Mastering PDEs: Techniques and Tricks for Solving with Separation of Variables

[spaghetti3451]

To solve a PDE, we can use the technique of separation of variables. However, this is not the most general solution.

But, the mathematical trick is that we can solve for the SoV solutions and then make a general solution by adding SoV solutions together. I don't understand what this means?

 

[HallsofIvy]

That's only true if the pde is linear.

Consider the equation
U_{xx}= U_{tt}
with boundary conditions U(0, t)= 0 and U(L, t)= 0 and initial conditions U(x, 0)= f(x), U_t(x, 0)= 0.
If U(x,t) and V(x,t) are two functions satisfying that equation and a and b are any two numbers, then (aU+ bV)_{xx}= aU_{xx}+ bV_{xx} and because U and V satisfy U_{xx}= U_{tt} and V_{xx}= V_{tt}, we can write that as aU_{tt}+ bV_{tt}= (aU+ bV)_{tt} so that (aU+ bV)_{xx}= (aU+ bV)_{tt}. That is, aU+ bV satisfies the same differential equation.


If we let U= T(t)X(x) (separate it into a function of t only and a function of x only) we get
TX''= T''X
which we can rewrite as
\frac{X''}{X}= \frac{T''}{T}

Since the left side depends only upon x and the right depends only on t, in order to be equal for all x and t, the two sides must be equal to the same constant.
That is,
\frac{X''}{X}= \frac{T''}{T}= \lambda
for some constant \lambda.

That gives us the two equations X''= \lambda X and T''= \lambda T.

Further, U(0, t)= X(0)T(t)= 0. Either T(t)= 0 which would mean U(x, t)= 0 for all x and t, or X(0)= 0. In order that this problem have a non-trivial solution, we must have X(0)= 0. Also U(L, t)= X(L)T(t)= 0 gives X(L)= 0.

To solve X''(0)= \lambda with boundary conditions X(0)= 0, X(L)= 0, consider possible values for \lambda
1) If \lambda= 0, we have X''= 0 which has general solution X(x)= Ax+ B. Then X(0)= B= 0 and X(L)= AL+ 0= 0 so A= 0. X must be identically 0 which means U must be identically 0 which means we cannot match the initial conditions.

2) If \lambda> 0, we can write \lambda= \alpha^2 with \alpha\ne 0. Now the equation is X''= \alpha^2X which has general solution
X(x)= Ce^{\alpha x}+ De^{-\alpha x}

X(0)= C+ D= 0
X(L)= Ce^{\alpha L}+ De^{-\alpha L}= 0

From the first equation, D= -C so the second equation becomes
Ce^{\alpha L}- Ce^{-\alpha L}= C(e^{\alpha L}- e^{-\alpha L})= 0

But one of those exponentials is greater than 1 while the other is less than 1. The difference cannot be 0 which means we must have C= 0 and then D= 0. Again, that is the trivial solution which we cannot use.

If \lambda< 0, we can write \lambda= -\alphaa^2 with \alpha\ne 0. Now the differential equation for X is X''= -\alpha^2X which has general solution
X(x)= C cos(\alpha x)+ D sin(\alpha x)

X(0)= C= 0
X(L)= D sin(\alpha L)
The only way to avoid D= 0 and another trivial solution is to have sin(\alpha L)= 0 which means \alpha L must be a multiple of \pi: \alpha L= k\pi so \alpha= k\pi/L for integer k.

But then
X(x)= C sin((k\pi/L)x)
satisfies the differential equation
X''= -(k\pi/L)^2X
and the boundary condition X(0)= 0, X(L)= 0 for any C or integer k.

Now that we know \lambda= -\alpha^2= -(k\pi/L)^2 we can write the differential equation of T as
T''(t)= -(k\pi/L)^2T(t)
which also has general solution
T(t)= Acos((k\pi/L)t)+ B sin((k\pi/L)t)

Now, one of the intial conditions is U_t(x, 0)= X(x)T'(0)= 0 which means that we must have
T'(0)= -A(k\pi/L)sin((k\pi/L)0)+ B(k\pi/L)cos((k\pi/L)0)= B(k\pi/L)= 0
so B= 0.

But the other intial condition, U(x, 0)= f(x) cannot be separated in that way.

What we have so far is
U(x,t)= X(x)T(t)= AC sin((k\pi/L)x)cos((k\pi/L)t)
and so
U(x, 0)= C' sin((k\pi/L)x)= f(x)
where I have written C' for AC.

Now for some functions, f(x), that can easily be done- if f(x)= 4sin((3\pi/P)x) it is easy to see that we can set C'= 4 and k= 3 to get
U(x, t)= 4sin((3\pi/L)x)cos((3\pi/L)t)

But what if f(x) were some more complicated function? Say, f(x)= x(L- x) (chosen so that f(0)= 0, f(L)= 0)? Because, as I showed above, sums of solutions to the equation also satisfy the equation, what we can do in that case is combine the solutions for all integer values of k:
U(x, t)= \sum_{k=0}^\infty C_ksin((k\pi/L)x)cos((k\pi/L)t)
and try to find C_k such that
U(x, 0)= \sum_{k=0}^\infty C_k sin((k\pi/L)x)= x(L- x)

That is precisely the "Fourier series" solution.