How do you solve for multiple constants given the wave functions at the boundary?

[Danielk010]

Homework Statement

In the region 0 ≤ x ≤ a, a particle is described by
the wave function ψ1(x) = $−b(x^2 − a^2 )$. In the region
a ≤ x ≤ w, its wave function is ψ2(x) = $(x − d)^2 − c$. For
x ≥ w, ψ3(x) = 0. (a) By applying the continuity conditions
at x = a, find c and d in terms of a and b. (b) Find w in terms of
a and b.

Relevant Equations

ψ1(x) = $−b(x^2 − a^2 )$. 0 ≤ x ≤ a
ψ2(x) = $(x − d)^2 − c$. a ≤ x ≤ w
ψ3(x) = 0 x ≥ w

From my understanding, you can equate ψ1(x) and ψ2(x) at the boundary of x = a, so I plugged in the values of a into x for both equations and I got ψ1(x) = 0 and ψ2(x) = $(a-d)^2-c$. I am a bit stuck on where to go from here.

 

[BvU]

and I got ψ1(x) = 0 and ψ2(x) = $(a-d)^2-c$. I am a bit stuck on where to go from here.

No. You got ψ1(a) = 0 and ψ2(a) = $(a-d)^2-c$. Equating ψ1 and ψ2 at x=a means you have ψ1(a) = ψ2(a)

Then: what about continuity of the first derivative ?

$\$

 

[Danielk010]

No. You got ψ1(a) = 0 and ψ2(a) = $(a-d)^2-c$. Equating ψ1 and ψ2 at x=a means you have ψ1(a) = ψ2(a)

Then: what about continuity of the first derivative ?

$\$

The continuity for the first derivative of which wave function?

 

[Danielk010]

No. You got ψ1(a) = 0 and ψ2(a) = $(a-d)^2-c$. Equating ψ1 and ψ2 at x=a means you have ψ1(a) = ψ2(a)

Then: what about continuity of the first derivative ?

$\$

I got it. Thank you so much for the help