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Use the Wronskian to find all solutions of a given second-order differential equation

The Wronskian W(x) is defined by

    \[ W(x) = u_1(x)u_2'(x) - u_2(x)u_1'(x) \]

for given functions u_1 and u_2.

Let v_1 and v_2 be two solutions of the second-order linear differential equation

    \[ y'' + ay' + by = 0, \]

such that \frac{v_2}{v_1} is not constant.

  1. Let y = f(x) be any solution of the given differential equation. Use the properties of the Wronskian proved in the previous two exercises (here and here) to prove that there exist constants c_1 and c_2 such that

        \[ c_1 v_1(0) + c_2 v_2(0) = f(0), \qquad c_1 v_1'(0)+ c_2 v_2'(0) = f'(0). \]

  2. Prove that every solution of the differential equation has the form y = c_1 v_1 + c_2 v_2.

  1. Proof. By the previous exercise, since \frac{v_2}{v_1} is not constant we know v_1(0) v_2'(0) - v_1'(0) v_2(0) \neq 0. So we may define,

        \[ c_1 = \frac{f(0) v_2'(0) - v_2(0) f'(0)}{v_1(0)v_2'(0) - v_1'(0) v_2(0)}, \qquad c_2 = \frac{-f(0)v_2'(0) + v_1(0) f'(0)}{v_1(0)v_2'(0) - v_2(0) v_1'(0)}. \]

    This implies

        \begin{align*}  c_1 v_1'(0) + c_2v_2'(0) &= \frac{1}{W(0)} \cdot \big( f(0) v_2'(0) v_1'(0) - v_2(0)v_1'(0) f'(0) - f(0) v_2'(0) v_1'(0) + v_1(0) v_2'(0) f'(0) \big) \\  &= \frac{1}{W(0)} \cdot \big( f'(0) \cdot(v_1(0) v_2'(0) - v_2(0)v_1'(0)) \big) \\  &= f'(0), \end{align*}


        \begin{align*}  c_1 v_1(0) + c_2 v_2(0) &= \frac{1}{W(0)} \cdot \big( f(0) (v_1(0) v_2'(0) - v_1'(0) v_2(0) \big) \\  &= f(0). \qquad \blacksquare \end{align*}

  2. Proof. Let W be the Wronskian of v_1 and v_2. If y = f(x) is any solution then the Wronskian of any pair of f, v_1, v_2 is a solution of W' + aW = 0. Hence,

        \[ W(x) = W(0) e^{-ax} \]

    This implies,

        \[ fv_1' - f'v_1 = (f(0)v_1'(0) - v_1(0) f'(0))e^{-ax} \]


        \[ fv_2' - f'v_2 = (f(0) v_2'(0) - f'(0) v_2(0)) e^{-ax}. \]

    Since \frac{v_2}{v_1} is constant we know W(x) \neq 0 for any x. Hence, e^{-ax} = \frac{W(x)}{W(0)}. Therefore, by part (a),

        \[ f = c_1 v_1 + c_2 v_2. \qquad \blacksquare \]


  1. Mohammad Azad says:

    You skipped lots of nontrivial details in (b). Here is my proof using minimal symbols:
    Define two new functions u1 and u2 just like in theorem 8.4, u1 for f(x) and u2 for the linear combination. If we manage to prove that u1=u2 then we are done. To do that we use part (a) to show that u1 and u2 have the same value and derivative at 0 and then apply theorem 8.5

  2. Anonymous says:

    In a, uniqueness of solutions looks to be assumed given that their value and derivative at some point (0) is equal. As far as I know Apostol only proved uniqueness for $\ddot{y} + by = 0$. It looks like its easy to extend that uniqueness prrof for the case $\ddot{y}+ a\dot{y} + by = 0$, but it wasn’t mentioned.

  3. Eiji says:

    For (b), I solved as follows: Let v1 = u1*e^(-ax/2). Let v2 = u2*e^(-ax/2). Let f=u3*e^(-ax/2). Set y” + ay’ + by = (u” + (4b-a^2)u/4)*e^(-ax/2). From part (a) derive c1u1(0) + c2u2(0) = u3(0) nad c1u1′(0) + c2u2′(0) = u3′(0). Then by using theorem 8.5 you can conclude that c1u1 + c2u2 = u3. Therefore, c1v1 + c2v2 = f

  4. tom says:

    Okay, one final attempt (sorry for all the posts). We started out needing to prove the existence of constants so that two solutions to a diff equation, of which the ratio is not constant, are in fact the basis for all solutions. These constants give us an equality at f(0). (I can’t explain why Apostol had us also find f'(0)). Then, using the properties of the Wronskian we find the existence of constants for arbitrary x, plug these into the equation in part a) for f and find a characterization for all solutions. I hope I haven’t missed something because then I would need to leave another post..

