I should determine the function $M(H,T)$ by only knowing the derivatives:
$$
\left ( \frac{\partial M}{\partial H } \right )_{T}=\frac{a}{1-T/T_c}+3bH^{2}
$$
where $a$ and $b$ are some real constants
$$
\left ( \frac{\partial M}{\partial T } \right )_{H}=
\frac{1}{T_{c}}\frac{f(H)}{(1-T/T_c)^{2}}-
\frac{1}{2}\frac{M_{0}}{T_{c}}\frac{1}{(1-T/T_c)^{1/2}}
$$
with $M_{0}$, $T_{c}$, $a$ and $b$ are constants and $f(H)$ is some function
with the property $f(H=0)=0$. The variable $T$ will be bound to the domain $0\le T\lt T_{c}$.
a) determine $f(H)$
b) determine $M(T,H)$
Now I would start by writing down
$$
dM = \left ( \frac{\partial M}{\partial H } \right )_{T}dH + \left ( \frac{\partial M}{\partial T } \right )_{H} dT
$$
This tells me how I would obtain the M in terms of the derivatives.
I tried now to do the integration what gives:
$$
M(T,H)=\frac{aH}{1-T/T_{c}}+bH^{3}+\frac{f(H)}{1-T/T_{c}}+M_{0}(1-T/T_{c})^{1/2}
$$
If I compute now again the derivative with respect to H I obtain:
$$
\left ( \frac{\partial M}{\partial H } \right )_{T}=\frac{a}{1-T/T_c}+3bH^{2}+\frac{1}{1-T/T_{c}}\frac{\partial f(H)}{\partial H}
$$
This is now not the same as at the beginning. From here I don't know how to go on. Is this already wrong or is it somehow possible from this equation to determine $f(H)$.
My first intuition was to say $f(H)$ is just zero, because then the derivative with respect to $H$ would be satisfied and so would be the $f(H=0)=0$ criterion. But this might also be a trivial solution.
Best Answer
Let us integrate the first equation with respect to $H$ while $T$ is seen as a constant: $$ M(T,H) = \frac{a H}{1-T/T_c} + bH^3 + \phi(T) $$ where $\phi(T)$ is an "integration constant" depending on $T$. Now, the partial derivative w.r.t. $T$ is $$ \frac{\partial M}{\partial T} = \frac{a H}{T_c}\frac{1}{(1-T/T_c)^2} + \phi'(T) $$ where $\phi'$ is the derivative of $\phi$. Therefore, $f(H)=aH$ and $\phi'(T) = -\frac{1}{2}\frac{M_{0}}{T_{c}}(1-T/T_c)^{-1/2}$. The integration of this last equation with respect to $T$ gives $\phi(T)$ and introduces an integration constant $C$. Finally, $M$ is defined up to a constant $C$ by $$ M(T,H) = \frac{a H}{1-T/T_c} + bH^3 + M_0\, (1-T/T_c)^{1/2} + C\, . $$