Solve the PDE of third order by using separation variables method

Question

u_tt+c^2 u_xxx=0 0<x<l t>0

the conditions

u(0,t)=u(l,t)=0

u_xx(0,t)=u_xx(l,t)=0

u(x,0)=f(x), u_t (x,0)=g(x)

Answer 1

To start, let's recall that separation of variables means using an ansatz of the form

u(t,x) = T(t)X(x)

Then we have u_xxx = T X''' and u_tt = T'' X. Since T and X are both functions of a single variable, I'm slightly abusing notation by using a prime for ordinary derivatives with respect to both t and x, depending on whether I'm differentiating T(t) or X(x). So our third-order PDE takes the form:

u_tt+c² u_xxx = 0 = T''(t)X(x) + c²T(t)X'''(x)

Dividing both sides by u = TX we find

0 = T''(t)/T(t) + c² X'''(x)/X(x)

Because the first term depends only on t and the second term only on x, the only way they can sum to zero is if they are both separately constant. Let's call the second term k² so that

T''(t) = -k² T(t)

and

c² X'''(x) = k² X(x)

Note that k² in principle can be negative; I'm writing it as k² rather than k so that I don't have to write square root signs later, and I'm choosing the RHS of the T''(t) equation to be negative in the hopes that the solution for T(t) is sines and cosines rather than exponentials. Of course I can always solve these equations with exponentials, which are equivalent to sin and cos if the argument is imaginary.

The solution for T(t) is straightforward,

T(t) = A sin(kx) + B cos(kx)

To solve for X(x) we can use the ansatz X(x) = e^λx; the three linearly independent solutions are then the three roots of

k² = c² λ³

These are given by (c/k)^2/3 λ = 1, -1/2 (1 ± i√3). We can write the last two more compactly as -(-1)^1/3 and (-1)^2/3, respectively. Note that two of the linearly independent terms can be written in the form

exp[-1/2 (k/c)^2/3 x (1 ± i√3)] = exp[-1/2 (k/c)^2/3 x][cos(√3/2 (k/c)^2/3 x) ± i sin(√3/2 (k/c)^2/3 x)]

so let's write X(x) about as nicely as we can for now,

X(x) = C exp[(k/c)^2/3 x] + exp[-1/2 (k/c)^2/3 x][D cos(√3/2 (k/c)^2/3 x) + E sin(√3/2 (k/c)^2/3 x)].

Our solution for u(t,x) now is determined up to the five constants A, B, C, D, and E. To constrain these let's turn to our boundary conditions. The easiest one is u(0,t) = 0 for all t, from which we can read off

C + D = 0

The next easiest is the condition u_xx(0,t) = 0, which we can evaluate (after setting D = -C and doing a bit of algebra) to

E = -i √3 C

We see that

1/C X(x) = exp[(k/c)^2/3 x] - exp[-1/2 (k/c)^2/3 x][cos(√3/2 (k/c)^2/3 x) - √3 sin(√3/2 (k/c)^2/3 x)].

Note that at this point X(x) is entirely fixed up to a multiplicative constant. This means that it is not possible to impose the conditions at x = l. This can be seen from the start by counting: there are three linearly independent solutions to a third-order equation, on which the problem as presented wants to impose four conditions.

The other two conditions are trivially satisfied (unless f(x) and g(x) are given, which is not in your post), because at fixed t, u(t,x) and its derivatives are functions of x.

I hope that helps. As you can see, something seems off in the formulation of the problem, but this should give you an idea how to go about solving higher-order PDEs using separation of variables.

For the record, when I feed this PDE and the given initial conditions into Mathematica, I don't obtain any solution. Interestingly, the solution it obtains ignoring the ICs at x = l are polynomials of x and t, rather than the separable form we've found here. Both are perfectly valid solutions.

Answer 2

We are given the third-order partial differential equation:

utt+c2uxxx=0,0<x<l,t>0u_{tt} + c^2 u_{xxx} = 0, \quad 0 < x < l, \quad t > 0utt​+c2uxxx​=0,0<x<l,t>0with the boundary conditions:

u(0,t)=u(l,t)=0(Dirichlet)u(0,t) = u(l,t) = 0 \quad \text{(Dirichlet)}u(0,t)=u(l,t)=0(Dirichlet) uxx(0,t)=uxx(l,t)=0(second derivative zero)u_{xx}(0,t) = u_{xx}(l,t) = 0 \quad \text{(second derivative zero)}uxx​(0,t)=uxx​(l,t)=0(second derivative zero)and the initial conditions:

u(x,0)=f(x),ut(x,0)=g(x)u(x,0) = f(x), \quad u_t(x,0) = g(x)u(x,0)=f(x),ut​(x,0)=g(x)We aim to solve this using separation of variables.

