trivia solution" so we cannot satisfy the condition that u(x, 0)= x.

If α<1\displaystyle \alpha< 1 then 1α\displaystyle 1- \alpha is negative. We can make that explicit by writing \(\displaystyle 1- \alpha= -\lambda^2['tex] where \(\displaystyle \lambda\) can be any non-zero number (and, without loss of generality, we can take it to be positive). Now our equation is Xλ2X=0\displaystyle X''- \lambda^2 X= 0. That is a "linear, homogenous, differential equation with constant coefficients". It "characteristic equation" is r2λ2=0\displaystyle r^2- \lambda^2= 0 which has roots r=±λ\displaystyle r= \pm\lambda and so general solution X(x)=Aeλx+Beλx\displaystyle X(x)= Ae^{\lambda x}+ Be^{-\lambda x}. Now we have X(0)=A+B=0\displaystyle X(0)= A+ B= 0 and X(1)=Aeλ+Beλ=0\displaystyle X(1)= Ae^{\lambda}+ Be^{-\lambda}= 0. From the first equation, B= -A, so the second equation becomes A(eλeλ)=0\displaystyle A(e^{\lambda}- e^{-\lambda})= 0. Now eλ\displaystyle e^{\lambda} is greater than 1 while eλ\displaystyle e^{-\lambda} is less than 1. They are NOT equal so eλeλ\displaystyle e^{\lambda}- e^{-\lambda} cannot be 0. That means we must have A= 0. But then B= 0 also so once again we have the trivial solution. There is no non-trivial solution in this case either.

So we must have α>1\displaystyle \alpha> 1 so that 1α>0\displaystyle 1- \alpha> 0. Write 1α=λ2\displaystyle 1- \alpha= \lambda^2 where, again, we can take λ\displaystyle \lambda to be positive. Now our equation is X''+ \lambda^2 X= 0. Now the "characteristic equation" is r2+λ2=0\displaystyle r^2+ \lambda^2= 0 which has imaginary roots, r=λi\displaystyle r= \lambda i and r=λi\displaystyle r= -\lambda i and so general solution X(x)=Acos(λx)+Bsin(λx)\displaystyle X(x)= A cos(\lambda x)+ B sin(\lambda x). X(0)=A=0\displaystyle X(0)= A= 0. X(1)=Bsin(λ)=0\displaystyle X(1)= B sin(\lambda)= 0. We must have B= 0 or sin(λ)=0\displaystyle sin(\lambda)= 0. We can have B non-zero, and so a non-trivial solution if and only if λ\displaystyle \lambda is a multiple of π\displaystyle \pi: λ=nπ\displaystyle \lambda= n\pi. That is, we have a non-trivial solution to X+(1α)X=0\displaystyle X''+ (1- \alpha)X= 0 if and only if 1α=n2π2\displaystyle 1- \alpha= -n^2\pi^2 so that α=1+n2π2\displaystyle \alpha= 1+ n^2\pi^2, and in that case, we have X(x)=Bsin(nπx)\displaystyle X(x)= B sin(n\pi x).

Now, look at the equation for T. We had TT=α=1+n2π2\displaystyle \frac{T'}{T}= \alpha= 1+ n^2\pi^2 for n any integer. That is T=(1+n2π2)T\displaystyle T'= (1+ n^2\pi^2)T which has general solution T=Ce(1+n2π2)t\displaystyle T= C e^{(1+ n^2\pi^2)t}.

Putting that together with X, u(x,t)=Ce(1+n2π2)tsin(nπx)\displaystyle u(x, t)= Ce^{(1+ n^2\pi^2)t}sin(n\pi x).

To find a solution to the entire equation, since the "non-homogeneous part" is xet\displaystyle xe^{-t}, look for a solution of the form u(x,t)=(ax+b)ect\displaystyle u(x, t)= (ax+ b)e^{ct}. Then ux=aect\displaystyle u_x= ae^{ct}, uxx=0\displaystyle u_xx= 0, and ut=c(ax+b)ect\displaystyle u_t= c(ax+ b)e^{ct}. The equation becomes c(a+b)ect=((aca)x2b))ectxet\displaystyle c(a+ b)e^{ct}= ((-ac- a)x- 2b))e^{ct}- xe^{-t}.
Clearly, we must have c=1\displaystyle c= -1. Then we can divide by ect=et\displaystyle e^{ct}= e^{-t} and have (a+b)=\displaystyle -(a+ b)= or \(\displaystyle adx^2+ bdx+ dc= -ax^2+ (b- 1)x+ (2a+ c) for all x so we must have ad= -a, bd= b- 1, and dc= 2a+ c. d= -1, -b= b- 1 so b= 1/2, and -c= 2a+ c so a= c. That is, \(\displaystyle u(x, t)= Ce^{(1+ n^2\pi^2)t} sin(n\pi x)+

Now, we must have u(x,0)=Csin(nπx)=x\displaystyle u(x, 0)= C sin(n\pi x)= x. There is, of course, no "C" that will make that true for all x but, since this is a linear equation, we can form new solution as linear combinations of solutions. That is we can have solutions of the form u(x,t)=n=0Cne(1+n2π2)tsin(nπx)\displaystyle u(x, t)= \sum_{n=0}^\infty C_n e^{(1+ n^2\pi^2)t} sin(n\pi x). That is a "Fourier sine series" solution. Now we have u(x,0)=n=0Cnsin(nπx)=x\displaystyle u(x, 0)= \sum_{n=0}^\infty C_n sin(n\pi x)= x. That is, we must find the coefficients, Cn\displaystyle C_n, for the Fourier series equal to x and we have found the general solution to the associated homogeneous equation.\)\)\)
 
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