Can the Non-Homogenous Heat Equation be Solved Using Eigenfunctions?

[JoernE]

Consider the following non-homogenous heat equation on 0 \leq x \leq \pi

u_t = k u_{xx} - 1 with u(x,0) = 0, u(0,t) = 0, u(\pi, t) = 0

Find a solution of the form

\displaystyle \sum_1^{\infty} b_n(t) \phi_n (x)

where \phi_n(x) are the eigenfunctions of an appropriate homogenous problem, and find explicit expressions for b_n(t)

So I think

\phi_n(x) = \sin \frac{n \pi x}{L}

so I find solutions in the form

\displaystyle u(x,t) = \sum_1^{\infty} b_n (t) \ \sin \frac{n \pi x}{L}

Am I on the right track? Is the eigenfunction correct?

 

[JoernE]

I have a different method...

Let u = w(x,t) + v(x)

Then

\frac{\partial u}{\partial t} = \frac{\partial w}{\partial t} and \frac{\partial^2 u}{\partial x^2} = \frac{\partial^2 w}{\partial x^2} + \frac{\partial^2 v}{\partial x^2}

Sub these back into the PDE to obtain

\frac{\partial w}{\partial t} = k \frac{\partial^2 w}{\partial x^2} + k \frac{\partial^2 v}{\partial x^2} - 1

Choose k \frac{\partial^2 v}{\partial x^2} - 1 = 0

So

V'' = \frac{1}{k}, \ V(0) = V(\pi) = 0

V' = \frac{1}{k} + A

V = \frac{1}{k} + Ax + B


V(0) = \frac{1}{k} + B = 0 \ \Rightarrow \ B = - \frac{1}{k}

V(\pi) = \frac{1}{k} + A \pi + B \ \Rightarrow \ A = -\frac{1}{k \pi} - \frac{B}{\pi}

V(x) = 0??

Also since k \frac{\partial^2 v}{\partial x^2} - 1 = 0 we have

\frac{\partial w}{\partial t} = k \frac{\partial^2 w}{\partial x^2} which is homogeneous.

u(0, t) = v(0) + w(0, t) = 0

\Rightarrow \ w(0,t) = 0

u(x, 0) = v(x) + w(x,0) = 0

\Rightarrow \ w(x,0) = 0

u(\pi, t) = v(\pi) + w(\pi, t) = 0
\Rightarrow \ w(\pi, t) = 0

Then we have

w_t = k w_{xx} with w(x,o) = 0, w(0,t) = 0, w(\pi, t) = 0.

Since this is homogenous, we can solve by letting

w = XT

\frac{T'}{kT} = \frac{X''}{X} = - \lambda

X = A \cos(\sqrt{\lambda}t) + B \sin(\sqrt{\lambda}t)

X(0) = A = 0

but now I get a little lost...I think I know where to go, just not exactly how to get there. Already have calculated that v(x) = 0, which will lead to what I need to prove (I think).

 

[HallsofIvy]

Joerne, on this board use "", not "" to show LaTex. I have editted your post to correct that.<br /> <br /> Your first method is perfectly correct. You will, of course, have to expand the constant, -1, in a Fourier sine series.

 

[JoernE]

Joerne, on this board use "", not "" to show LaTex. I have editted your post to correct that.

<br /> <br /> Thanks.<br /> <br /> <blockquote data-attributes="" data-quote="HallsofIvy" data-source="post: 2951688" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-title"> HallsofIvy said: </div> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> Your first method is perfectly correct. You will, of course, have to expand the constant, -1, in a Fourier sine series. </div> </div> </blockquote><br /> Are you referring to my first post here?<br /> <br /> Are you saying that what I wrote in the second post (under "I have a different method") is not correct?

 

[MasterX]

w = XT

\frac{T'}{kT} = \frac{X''}{X} = - \lambda

X = A \cos(\sqrt{\lambda}t) + B \sin(\sqrt{\lambda}t)

X(0) = A = 0

but now I get a little lost...I think I know where to go, just not exactly how to get there. Already have calculated that v(x) = 0, which will lead to what I need to prove (I think).

You have a typo, X is a function of x, and T is a function of t.

You need to find the value of A,B, and lamda.
The first b.c. gave that A=0. The second b.c. gives X(Pi)=B Sin(sqrt(lamda)*Pi), which is equal to 0. Unless, sqrt(lamda)=n/2 where n is an integer number . n=1,2,3,4,5,6...

Now that you know what lamda is equal to , solve for T, but do not apply the initial condition,yet!

Once you have the expressions for X and T, take their product (which is u), and apply the initial condition.

 

[JoernE]

Thanks MasterX!


This should be \displaystyle X = A \cos(\sqrt{\lambda}x) + B \sin(\sqrt{\lambda}x)

so

\displaystyle X(\pi) = B \sin (\sqrt{\lambda} \pi) = 0

\displaystyle \ \sqrt{\lambda} \pi = \frac{n \pi }{2}

\displaystyle \ \sqrt{\lambda} = \frac{n}{2}

\displaystyle \ \lambda_n = \big( \frac{n}{2} \big)^2 for n = 1,2,...

So

\displaystyle \phi_n(x) = \sin \big(\frac{n \pi x}{2} \big)

and so

\displaystyle u(x,t) = v(x) + \sum_1^{\infty} b_n \sin \big(\frac{n \pi}{2} \big)

\displaystyle \ = \displaystyle \sum_1^{\infty} b_n(t) \phi_n (x)<br />

and I think

\displaystyle \ b_n = \frac{1}{\pi} \int_0^{\pi} \sin(\sqrt{\lambda} x) dx

 

[MasterX]

Do not forget that u=X*T and T' = -lamda * k*T => T=C*Exp(-k *lamda*t)

Therefore b_n=D*Exp(-k*lamda*t), where D=C*B, but this is not important.

Substitute this equation into u, and apply the initial condition to get D.

EDIT:
Since V(x)=0, the above analysis will give you that D=0, which is wrong! Also, V(x) can not be zero.

I do not think this is the way to solve non homogeneous equations. I remember first solving the homogeneous equation (exactly as you did for w), then I assumed that A,B, C are function of x,t and I solved the non-homogeneous. In the end, I used the b.c and i.c. to compute the value of the constants.