    • Evangelos says:

      In the first part of the problem, what Apostol wanted was to solve the system of equations algebraically. It’s quite a bit of symbolic manipulation, but you don’t necessarily need any techniques from linear algebra here. Hence the reason for the two systems (one for f and one for f’), you solve for c1 in the first equation in terms of c2, plug it into the second, solve for c2, and then take c2 and plug it back into the equation for c1. It’s just an extension of the initial value problems that we would do earlier in the chapter. The constants c1 and c2 don’t change with x, the values f(0) and f'(0) were probably chosen for simplicity. It just as easily could have been f(a) and f'(a), where a is some arbitrary point on the interval we’re working with.

      You had it right when you said that the Wronskian of v1 and v2 is not 0 since v2/v1 is not constant (the solution of a previous exercise), which is why constants c1 and c2 exist, since f, f’, v1, v1′, v2, and v2′ are all assumed to be continuous and differentiable on the interval of choice (i.e., c1 and c2 exist since there exist values for f(0), f'(0), v1(0), v1′(0), v2(0), and v2′(0), and that the denominator is never zero)

      As for the second part of the problem, I did it a different (possibly wrong) way, as follows

          \begin{align*} y &= c_1v_1 + c_2v_2\\ y' &= c_1v_1' + c_2v_2'\\ y'' &= c_1v_1'' + c_2v_2''\\ \end{align*}

      And since y, v_1 and v_2 all satisfy the equation

          \begin{align*} y'' + ay' + by &= 0\\ \end{align*}

          \begin{align*} y'' + ay' + by &= [c_1v_1'' + c_2v_2''] + a[c_1v_1' + c_2v_2'] + b[c_1v_1 + c_2v_2]\\ &= [c_1v_1'' + ac_1v_1' + bc_1v_1] + [c_2v_2'' + ac_2v_2' + bc_2v_2]\\ &= c_1[v_1'' + av_1' + bv_1] + c_2[v_2'' + av_2' + bv_2]\\ &= 0 \quad \blacksquare \\  \end{align*}

      Could be wrong, but there it is.

    • hteica says:

      We can show (b) by using (a) and the uniqueness theorem for the solution of y’+by=0. Let g= c1v1 + c2v2 with the property of (a), namely g(0)=f(0) and g'(0)=f'(0), by linearity it fulfills the property g”+ag’+bg=0, we can let g(x)=u1(x)*e^(-ax/2) and f(x)=u2(x)*e^(-ax/2) for some u1 and u2. Then proceeding as in the theorem of uniqueness, we see that g”(x)+ag'(x)+bg(x)=e^(-ax/2) * (u1″(x)- D*u1(x)) and similarly with f where u2 is replaced with u1. Then since the expression above is equal to 0, we know that it is not the case that e^(-ax/2)=0, so the (uj”(x)- D*uj(x)) must equal to 0, for j=1,2, moreover, in order to be able to appeal to the theorem, we need to make sure that u1(0)=u2(0) and u1′(0)=u2′(0), which is direct, we just need to plug in 0 to g and f and use the fact that g(0)=f(0) and g'(0)=f'(0). Hence we have u1=u2, but this just means that g=f. I don’t know if I made an error or not in this proof, if there’s any fault, please point it out. I hope this helps.

  5. tom says:

    Some more thoughts: The problem asks to prove f(0) and f'(0) are equal to the corresponding values at zero of a linearly independent combination of solutions. Since f is a solution of y” + by = 0, it seems the problem then becomes only to prove the basis is also a solution of y” + by = 0. So the fact that we are asked to find f'(0) seems to be relevant yet your solution doesn’t seem to require that information. Also, since W(x) is guaranteed to not be zero, could we have not skipped part a) altogether and just found the constants with W(x) instead of W(0)? I have only briefly looked at linear algebra so I can wait for those studies to get a better proof of these matters, or perhaps Apostol will give me some clues in the next chapter. But I would very much appreciate it if you could elucidate the strategy you took; ie how the basis is proven to be a solution.

    • tom says:

      Correction: y = e^(-ax/2)*u, where u is a solution of y” + by = 0. And so the purpose of calculating y'(0) in part a) is even more confusing now.

  6. tom says:

    Might need a hint for the general part where you stated ‘Therefore, by part (a),’. It appears you equated the constants c1 & c2, which use values at zero, with similar versions of themselves but with arbitrary x’s, ie W(x) in the denominator rather than W(0). Is this the crux of the proof?

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