Exercise I: Use the eigenfunctions expansion method to solve the non-homogeneous

Question

Exercise I: Use the eigenfunctions expansion method to solve the non-homogeneous heat equation au = arru+g(x, t), with the initial condition u(x,0) = f(x) and the boundary condition aru(0,t) = A(t) and aru(1,t) = B(t) for x E [0,1]. Do not make the BC homogeneous. Note that first you have to find the eigenfunctions and eigenvalues, and that we cannot diff , An(t)d, (x)) since u has erentiate term-by-term versus x the infinite series (u(z,t)= non-homogeneous BC, so that we have to use integration by parts or the Green's formula. You have to find the equation for An and give its general solution.

Explanation / Answer

Given non-homogeneous differential equation is

dtu=dxxu+q(x,t).

We know that dt=du/dt and dx=du/dx.

Now we can write above equation in homogeneous form by setting q(x,y)=0.

t'x=x"t

X"/X=T'/T=k. Here let us take k is constant

By variable separation method differential equations become

X"+kX=0 and T'+KT=0.

and general solutions will come from roots

(D2+k)x=0 and (D+k)T=0.

First one has imaginary roots and t got real root

and General solution will be and let us assume k=p2

u(x,t)=(c1cospt+c2sinpt)c3e-kt

Now,given initial conditions are u(x,0)=f(x) and dxu(0,t)=A(t),dxu(1,t)=B(t).

Given non homogenous initial condition

u(x,0)=f(x).

u(x,0)=(c1cospx+c2sinpx)=f(x).---.(1)

X(x)=(c1cospx+c2sinpx)

dxu(x,t)=(-c1sinpx+c2cospx)c3e-kt.

apply boundary conditions

dxu(0,t)=c2c3e-kt=A(t). By taking coefficients as 1.A(t)=e-kt.---->(2). This function is in the form of exponential only

dxu(1,t)=(-c1cosp+c2sinp)e-kt=B(t).

Finally from above calculations of eigen functions and eigen values p,k. from (1) and (2), we will get the infinite series for u(x,t) is

u(x,t)=£n An(t)fn(x).

Now,u has non homogenous form f(x).

u(x,t)=£nAn(t).fn(x).=q(x,t)

An(t)=Ane-kt.

f(x)=(c1cospx+c2sinpx).

This non homogeneous function will be in the form

u(x,t)=Ane-kt(c1cospx+c2sinpx).

q(x,t)=£ne-t(cosx+sinx)

Now,take given differential equation

dtu=dxxu+q(x,t).

£n(-kAne-kt)(c1cospx+c2sinpx)+£nAne-kt(c1cospx(p2)+c2sinpx(p2))=£ne-t(cosx+sinx).

From the above equation and equating both

An(p2-k)=1. and c1=c2=1.

From the above equation An=1/(p2-k).

After substituting above values Finally General solution will be

u(x,t)=(1/(p2-k))e-kt(cospx+sinpx).

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