Step 1: Separation of Variables

Assume a solution of the form:

u(x,t)=X(x)T(t)u(x,t) = X(x) T(t)u(x,t)=X(x)T(t)Substitute into the PDE:

X(x)T′′(t)+c2X′′′(x)T(t)=0X(x) T''(t) + c^2 X'''(x) T(t) = 0X(x)T′′(t)+c2X′′′(x)T(t)=0Divide both sides by X(x)T(t)X(x)T(t)X(x)T(t) (assuming non-zero):

T′′(t)T(t)+c2X′′′(x)X(x)=0\frac{T''(t)}{T(t)} + c^2 \frac{X'''(x)}{X(x)} = 0T(t)T′′(t)​+c2X(x)X′′′(x)​=0Separate variables:

T′′(t)T(t)=−λ,c2X′′′(x)X(x)=λ\frac{T''(t)}{T(t)} = -\lambda, \quad c^2 \frac{X'''(x)}{X(x)} = \lambdaT(t)T′′(t)​=−λ,c2X(x)X′′′(x)​=λWe now solve:

1. T′′(t)+λT(t)=0T''(t) + \lambda T(t) = 0T′′(t)+λT(t)=0
2. X′′′(x)=λc2X(x)X'''(x) = \frac{\lambda}{c^2} X(x)X′′′(x)=c2λ​X(x)

Let’s define μ=λc2\mu = \frac{\lambda}{c^2}μ=c2λ​, then:

X′′′(x)=μX(x)X'''(x) = \mu X(x)X′′′(x)=μX(x) Step 2: Solve the Spatial Part X(x)X(x)X(x)

We solve the third-order ODE:

X′′′(x)=μX(x)X'''(x) = \mu X(x)X′′′(x)=μX(x)This is a third-order linear ODE with constant coefficients.

Let’s solve the characteristic equation:

r3−μ=0⇒r=μ1/3e2πin/3,n=0,1,2r^3 - \mu = 0 \quad \Rightarrow \quad r = \mu^{1/3} e^{2\pi i n / 3}, \quad n=0,1,2r3−μ=0⇒r=μ1/3e2πin/3,n=0,1,2So, the general solution for X(x)X(x)X(x) is:

1. If μ>0\mu > 0μ>0, solutions are exponential.
2. If μ<0\mu < 0μ<0, solutions are in terms of sines and cosines.

Since we have boundary conditions involving sines and zeros, we expect a trigonometric basis, so try:

Let’s suppose:

X(x)=Asin⁡(kx)+Bcos⁡(kx)+CxX(x) = A \sin(kx) + B \cos(kx) + C xX(x)=Asin(kx)+Bcos(kx)+CxBut this does not solve a third-order ODE of the form X′′′=μXX''' = \mu XX′′′=μX. Instead, let’s directly solve the third-order equation for XXX using exponential form:

Let’s set:

X(x)=erx⇒X′′′=r3erx=μerx⇒r3=μX(x) = e^{rx} \Rightarrow X''' = r^3 e^{rx} = \mu e^{rx} \Rightarrow r^3 = \muX(x)=erx⇒X′′′=r3erx=μerx⇒r3=μThus, the general solution is:

X(x)=Aer1x+Ber2x+Cer3xX(x) = A e^{r_1 x} + B e^{r_2 x} + C e^{r_3 x}X(x)=Aer1​x+Ber2​x+Cer3​xwhere rir_iri​ are the roots of r3=μr^3 = \mur3=μ

These roots are complex unless μ=0\mu = 0μ=0. Let’s proceed with the trigonometric version by assuming that eigenfunctions take the form:

Xn(x)=sin⁡(nπxl),n=1,2,3,…X_n(x) = \sin\left(\frac{n\pi x}{l}\right), \quad n = 1,2,3,\dotsXn​(x)=sin(lnπx​),n=1,2,3,…This satisfies the Dirichlet boundary conditions:

1. X(0)=X(l)=0X(0) = X(l) = 0X(0)=X(l)=0

Now, compute X′′′(x)X'''(x)X′′′(x):

X(x)=sin⁡(nπxl)⇒X′′′(x)=−(nπl)3cos⁡(nπxl)X(x) = \sin\left(\frac{n\pi x}{l}\right) \Rightarrow X'''(x) = -\left(\frac{n\pi}{l}\right)^3 \cos\left(\frac{n\pi x}{l}\right)X(x)=sin(lnπx​)⇒X′′′(x)=−(lnπ​)3cos(lnπx​)So X′′′(x)≠μX(x)X'''(x) \neq \mu X(x)X′′′(x)=μX(x), hence sine functions are not eigenfunctions of X′′′=μXX''' = \mu XX′′′=μX

Thus, standard separation of variables fails unless we impose additional conditions.

