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How to solve the following mixed recurrence relations:

$$T(n)= \max_{0

Here $a,b \ge 1$ are constants.

I could solve some special cases such as when $a=b=2$ then by guessing method, $T(n) = \Theta (\dfrac n {\lg n})$. However, I could not generalize for all values of $a,b$.

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    What do you mean when you say "solve"? Do you only want to find a lower bound? – 2017-01-23
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    By solve I mean to obtain asymptotically very close upper and lower bound (say the upper and lower bound differ by a constant factor). For example, $a=b=2$ will imply $T(n)=\Theta(\frac{n}{\log n})$. – 2017-01-24

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Since you did not define starting values of $T$ and didn’t care about divisibility $n-k$ by $b$, I think a solution by asymptotical handwaving is OK for your needs. Assume that $T$ increases. If $a=1$ then the right hand side of the equality for $T(n)$ is less than $T(n)$, so from now we shall assume that $a>1$. Since when $k$ increases $ aT\left(\frac{n-k}{b}\right)$ decreases, we have $T(n)=k=aT\left(\frac{n-k}{b}\right)$ for some $k\in (0,n)$. Let’s look for a solution of this equality with a simple form. The following cases are possible.

$a>b$. Assume that $T(n)\simeq cn$ for some constant $c<1$. Then $$k\simeq cn \simeq ac\frac{n-cn}{b},$$ so we can put $c=1-b/a$.

$a=b$. Assume that $T(n)\simeq \frac{n}{\ln n}$. Then $$k\simeq \frac{n}{\ln n}\simeq a\frac{\frac{n-\frac{n}{\ln n}}{b}}{\ln \left(n-\frac n{\ln n}\right)}\simeq \frac{n}{\ln n}.$$

$a. Assume that $T(n)\simeq n^c$ for some constant $c<1$. Then $$k\simeq n^c \simeq a\left(\frac{n-n^c}{b}\right)^c\simeq a \left(\frac{n}{b}\right)^c,$$ so we can put $c=\log_b a$.