Step 3: Alternative Approach (Fourier Series Method)

Given the complexity of the third-order spatial derivative, it's more productive to expand u(x,t)u(x,t)u(x,t) in a Fourier sine series that satisfies the boundary conditions:

Assume:

u(x,t)=∑n=1∞An(t)sin⁡(nπxl)u(x,t) = \sum_{n=1}^\infty A_n(t) \sin\left(\frac{n\pi x}{l}\right)u(x,t)=n=1∑∞​An​(t)sin(lnπx​)Then:

ut=∑An′(t)sin⁡(nπxl),utt=∑An′′(t)sin⁡(nπxl)u_t = \sum A_n'(t) \sin\left(\frac{n\pi x}{l}\right), \quad u_{tt} = \sum A_n''(t) \sin\left(\frac{n\pi x}{l}\right)ut​=∑An′​(t)sin(lnπx​),utt​=∑An′′​(t)sin(lnπx​) uxxx=∑An(t)(−(nπl)3cos⁡(nπxl))u_{xxx} = \sum A_n(t) \left( -\left(\frac{n\pi}{l} \right)^3 \cos\left(\frac{n\pi x}{l} \right) \right)uxxx​=∑An​(t)(−(lnπ​)3cos(lnπx​))So we see that the trigonometric sine expansion is incompatible with the third derivative in xxx. To make progress, we instead use an eigenfunction expansion that satisfies:

1. u(0,t)=u(l,t)=0u(0,t) = u(l,t) = 0u(0,t)=u(l,t)=0
2. uxx(0,t)=uxx(l,t)=0u_{xx}(0,t) = u_{xx}(l,t) = 0uxx​(0,t)=uxx​(l,t)=0

Try a general form:

u(x,t)=∑n=1∞An(t)ϕn(x)u(x,t) = \sum_{n=1}^\infty A_n(t) \phi_n(x)u(x,t)=n=1∑∞​An​(t)ϕn​(x)where ϕn(x)\phi_n(x)ϕn​(x) satisfies:

ϕn(0)=ϕn(l)=0,ϕn′′(0)=ϕn′′(l)=0\phi_n(0) = \phi_n(l) = 0, \quad \phi_n''(0) = \phi_n''(l) = 0ϕn​(0)=ϕn​(l)=0,ϕn′′​(0)=ϕn′′​(l)=0Let’s suppose:

ϕn(x)=sin⁡((2n−1)πxl)\phi_n(x) = \sin\left( \frac{(2n-1)\pi x}{l} \right)ϕn​(x)=sin(l(2n−1)πx​)Then:

1. ϕn(0)=ϕn(l)=0\phi_n(0) = \phi_n(l) = 0ϕn​(0)=ϕn​(l)=0
2. ϕn′′(x)=−((2n−1)πl)2sin⁡((2n−1)πxl)⇒ϕn′′(0)=ϕn′′(l)=0\phi_n''(x) = -\left( \frac{(2n-1)\pi}{l} \right)^2 \sin\left( \frac{(2n-1)\pi x}{l} \right) \Rightarrow \phi_n''(0) = \phi_n''(l) = 0ϕn′′​(x)=−(l(2n−1)π​)2sin(l(2n−1)πx​)⇒ϕn′′​(0)=ϕn′′​(l)=0

So these are suitable basis functions.

Now write:

u(x,t)=∑n=1∞An(t)sin⁡((2n−1)πxl)u(x,t) = \sum_{n=1}^\infty A_n(t) \sin\left( \frac{(2n-1)\pi x}{l} \right)u(x,t)=n=1∑∞​An​(t)sin(l(2n−1)πx​)Substitute into the PDE:

utt+c2uxxx=∑n=1∞An′′(t)ϕn(x)+c2An(t)ϕn′′′(x)=0u_{tt} + c^2 u_{xxx} = \sum_{n=1}^\infty A_n''(t) \phi_n(x) + c^2 A_n(t) \phi_n'''(x) = 0utt​+c2uxxx​=n=1∑∞​An′′​(t)ϕn​(x)+c2An​(t)ϕn′′′​(x)=0Note:

ϕn′′′(x)=−((2n−1)πl)3cos⁡((2n−1)πxl)\phi_n'''(x) = -\left( \frac{(2n-1)\pi}{l} \right)^3 \cos\left( \frac{(2n-1)\pi x}{l} \right)ϕn′′′​(x)=−(l(2n−1)π​)3cos(l(2n−1)πx​)Again, cosine is orthogonal to sine over (0,l)(0, l)(0,l), which implies the Fourier sine basis is not closed under third derivatives. So no standard sine or cosine basis gives a closed system under uxxxu_{xxx}uxxx​.